linear programming: the simplex method ... 4 linear programming: the simplex method 4.1 slack...

Copyright © 2016 Pearson Education, Inc. 217

Chapter 4

LINEAR PROGRAMMING: THE SIMPLEX METHOD

4.1 Slack Variables and the Pivot

Your Turn 1

The two new equations are:

1 2 3

1 2 3

300 60 180 20,0005 10 15 900x x x

x x x+ + £

+ + £

The new answer tableau is:

1 2 3 1 2 3

1 1 1 1 0 0 0 10015 3 9 0 1 0 0 1000

1 2 3 0 0 1 0 180120 40 60 0 0 0 1 0

x x x s s s zé ùê úê úê úê úê úê ú- - -ê úë û

Your Turn 2

Pivot around the indicated 6.

1 2 3 1 2 3

3 6 2 1 0 0 0 608 5 4 0 1 0 0 803 6 7 0 0 1 0 120

30 50 15 0 0 0 1 0


The result is:

1 2 3 1 2 3

1 2 2

1 3 3

1 4 4

3 6 2 1 0 0 0 60

33 0 14 5 6 0 0 1805R 6R R

0 0 5 1 0 1 0 60R R R

15 0 5 25 0 0 3 150025R 3R R

x x x s s s z

-- +

-- +

-+

é ùê úê úê úê úê úê úê úë û

1 2 3 1 2 3

1 1

2 2

4 4

1 R R6 1/2 1 1/3 1/6 0 0 0 101 11 0 7/3 5/6 1 0 0 30R R6

0 0 5 1 0 1 0 60

5 0 5/3 25/3 0 0 1 5001 R R3

x x x s s s z

-

-

-


The solution given by this tableau is:

1 2 3 1

2 3

0, 10, 0, 0,30, 60, 500

x x x ss s z

= = = =

= = =

4.1 Exercises

1. 1 22 6x x+ £

Add 1s to the given inequality to obtain

1 2 12 6.x x s+ + =

2. 1 26 2 50x x+ £


1 2 16 2 50.x x s+ + =

3. 1 2 32.3 5.7 1.8 17x x x+ + £


1 2 3 12.3 5.7 1.8 17.x x x s+ + + =

4. 1 2 38 6 5 250x x x+ + £


1 2 3 18 6 5 250.x x x s+ + + =

5. (a) Since there are three constraints to be converted into equations we need three slack variables.

(b) We use 1,s 2,s and 3s for the slack variables.

(c) The equations are

1 2 1

1 2 2

1 2 3

2 3 154 5 35

6 20.

x x sx x sx x s

+ + =

+ + =

+ + =

6. Maximize 1 21.2 3.5z x x= +

subject to: 1 2

1 2

2.4 1.5 101.7 1.9 15

x xx x

+ £

+ £

with 1 20, 0.x x³ ³

(a) We need one slack variable for each inequality. Thus, 2 are needed.

(b) We will use 1s and 2s for the slack variables.

(c) 1 22.4 1.5 10x x+ £ becomes

1 2 12.4 1.5 10.x x s+ + =

1 21.7 1.9 15x x+ £ becomes

218 Chapter 4 LINEAR PROGRAMMING: THE SIMPLEX METHOD

Copyright © 2016 Pearson Education, Inc.

1 2 21.7 1.9 15.x x s+ + =

7. (a) There are two constraints to be converted into equations, so we must introduce two slack variables.

(b) Call the slack variables 1s and 2.s

(c) The equations are

1 2 3 1

1 2 3 2

7 6 8 1184 5 10 220.

x x x sx x x s

+ + + =

+ + + =

8. Maximize 1 2 312 15 10z x x x= + +

subject to: 1 2 3

1 2 3

2 2 84 3 12

x x xx x x

+ + £

+ + £

with 1 2 30, 0, 0.x x x³ ³ ³

(a) There are two inequalities, so 2 slack variables are needed.

(b) Use 1s and 2s for the slack variables.

(c) 1 2 3 1

1 2 3 2

2 2 84 3 12

x x x sx x x s

+ + + =

+ + + =

9. Find 1 0x ³ and 2 0x ³ such that

1 2

1 2

4 2 52 4

x xx x

+ £

+ £

and 1 27z x x= + is maximized.

We need two slack variables, 1s and 2.s Then the problem can be restated as:

Find 1 2 10, 0, 0,x x s³ ³ ³ and 2 0s ³ such that

1 2 1

1 2 2

4 2 52 4.

x x sx x s

+ + =

+ + =


Rewrite the objective function as

1 27 0.x x z- - + =

The initial simplex tableau is

1 2 1 2

4 2 1 0 0 51 2 0 1 0 4 .7 1 0 0 1 0

x x s s zé ùê úê úê úê ú- -ê úë û


1 2

1 2

2 3 1005 4 200

x xx x

+ £

+ £

and 1 23z x x= + is maximized. We need two slack variables. Add 1s and 2s to get the system

1 2 1

1 2 2

1 2

2 3 1005 4 200

3 0.

x x sx x sx x z

+ + =

+ + =

- - + =


1 2 1 2

2 3 1 0 0 1005 4 0 1 0 200 .1 3 0 0 1 0

x x s s z

é ùê úê úê úê ú- -ê úë û


1 2

1 2

1 2

105 2 20

2 36

x xx xx x

+ £

+ £

+ £


Using slack variables 1 2, ,s s and 3,s the problem can be restated as:

Find 1 2 10, 0, 0,x x s³ ³ ³ 2 0,s ³ and

3 0s ³ such that

1 2 1

1 2 2

1 2 3

105 2 20

2 36

x x sx x sx x s

+ + =

+ + =

+ + =



1 23 0.x x z- - + =


1 2 1 2 3

1 1 1 0 0 0 105 2 0 1 0 0 20

.1 2 0 0 1 0 361 3 0 0 0 1 0

x x s s s zé ùê úê úê úê úê úê ú- -ê úë û


1 2

1 2

25

4 3 48

x x

x x

+ £

+ £

and 1 25 3z x x= + is maximized.

We add 1s and 2s to get the system

1 2 1

1 2 2

1 2

254 3 485 3 0.

x x sx x sx x z

+ + =

+ + =

- - + =

Section 4.1 219



1 2 1 2

1 1 1 0 0 254 3 0 1 0 48 .5 3 0 0 1 0



1 2

1 2

3 1215

x xx x

+ £

+ £


Using slack variables 1s and 2,s the problem can be restated as:

Find 1 2 10, 0, 0,x x s³ ³ ³ and 2 0s ³ such that

1 2 1

1 2 2

3 1215

x x sx x s

+ + =

+ + =



1 22 0.x x z- - + =


1 2 1 2

3 1 1 0 0 121 1 0 1 0 15 .2 1 0 0 1 0



1 2

1 2

10 4 10020 10 150

x xx x

+ £

+ £


Add 1s and 2s to get

1 2 1

1 2 2

1 2

10 4 10020 10 150

4 5 0.

x x sx x sx x z

+ + =

+ + =

- - + =


1 2 1 2

10 4 1 0 0 10020 10 0 1 0 150 .4 5 0 0 1 0


15. 1 2 3 1 2

1 0 4 5 1 0 83 1 1 2 0 0 42 0 2 3 0 1 28

x x x s s zé ùê úê úê úê ú-ê úë û

The variables 2x and 2s are basic variables, because the columns for these variables have all zeros except for one nonzero entry. If the remaining variables 1,x 3,x and 1s are zero, then 2 4x = and 2 8.s = From the bottom row, 28.z = The basic feasible solution is 1 0,x = 2 4,x =

3 0,x = 1 0,s = 2 8,s = and 28.z =

16. 1 2 3 1 2

1 5 0 1 2 0 60 2 1 2 3 0 150 4 0 1 2 1 64

x x x s s zé ùê úê úê úê ú-ê úë û

1x and 3x are the basic variables. The solution is 1 6,x = 2 0,x = 3 15,x = 1 0,s = 2 0,s =

and 64.z =

17. 1 2 3 1 2 3

6 2 2 3 0 0 0 162 2 0 1 0 5 0 352 1 0 3 1 0 0 63 2 0 2 0 0 3 36

x x x s s s zé ùê úê úê úê úê úê ú- -ê úë û

The basic variables are x3, s2, and s3. If x1, x2, and s1 are zero, then 32 16,x = so 3 8.x = Similarly, s2 = 6 and 35 35,s = so 3 7.s = From the bottom row, 3z = 36, so z = 12. The basic feasible solution is x1 = 0, x2 = 0, x3 = 8, s1 = 0, s2 = 6, s3 = 7, and z = 12.

18. 1 2 3 1 2 3

0 2 0 5 2 2 0 150 3 1 0 1 2 0 27 4 0 0 3 5 0 350 4 0 0 4 3 2 40

x x x s s s zé ùê úê úê úê úê úê ú-ê úë û

x1, x3, and s1 are the basic variables. The solution is x1 = 5, x2 = 0, x3 = 2, s1 = 3, s2 = 0, s3 = 0, and z = 20.

19. 1 2 3 1 2

1 2 4 1 0 0 56

2 2 1 0 1 0 401 3 2 0 0 1 0

x x x s s zé ùê úê úê úê ú- - -ê úë û

Clear the x2 column.



1 2 3 1 2

2 1 1 1 0 3 1 1 0 16R R R2 2 1 0 1 0 401 3 2 0 0 1 0

x x x s s zé ù- -- + ê úê úê úê ú- - -ê úë û

1 2 3 1 2

2 3 3

1 0 3 1 1 0 162 2 1 0 1 0 40

3R 2R R 4 0 1 0 3 2 120

x x x s s zé ù- -ê úê úê úê ú+ -ê úë û

x2 and s1 are now basic. The solution is x1 = 0, x2 = 20, x3 = 0, s1 = 16, s2 = 0, and z = 60.

20. 1 2 3 1 2

2 3 4 1 0 0 18

6 3 2 0 1 0 151 6 2 0 0 1 0



1 2 3 1 2

2 1 1

2 3 3

4 0 2 1 1 0 3R R R6 3 2 0 1 0 15

2R R R 11 0 2 0 2 1 30

x x x s s z

é ù- -- + ê úê úê úê ú+ ê úë û

x2 and s1 are now basic. Thus, the solution is 1 0,x = 2 5,x = 3 0,x = 1 3,s = 2 0,s = and

30.z =

21. 1 2 3 1 2 3

2 2 1 1 0 0 0 121 2 3 0 1 0 0 453 1 1 0 0 1 0 202 1 3 0 0 0 1 0



1 2 3 1 2 3

1 2 2

1 3 3

1 4 4

2 2 1 1 0 0 0 125 4 0 3 1 0 0 93R R R1 1 0 1 0 1 0 8R R R4 5 0 3 0 0 1 363R R R

x x x s s s zé ùê úê ú- - -- + ê úê ú- -- + ê úê ú+ ê úë û

3 2, ,x s and 3s are now basic. The solution is

1 0,x = 2 0,x = 3 12,x = 1 0,s = 2 9,s =

3 8,s = and 36.z =

22. 1 2 3 1 2 3

4 2 3 1 0 0 0 22

2 2 5 0 1 0 0 281 3 2 0 0 1 0 453 2 4 0 0 0 1 0



1 2 3 1 2 3

2 1 1

2 3 3

2 4 4

14 4 0 5 3 0 0 263R 5R R2 2 5 0 1 0 0 281 11 0 0 2 5 0 1692R 5R R7 2 0 0 4 0 5 1124R 5R R

x x x s s s zé ù-- + ê úê úê úê ú-- + ê úê ú- -+ ê úë û

3,x 1,s and 3s are now basic. Thus, the solution is

1 0,x = 2 0,x = 283 5 ,x = 26

1 5 ,s = 2 0,s = 169

3 5 ,s = and 1125 .z =

23. 1 2 3 1 2 3

2 2 3 1 0 0 0 5004 1 1 0 1 0 0 3007 2 4 0 0 1 0 7003 4 2 0 0 0 1 0



1 2 3 1 2 3

1 2 2

1 3 3

1 4 4

2 2 3 1 0 0 0 5006 0 1 1 2 0 0 100R 2R R5 0 1 1 0 1 0 200R R R1 0 4 2 0 0 1 10002R R R

x x x s s s zé ùê úê ú- -- + ê úê ú-- + ê úê ú

+ ê úë û

2 2, ,x s and 3s are now basic. Thus, the solution is

1 0,x = 2 250,x = 3 0,x = 1 0,s = 2 50,s =

3 200,s = and 1000.z =

24. 1 2 3 4 1 2 3

1 2 3 1 1 0 0 0 1152 1 8 5 0 1 0 0 200

1 0 1 0 0 0 1 0 502 1 1 1 0 0 0 1 0

x x x x s s s zé ùê úê úê úê úê úê ú- - - -ê úë û


Section 4.1 221


1 2 3 4 1 2 3

3 1 1

3 2 2

3 4 4

0 2 2 1 1 0 1 0 65R R R0 1 6 5 0 1 2 0 1002 R R R1 0 1 0 0 0 1 0 500 1 1 1 0 0 2 1 1002 R R R

x x x x s s s z

-- +

-- +

- -+


1 1, ,x s and 2s are now basic. Thus, the solution is

1 50,x = 2 0,x = 3 0,x = 4 0,x = 1 65,s =

2 100,s = 3 0,s = and 100.z =

25. A slack variable (a nonnegative quantity), converts a linear inequality into a linear equation by adding the amount needed in an expression to be equal to a specific value.

26. The number of slack variables must be the same as the number of constraints in a linear programming problem.

27. Let x1 represent the number of simple figures, x2 the number of figures with additions, and x3 the number of computer-drawn sketches. Organize the information in a table.

Simple Figures

Figures with

Additions

Computer-Drawn Sketches

Maximum Allowed

Cost 20 35 60 2200 Royalties 95 200 325

The cost constraint is

1 2 320 35 60 2200.x x x+ + £

The limit of 400 figures leads to the constraint

1 2 3 400.x x x+ + £

The other stated constraints are

3 1 2x x x£ + and 1 22 ,x x³

and these can be rewritten in standard form as

1 2 3 0x x x- - + £ and 1 22 0x x- + £

respectively. The problem may be stated as:

Find 1 0,x ³ 2 0,x ³ and 3 0x ³ such that

1 2 3

1 2 3

1 2 3

1 2

20 35 60 2200400

02 0

x x xx x xx x xx x

+ + £

+ + £

- - + £

- + £

and 1 2 395 200 325z x x x= + + is maximized.

Introduce slack variables s1, s2, s3, and s4, and the problem can be restated as:

Find 1 0,x ³ 2 0,x ³ 3 0,x ³ 1 0,s ³ 2 0,s ³

3 0,s ³ and 4 0s ³ such that

1 2 3 1

1 2 3 2

1 2 3 3

1 2 4

20 35 60 2200400

02 0

x x x sx x x sx x x sx x s

+ + + =

+ + + =

- - + + =

- + + =



1 2 395 200 325 0.x x x z- - - + =

The initial simlpex tableau is

1 2 3 1 2 3 4

20 35 60 1 0 0 0 0 22001 1 1 0 1 0 0 0 4001 1 1 0 0 1 0 0 0 .1 2 0 0 0 0 1 0 0

95 200 325 0 0 0 0 1 0

x x x s s s s zé ùê úê úê úê ú- -ê úê ú-ê úê ú- - -ê úë û

28. Let x1 represent the number of racing bicycles, x2 the number of touring bicycles, and x3 the number of mountain bicycles. Organize the information in a table.

Racing Touring MountainAmount

AvailableSteel 17 27 34 91,800 Aluminum 12 21 15 42,000 Profit $8 $12 $22

Using this information, the problem may be stated as:


1 2 3

1 2 3

17 27 34 91,80012 21 15 42,000

x x xx x x

+ + £

+ + £


Introduce slack variables 1s and 2,s and the problem can be restated as:

Find 1 0,x ³ 2 0,x ³ 3 0,x ³ 1 0,s ³ and 2 0s ³ such that

1 2 3 1

1 2 3 2

17 27 34 91,80012 21 15 42,000

x x x sx x x s

+ + + =

+ + + =



1 2 38 12 22 0.x x x z- - - + =




1 2 3 1 2

17 27 34 1 0 0 91,00012 21 15 0 1 0 42,000 .

8 12 22 0 0 1 0


29. Let 1 2 3, , andx x x represent the number produced of each of the three styles of jackets. Organize the information in a table.

Style 1 Style 2 Style 3 Sq ft or $ available

Nylon 6 4 3 300 sq ft

used Fleece 2 3 5 150 sq ft used Cost $20 $18 $17 $600

The limit of 300 sq ft for nylon leads to the constraint

1 2 36 4 3 300.x x x+ + £

The limit of 150 sq ft for fleece leads to the constraint

1 2 32 3 5 150.x x x+ + £


1 2 320 18 17 600.x x x+ + £

The problem may be stated as: Find 1 0,x ≥ 2 0,x ≥ and 3 0x ≥ such that

1 2 3

1 2 3

1 2 3

6 4 3 3002 3 5 150

20 18 17 600

x x xx x xx x x

+ + £

+ + £

+ + £ and 1 2 3z x x x= + + is maximized.

Introduce slack variables 1,s 2,s and 3,s and the problem can be restated as:

Find 1 0,x ³ 2 0,x ³ 3 0,x ³ 1 0,s ³


1 2 3 1

1 2 3 2

1 2 3 3

6 4 3 3002 3 5 150

20 18 17 600

x x x sx x x s

x x x s

+ + + =

+ + + =

+ + + =

and 1 2 3z x x x= + + is maximized.


1 2 3 0.x x x z- - - + = The initial simplex tableau is

1 2 3 1 2 3

6 4 3 1 0 0 0 3002 3 5 0 1 0 0 150

.20 18 17 0 0 1 0 6001 1 1 0 0 0 1 0


30. Let 1x represent the number of Basic sets, 2x the number of Regular sets, and 3x the number of Deluxe sets. Organize the information in a table.

Basic Set

Regular Set

Deluxe Set

Number Available

Utility 2 2 3 800 Knife Chef’s 1 1 1 400 Knife Slicer 0 1 1 200 Profit $30 $40 $60

Using this information, the problem may be stated as:


1 2 3

1 2 3

2 3

2 2 3 800

400200

x x x

x x xx x

+ + £

+ + £

+ £


Introduce slack variables 1,s 2,s and 3,s and the problem can be restated as:

Find 1 0,x ³ 2 0,x ³ 3 0,x ³ 1 0,s ³ 2 0,s ³ and 3 0s ³ such that

1 2 3 1

1 2 3 2

2 3 3

2 2 3 800400200

x x x sx x x s

x x s

+ + + =

+ + + =

+ + =



1 2 330 40 60 0.x x x z- - - + =


1 2 3 1 2 3

2 2 3 1 0 0 0 8001 1 1 0 1 0 0 400

.0 1 1 0 0 1 0 20030 40 60 0 0 0 1 0


Section 4.2 223


31. Let 1x = the number of newspaper ads, 2x = the number of internet banners, and 3x = the number of TV ads.

Organize the information in a table.

Newspaper Ads

Internet Banners

TV Ads

Cost per Ad 400 20 2000 Maximum Number 30 60 10Women Seeing Ad 4000 3000 10,000


1 2 3400 20 2000 8000x x x+ + £

The constraints on the numbers of ads is

1

2

3

306010.

xxx

£

£

£

The problem may be stated as:


1 2 3

1

2

3

400 20 2000 8000306010

x x xx

xx

+ + £

£

£

£

and 1 2 34000 3000 10,000z x x x= + + is maximized.

Introduce slack variables 1 2 3, ,s s s and 4,s and the problem can be restated as:

Find 1 0,x ³ 2 0,x ³ 3 0,x ³ 1 0,s ³ 2 0,s ³


1 2 3 1

1 2

2 3

3 4

400 20 2000 8000306010

x x x sx s

x sx s

+ + + =

+ =

+ =

+ =

and 1 2 34000 3000 10,000z x x x= + + is maximized.


1 2 34000 3000 10,000 0.x x x z- - - + =


1 2 3 1 2 3 4

400 20 2000 1 0 0 0 0 80001 0 0 0 1 0 0 0 300 1 0 0 0 1 0 0 600 0 1 0 0 0 1 0 10

4000 3000 10,000 0 0 0 0 1 0

x x x s s s s zé ùê úê úê úê úê úê úê úê ú- - -ê úë û

4.2 Maximization Problems Your Turn 1

Example 2 of Section 3.3 yields the following linear programming problem, where we have renamed x and y as 1x and 2.x

Maximize 1 212 40z x x= +

subject to: 1 2

1 2

1

1

2

163 36

1000

x xx x

xxx

+ £

+ £

£

³

³

Add a slack variable to each of the first three constraints:

1 2 1

1 2 2

1 3

163 36

10

x x sx x s

x s

+ + £

+ + £

+ £

with 1 2 1 2 30, 0, 0, 0, 0x x s s s³ ³ ³ ³ ³

The corresponding initial tableau is

1 2 1 2 3

1 1 1 0 0 0 16

1 3 0 1 0 0 361 0 0 0 1 0 10

12 40 0 0 0 1 0


Since the most negative indicator is 40- and the quotient 36/3 is smaller than 16/1, we pivot on the 3 in column 2:

1 2 1 2 3

2 1 1

2 4 4

2 0 3 1 0 0 12R 3R R1 3 0 1 0 0 361 0 0 0 1 0 104 0 0 40 0 3 144040R 3R R

x x s s s zé ù-- + ê úê úê úê úê úê ú+ ê úë û

There are now no negative indicators, so we can read the solution:

1 236 14400, 12, 4803 3

x x z= = = = =



Your Turn 2

Pivot on the 4 in column 2 of the following tableau.

1 2 1 2 3

1 2 1 0 0 0 100

3 4 0 1 0 0 2005 0 0 0 1 0 150

10 25 0 0 0 1 0

x x s s s zé ù-ê úê úê úê úê úê ú- -ê úë û

1 2 1 2 3

2 1 1

2 4 4

5 0 2 1 0 0 400R 2R R3 4 0 1 0 0 2005 0 0 0 1 0 150

35 0 0 25 0 4 500025R 4R R

x x s s s zé ù+ ê úê úê úê úê úê ú+ ê úë û

There are no negative indicators, so the optimal solution is:

1 2 1

2000, 50, 200,4

x x s= = = =

2 350000, 150, 1250

4s s z= = = =

4.2 Warmup Exercises W1. Maximize 1 23 2z x x= +

subject to:

1 2

1 2

1

2

3 4 122 5 10

00

x xx x

xx

+ ≤+ ≤

≥≥

Introduce slack variables to change the constraints into equalities.

1 2 1

1 2 2

3 4 122 5 10x x sx x s

+

+

+ =+ =

Write the objective function with all variables on the left side of the equals sign.

1 23 2 0z x x− − =


1 2 1 2

3 4 1 0 0 122 5 0 1 0 103 2 0 0 1 0

x x s s z

W2. Maximize 1 2 314 16 12z x x x= + +

subject to:

1 2 3

1 2 3

1

2

3

3 8 2 245 6 3 30

000

x x xx x x

xxx

+ + ≤+ + ≤

≥≥≥


1 2 3 1

1 2 3 2

3 8 2 245 6 3 30

x x x sx x x s

+ + + =+ + + =


1 2 314 16 12 0z x x x− − − =


1 2 3 1 2

3 8 2 1 0 0 245 6 3 0 1 0 3014 16 12 0 0 1 0

x x x s s z

W3. The initial tableau from W1 is

1 2 1 2

3 4 1 0 0 122 5 0 1 0 103 2 0 0 1 0

x x s s z

1 2 1 2

1 2 2

1 3 3

3 4 1 0 0 12( 2)R 3R R 0 7 2 3 0 6

R R R 0 2 1 0 1 12

x x s s z

The basic variables are 1x and 2s , while

2x and 1s are 0.

1

2

12 436 23

x

s

= =

= =

For these values of the basic variables, 12.z =

W4. The initial tableau from W2 is

1 2 3 1 2

3 8 2 1 0 0 245 6 3 0 1 0 3014 16 12 0 0 1 0

x x x s s z

1 2 3 1 2

1 2 2

1 3 3

3 8 2 1 0 0 24( 3)R 4R R 11 0 6 3 4 0 48

R R R 8 0 8 2 0 1 48

x x x s s z

The basic variables are 2x and 2s , while 1x , 3x ,

and 1s are 0.

Section 4.2 225


2

2

24 38

48 124

x

s

= =

= =

For these values of the basic variables, 48.z =

4.2 Exercises

1. 1 2 3 1 2

1 4 4 1 0 0 162 1 5 0 1 0 203 1 2 0 0 1 0


The most negative indicator is 3,- in the first column. Find the quotients 16

1 16= and 202 10;=

since 10 is the smaller quotient, 2 in row 2, column 1 is the pivot.

1 2 3 1 2161202

1 4 4 1 0 0 1616

10 2 1 5 0 1 0 203 1 2 0 0 1 0

x x x s s zé ù= ê úê ú= ê úê ú- - -ê úë û

Performing row transformations, we get the following tableau.

1 2 3 1 2

2 1 1

2 3 3

R 2R R 0 7 3 2 1 0 122 1 5 0 1 0 20

3R 2R R 0 1 11 0 3 2 60

x x x s s zé ù- + -ê úê úê úê ú+ ê úë û

All of the numbers in the last row are nonnegative, so we are finished pivoting. Create a 1 in the columns corresponding to 1 1,x s and .z

1 2 3 1 2

7 3 111 1 2 2 22

51 1 12 22 2 2 2

1 31 113 32 2 2 2

0 1 0 6R R

R R 1 0 0 10

R R 0 0 1 30

x x x s s zé ù- ê úê úê ú ê úê úê ú ê úë û

The maximum value is 30 and occurs when 1 10,x = 2 0,x = 3 0,x = 1 6,s = and

2 0.s =

2. 1 2 3 1 2

3 3 2 1 0 0 182 2 3 0 1 0 164 6 2 0 0 1 0


The most negative indicator is - 6, in the second column. Find the quotients 18

3 6= and 162 8;=

since 6 is the smaller quotient, 3 in row 1, column 2 is the pivot.

1 2 3 1 2183

162

6 3 3 2 1 0 0 182 2 3 0 1 0 1684 6 2 0 0 1 0

x x x s s zé ù= ê úê ú= ê úê ú- - -ê úë û

1 2 3 1 2

1 2 2

1 3 3

3 3 2 1 0 0 182R 3R R 0 0 5 2 3 0 122R R R 2 0 2 2 0 1 36

x x x s s zé ùê úê ú- + -ê úê ú+ ê úë û

All of the numbers in the last row are nonnegative, so we are finished pivoting. Create a 1 in the columns corresponding to 2x and 2.s

1 2 3 1 21 2 1

1 13 3 351 2

2 23 3 3

R R 1 1 0 0 6

R R 0 0 1 0 4

2 0 2 2 0 1 36

x x x s s zé ù ê úê ú

-ê úê úê úê úë û

The maximum value is 36 when 1 0,x = 2 6,x =

3 0,x = 1 0,s = and 2 4.s =

3. 1 2 1 2 3

1 3 1 0 0 0 122 1 0 1 0 0 101 1 0 0 1 0 42 1 0 0 0 1 0



1 12,= 102 5,= and

41 4;= since 4 is the smallest quotient, 1 in row

3, column 1 is the pivot.

1 2 1 2 3

1 3 1 0 0 0 122 1 0 1 0 0 10

1 1 0 0 1 0 42 1 0 0 0 1 0




1 2 1 2 3

3 1 1

3 2 2

3 4 4

0 2 1 0 1 0 8R R R0 1 0 1 2 0 22R R R1 1 0 0 1 0 4

2R R R 0 1 0 0 2 1 8

x x s s s zé ù-- + ê úê ú- -- + ê úê úê úê ú+ ê úë û

This is a final tableau since all of the numbers in the last row are nonnegative. The maximum value is 8 when 1 4,x = 2 0,x = 1 8,s = 2 2,s = and 3 0.s =

4. 1 2 3 1 2 3

2 1 2 1 0 0 0 254 3 2 0 1 0 0 403 1 6 0 0 1 0 604 2 3 0 0 0 1 0


The most negative indicator is - 4, in the first column. Find the quotients 25

2 12.5,= 404 10,=

and 603 20;= since 10 is the smallest quotient, 4

in row 2, column 1 is the pivot.

1 2 3 1 2 3

2 1 2 1 0 0 0 25

4 3 2 0 1 0 0 403 1 6 0 0 1 0 604 2 3 0 0 0 1 0

x x x s s s z

é ùê úê úê úê úê úê ú- - -ê úë û


1 2 3 1 2 3

2 1 1

2 3 3

2 4 4

0 1 2 2 1 0 0 10R 2R R4 3 2 0 1 0 0 400 5 18 0 3 4 0 1203R 4R R

R R R 0 1 1 0 1 0 1 40

x x x s s s zé ù- -- + ê úê úê úê ú- -- + ê úê ú+ -ê úë û

Since there is still a negative indicator, we must repeat the process. Find the quotients 10

2 5,= 402 20,= and 120

18 6.7;» since 5 is the smallest quotient, 2 in row 1, column 3 is the pivot.

1 2 3 1 2 3

0 1 2 2 1 0 0 104 3 2 0 1 0 0 400 5 18 0 3 4 0 1200 1 1 0 1 0 1 40

x x x s s s zé ù- -ê úê úê úê ú- -ê úê ú

-ê úë û


1 2 3 1 2 3

1 2 2

1 3 3

1 4 4

0 1 2 2 1 0 0 104 4 0 2 2 0 0 30R R R0 4 0 18 6 4 0 309R R R0 1 0 2 1 0 2 90R 2R R

x x x s s s zé ù- -ê úê ú-- + ê úê ú-- + ê úê ú+ ê úë û

All of the numbers in the last row are nonnegative, so we are finished pivoting. Create a 1 in the columns corresponding to x1, x3, s3, and z.

1 2 3 1 2 31 11

1 1 2 22151 11

2 2 2 2 249 3 151

3 3 2 2 241 11

4 4 2 22

0 1 1 0 0 5R R

1 1 0 0 0R R

0 1 0 1 0R R

0 0 1 0 1 45R R

x x x s s s zé ù- - ê úê úê ú- ê úê úê ú- ê úê úê ú ê úë û

The maximum is 45 when 151 2 ,x = 2 0,x =

3 5,x = 1 0,s = 2 0,s = and 153 2 .s =

5. 1 2 3 1 2 3

2 2 8 1 0 0 0 404 5 6 0 1 0 0 602 2 6 0 0 1 0 24

14 10 12 0 0 0 1 0

x x x s s s zé ùê úê ú-ê úê ú-ê úê ú- - -ê úë û


2 20,= 604 15,=

and 242 12;= since 12 is the smallest quotient, 2

in row 3, column 1 is the pivot.

1 2 3 1 2 3

2 2 8 1 0 0 0 404 5 6 0 1 0 0 60

2 2 6 0 0 1 0 2414 10 12 0 0 0 1 0

x x x s s s zé ùê úê ú-ê úê ú-ê úê ú- - -ê úë û


1 2 3 1 2 3

3 1 1

3 2 2

3 4 4

R R R 0 4 2 1 0 1 0 162R R R 0 1 6 0 1 2 0 12

2 2 6 0 0 1 0 247R R R 0 24 30 0 0 7 1 168

x x x s s s z

- + -

- + - - -

-

+ -


Since there is still a negative indicator, we must repeat the process. The second pivot is the 4 in column 2, since 16

4 is the only nonnegative quotient in the only column with a negative

Section 4.2 227


indicator. Performing row transformations again, we get the following tableau.

1 2 3 1 2 3

1 2 2

1 3 3

1 4 4

0 4 2 1 0 1 0 160 0 22 1 4 9 0 64R 4R R4 0 14 1 0 1 0 64R 2R R0 0 42 6 0 1 1 2646R R R

x x x s s s zé ù-ê úê ú- -+ ê úê ú+ ê úê ú+ ê úë û

All of the numbers in the last row are nonnegative, so we are finished pivoting. Create a 1 in the columns corresponding to 1 2, ,x x and 2.s

1 2 3 1 2 31 1 11

1 1 2 4 44911 11

2 2 2 4 447 1 11

3 3 2 4 44

0 1 0 0 4R R

0 0 1 0 16R R

1 0 0 0 16R R0 0 42 6 0 1 1 264

x x x s s s zé ù- ê úê úê ú- - ê úê úê ú ê úê úê úë û

The maximum value is 264 and occurs when 1 16,x = 2 4,x = 3 0,x = 1 0,s = 2 16,s =

and 3 0.s =

6. 1 2 3 1 2

3 2 4 1 0 0 18

2 1 5 0 1 0 81 4 2 0 0 1 0


The most negative indicator is 4.- Of the quotients 182 9= and 8

1 8,= the smallest is 8, so pivot on the 1 in row 2, column 2

1 2 3 1 2

2 1 1

2 3 3

2R R R 1 0 6 1 2 0 22 1 5 0 1 0 8

4R R R 7 0 18 0 4 1 32

x x x s s zé ù- + - - -ê úê úê úê ú+ ê úë û

This solution is optimal. The basic variables are x2 and 1.s The maximum is 32 when 1 0,x =

2 8,x = 3 0,x = 1 2,s = and 2 0.s =

7. Maximize 1 23 5z x x= +

subject to: 1 2

1 2

4 252 3 15

x xx x

+ £

+ £

with 1 20, 0.x x³ ³

Two slack variables, 1s and 2,s need to be introduced. The problem can be restated as:


subject to: 1 2 1

1 2 2

4 252 3 15

x x sx x s

+ + =

+ + =

with 1 2 1 20, 0, 0, 0.x x s s³ ³ ³ ³


1 23 5 0.x x z- - + =

The initial simplex tableau follows.

1 2 1 2

4 1 1 0 0 252 3 0 1 0 153 5 0 0 1 0


The most negative indicator is 5,- in the second column. To select the pivot from column 2, find the quotients 25

1 25= and 153 5.= The smaller

quotient is 5, so 3 is the pivot.

1 2 1 2

4 1 1 0 0 25

2 3 0 1 0 153 5 0 0 1 0


1 2 1 2

2 1 1

2 3 3

R 3R R 10 0 3 1 0 602 3 0 1 0 15

5R 3R R 1 0 0 5 3 75

x x s s zé ù- + -ê úê úê úê ú+ ê úë û

All of the indicators are nonnegative. Create a 1in the columns corresponding to 2 1, ,x s and .z

1 2 1 2

101 11 13 3 3

1 2 12 23 3 3

1 513 33 3 3

R R 0 1 0 20

R R 1 0 0 5

R R 0 0 1 25

x x s s zé ù -ê úê úê ú ê úê úê ú ê úë û

The maximum value is 25 when 1 0,x = 2 5,x =

1 20,s = and 2 0.s =


subject to: 1 2

1 2

2 4 153 10

x xx x

+ £

+ £

with 1 20, 0.x x³ ³

Two slack variables s1 and s2 need to be introduced. The problem can be restated as:




subject to: 1 2 1

1 2 2

2 4 153 10

x x sx x s

+ + =

+ + =

with 1 2 1 20, 0, 0, 0.x x s s³ ³ ³ ³


1 25 2 0.x x z- - + =

The initial simplex tableau as follows.

1 2 1 2

2 4 1 0 0 153 1 0 1 0 105 2 0 0 1 0


The most negative indicator is 5- in the first column. To select the pivot from column 1, find the quotients 15

2 and 103 , so 3 is the pivot.

1 2 1 2

2 4 1 0 0 15

3 1 0 1 0 105 2 0 0 1 0


1 2 1 2

2 1 1

2 3 3

2R 3R R 0 10 3 2 0 253 1 0 1 0 10

5R 3R R 0 1 0 5 3 50

x x s s zé ù- + -ê úê úê úê ú+ -ê úë û

Since there is still a negative indicator, we must repeat the process. Find the quotients 25

10 and 101 .

Since the smaller quotient is 2510 , the 10 in row 1,

column 2 is the pivot.

1 2 1 2

0 10 3 2 0 253 1 0 1 0 100 1 0 5 3 50

x x s s zé ù-ê úê úê úê ú-ê úë û

1 2 1 2

1 2 2

1 3 3

0 10 3 2 0 25R 10R R 30 0 3 12 0 75R 10R R 0 0 3 48 30 525

x x s s zé ù-ê úê ú- + -ê úê ú+ ê úë û

All of the indicators are nonnegative. Create a 1in the columns corresponding to 1 2, ,x x and z.

1 2 1 2

3 5111 1 10 5 210

51 1 22 230 10 5 2

1 8 3513 330 10 5 2

0 1 0R R

R R 1 0 0

R R 0 0 1

x x s s zé ù- ê úê úê ú -ê úê úê ú ê úë û

The maximum value is 17.5 when 1 2.5,x =

2 12.5, 0,x s= = and 2 0.s =

9. Maximize 1 210 12z x x= +

subject to: 1 2

1 2

1 2

4 2 205 502 2 24

x xx xx x

+ £

+ £

+ £

with 1 20, 0.x x³ ³

Three slack variables, 1 2, ,s s and 3,s need to be introduced. The initial tableau is as follows.

1 2 1 2 3

4 2 1 0 0 0 205 1 0 1 0 0 502 2 0 0 1 0 24

10 12 0 0 0 1 0


The most negative indicator is 12,- in column 2.

The quotients are 20 502 110, 50,= = and

242 12;= the smallest is 10, so 2 in row 1,

column 2 is the pivot.

1 2 1 2 3

4 2 1 0 0 0 205 1 0 1 0 0 502 2 0 0 1 0 24

10 12 0 0 0 1 0


1 2 1 2 3

1 2 2

1 3 3

1 4 4

4 2 1 0 0 0 206 0 1 2 0 0 80R 2R R2 0 1 0 1 0 4R R R

14 0 6 0 0 1 1206R R R

x x s s s zé ùê úê ú-- + ê úê ú- -- + ê úê ú+ ê úë û

All of the indicators are nonnegative, so we are finished pivoting. Create a 1 in the columns corresponding to 2x and 2.s

Section 4.2 229


1 2 1 2 312

1 11 12 21

2 22

2 1 0 0 0 10R R 3 0 1 0 0 40R R 2 0 1 0 1 0 4

14 0 6 0 0 1 120

x x s s s zé ùê úê ú ê ú-ê úê ú - -ê úê úê úë û

The maximum value is 120 when 1 0,x =

2 10,x = 1 0,s = 2 40,s = and 3 4.s =

10. Maximize 1 21.5 4.2z x x= +

subject to: 1 2

1 2

2.8 3.4 211.4 2.2 11

x xx x

+ £

+ £

with 1 20, 0.x x³ ³

The initial tableau follows.

1 2 1 2

2.8 3.4 1 0 0 21

1.4 2.2 0 1 0 111.5 4.2 0 0 1 0


Pivot on the 2.2 in row 2, column 2.

1 2 1 2

2 1 1

2 3 3

1.7R 1.1R R 0.7 0 1.1 1.7 0 4.41.4 2.2 0 1 0 11

2.1R 1.1R R 1.29 0 0 2.1 1.1 23.1

x x s s zé ù- + -ê úê úê úê ú+ ê úë û

All of the indicators are nonnegative. Create a 1 in the columns corresponding to 2 2, ,x s and .z

1 2 1 2

7 1711 11.1 11 11

7 512 22.2 11 11

1291 213 31.1 110 11

R R 0 1 0 4

R R 1 0 0 5

R R 0 0 1 21

x x s s zé ù -ê úê úê ú ê úê úê ú ê úë û

The maximum is 21 when 1 0,x = 2 5,x =

1 4,s = and 2 0.s =

11. Maximize 1 2 38 3z x x x= + +

subject to: 1 2 3

1 2 3

6 8 1185 10 220

x x xx x x+ + £

+ + £

with 1 2 30, 0, 0.x x x³ ³ ³

Two slack variables, s1 and s2, need to be introduced. The initial simplex tableau is as follows.

1 2 3 1 2

1 6 8 1 0 0 1181 5 10 0 1 0 2208 3 1 0 0 1 0


The most negative indicator is 8,- in the first column. The quotients are 118

1 118= and 2201 220;= since 118 is the smaller, 1 in row 1,

column 1 is the pivot. Performing row transformations, we get the following tableau.

1 2 3 1 2

1 2 2

1 3 3

1 6 8 1 0 0 118R R R 0 1 2 1 1 0 102

8R R R 0 45 63 8 0 1 944

x x x s s zé ùê úê ú- + - -ê úê ú+ ê úë û

All of the indicators are nonnegative, so we are finished pivoting. The maximum value is 944 when

1 118,x = 2 0,x = 3 0,x = 1 0,s = and

2 102.s =

12. Maximize 1 2 38 10 7z x x x= + +

subject to: 1 2 3

1 2 3

3 2 105 8

x x xx x x

+ + £

+ + £

with 1 2 30, 0, 0.x x x³ ³ ³

The initial tableau is as follows.

1 2 3 1 2

1 3 2 1 0 0 10

1 5 1 0 1 0 88 10 7 0 0 1 0


Pivot on the 5 in row 2, column 2.

1 2 3 1 2

2 1 1

2 3 3

2 0 7 5 3 0 263R 5R R1 5 1 0 1 0 8

2R R R 6 0 5 0 2 1 16

x x x s s zé ù-- + ê úê úê úê ú+ - -ê úë û


1 2 3 1 2

2 1 1

2 3 3

2R R R 0 10 5 5 5 0 101 5 1 0 1 0 8

6R R R 0 30 1 0 8 1 64

x x x s s zé ù- + - -ê úê úê úê ú+ ê úë û



Create a 1 in the column corresponding to 1.s

1 2 3 1 21

1 15 R R 0 2 1 1 1 0 21 5 1 0 1 0 81 30 1 0 8 1 64

x x x s s zé ù - -ê úê úê úê úê úë û

The maximum value is 64 when 1 8,x = 2 0,x = 3 0,x = 1 2,s = and 2 0.s =

13. Maximize 1 2 3 410 15 10 5z x x x x= + + +

subject to: 1 2 3 4

1 2 3 4

3002 3 360

x x x xx x x x

+ + + £

+ + + £

with 1 2 3 40, 0, 0, 0.x x x x³ ³ ³ ³


1 2 3 4 1 2

1 1 1 1 1 0 0 300

1 2 3 1 0 1 0 36010 15 10 5 0 0 1 0

x x x x s s zé ùê úê úê úê ú- - - -ê úë û

In the column with the most negative indicator, 15,- the quotients are 300

1 300=

and 3602 180.= The smaller quotient is 180,

so the 2 in row 2, column 2, is the pivot.

1 2 3 4 1 2

2 1 1

2 3 3

R 2R R 1 0 1 1 2 1 0 2401 2 3 1 0 1 0 360

15R 2R R 5 0 25 5 0 15 2 5400

x x x x s s zé ù- + - -ê úê úê úê ú+ -ê úë û


1 2 3 4 1 2

1 2 2

1 3 3

1 0 1 1 2 1 0 240R R R 0 2 4 0 2 2 0 120

5R R R 0 0 20 10 10 10 2 6600

x x x x s s z

- -

- + -

+

é ùê úê úê úê úê úë û

Create a 1 in the columns corresponding to x2 and z.

1 2 3 4 1 2

12 22

13 32

1 0 1 1 2 1 0 240R R 0 1 2 0 1 1 0 60

0 0 10 5 5 5 1 3300R R

x x x x s s zé ù- -ê úê ú -ê úê úê ú ë û

The maximum value is 3300 when 1 240,x =

2 60,x = 3 0,x = 4 0,x = 1 0,s = and

2 0.s =

14. Maximize 1 2 3 44 5z x x x x= + + +

subject to: 1 2 3 4

1 2 3 4

1 3

2 3 1152 8 5 200

50

x x x xx x x xx x

+ + + £

+ + + £

+ £

with 1 2 3 40, 0, 0, 0.x x x x³ ³ ³ ³

1 2 3 4 1 2 3

1 2 3 1 1 0 0 0 115

2 1 8 5 0 1 0 0 2001 0 1 0 0 0 1 0 501 1 4 5 0 0 0 1 0


Pivot on the 5 in row 2, column 4

1 2 3 4 1 2 3

2 1 1

2 4 4

3 9 7 0 5 1 0 0 375R 5R R2 1 8 5 0 1 0 0 2001 0 1 0 0 0 1 0 50

R R R 1 0 4 0 0 1 0 1 200

x x x x s s s z

é ù-- + ê úê úê úê úê úê ú+ ê úë û

Create a 1 in the columns corresponding to 4x and 1.s

1 2 3 4 1 2 3

3 9 7 111 1 5 5 5 55

2 1 8 112 2 5 5 5 55

0 1 0 0 75R R

1 0 0 0 40R R

1 0 1 0 0 0 1 0 501 0 4 0 0 1 0 1 200

x x x x s s s zé ùê ú- ê úê úê ú ê úê úê úê úê úê úë û

This solution is optimal. The maximum is 200 when 1 0,x = 2 0,x = 3 0,x = 4 40,x =

1 75,s = 2 0,s = and 3 50.s =


subject to: 1 2

1 2

5 254 3 12x xx x- £

- £

with 1 20, 0.x x³ ³

1 2 1 2

1 5 1 0 0 254 3 0 1 0 124 6 0 0 1 0

x x s s zé ù-ê úê ú-ê úê ú- -ê úë û

The most negative indicator is 6.- The negative quotients 25/( 5)- and 12/( 3)- indicate an unbounded feasible region, so there is no unique optimum solution.

Section 4.2 231


16. Maximize 1 2 32 5z x x x= + +

subject to: 1 2 3

1 2 3

5 2 304 3 6 72

x x xx x x

- + £

- + £

with 1 2 30, 0, 0.x x x³ ³ ³

1 2 3 1 2

1 5 2 1 0 0 304 3 6 0 1 0 722 5 1 0 0 1 0

x x x s s zé ù-ê úê ú-ê úê ú- - -ê úë û

The most negative indicator is 5.- The negative quotients 30/( 5)- and 72/( 3)- indicate an unbounded feasible region, so there is no unique optimum solution.

17. Maximize 1 2 3 4 537 34 36 30 35z x x x x x= + + + +

subject to:

1 2 3 4 5

1 2 3 4 5

1 2 3 4 5

1 2 3 4 5

16 19 23 15 21 42,00015 10 19 23 10 25,000

9 16 14 12 11 23,00018 20 15 17 19 36,000

x x x x xx x x x xx x x x xx x x x x

+ + + + £

+ + + + £

+ + + + £

+ + + + £

with 1 2 3 4 50, 0, 0, 0, 0.x x x x x³ ³ ³ ³ ³

Four slack variables, 1,s 2,s 3,s and 4,s need to be introduced. The initial simplex tableau follows.

1 2 3 4 5 1 2 3 4

16 19 23 15 21 1 0 0 0 0 42,00015 10 19 23 10 0 1 0 0 0 25,0009 16 14 12 11 0 0 1 0 0 23,000

18 20 15 17 19 0 0 0 1 0 36,00037 34 36 30 35 0 0 0 0 1 0

x x x x x s s s s zé ùê úê úê úê úê úê úê úê ú- - - - -ê úë û

Using a graphing calculator or computer program, the maximum value is found to be 70,818.18 when

1 181.82,x = 2 0,x = 3 454.55,x = 4 0,x =

5 1363.64,x = 1 0,s = 2 0,s = 3 0,s = and

4 0.s =

18. Maximize 1 2 3 4 52.0 1.7 2.1 2.4 2.2z x x x x x= + + + +

subject to:

1 2 3 4 5

1 2 3 4 5

1 2 3 4 5

1 2 3 4 5

12 10 11 12 13 42508 8 7 18 5 41309 10 12 11 8 35005 3 4 5 4 1600

x x x x xx x x x xx x x x xx x x x x

+ + + + £

+ + + + £

+ + + + £

+ + + + £

with 1 2 3 4 50, 0, 0, 0, 0.x x x x x³ ³ ³ ³ ³

Four slack variables, 1 2 3, , ,s s s and 4,s need to be introduced. The initial simplex tableau follows.

1 2 3 4 5 1 2 3 4

12 10 11 12 13 1 0 0 0 0 42508 8 7 18 5 0 1 0 0 0 41309 10 12 11 8 0 0 1 0 0 35005 3 4 5 4 0 0 0 1 0 1600

2.0 1.7 2.1 2.4 2.2 0 0 0 0 1 0

x x x x x s s s s zé ùê úê úê úê úê úê úê úê ú- - - - -ê úë û

Using a graphing calculator or computer program, the maximum value is found to be 795.68 when

1 0,x = 2 0,x = 3 46.97,x = 4x = 176.72,

5x = 124.05, 1s = 0, 2s = 0, 3s = 0, and

4s = 32.31.

23. Organize the information in a table.

Church Group

Labor Union

Maximum Time Available

Letter Writing 2 2 16

Follow-up 1 3 12 Money Raised $100 $200

Let x1 and x2 be the number of church groups and labor unions contacted respectively. We need two slack variables, s1 and s2.


subject to: 1 2 1

1 2 2

2 2 163 12

x x sx x s

+ + =

+ + =

with 1 2 1 20, 0, 0, 0.x x s s³ ³ ³ ³

The initial simplex tableau is as follows.

1 2 1 2

2 2 1 0 0 16

1 3 0 1 0 12100 200 0 0 1 0





1 2 1 2

2 1 1

2 3 3

2R 3R R 4 0 3 2 0 241 3 0 1 0 12

200R 3R R 100 0 0 200 3 2400



1 2 1 2

1 2 2

1 3 3

4 0 3 2 0 24R 4R R 0 12 3 6 0 24

25R R R 0 0 75 150 3 3000


This is a final tableau, since all of the indicators are nonnegative. Create a 1 in the columns corresponding to 1 2, ,x x and .z

1 2 1 23 11

1 1 4 241 1 1

2 212 4 21

3 33

1 0 0 6R R

R R 0 1 0 2

0 0 25 50 1 1000R R

x x s s zé ù- ê úê úê ú -ê úê úê úê úë û

The maximum amount of money raised is $1000/mo when 1 6x = and 2 2,x = that is, when 6 churches and 2 labor unions are contacted.

24. (a) Let x1 be the number of Flexscan sets and x2 be the number of Panoramic I sets. The problem can be stated as follows.


subject to: 1 2

1 2

1 2

5 7 36002 900

4 4 2600

x xx xx x

+ £

+ £

+ £

with 1 20, 0.x x³ ³

Since there are three constraints, introduce slack variables 1 2, ,s s and 3s and set up the initial tableau.

1 2 1 2 3

5 7 1 0 0 0 3600

1 2 0 1 0 0 9004 4 0 0 1 0 2600350 500 0 0 0 1 0



1 2 1 2 3

2 1 1

2 3 3

2 4 4

3 0 2 7 0 0 9007R 2R R1 2 0 1 0 0 9002 0 0 2 1 0 8002R R R

100 0 0 250 0 1 225,000250R R R

x x s s s z

é ù-- + ê úê úê úê ú-- + ê úê ú-+ ê úë û


1 2 1 2 3

1 2 2

1 3 3

1 4 4

3 0 2 7 0 0 9000 6 2 10 0 0 1800R 3R R0 0 4 8 3 0 6002R 3R R

100R 3R R 0 0 200 50 0 3 765,000

x x s s s z

é ù-ê úê ú-- + ê úê ú-- + ê úê ú+ ê úë û

Create a 1 in the columns corresponding to 1 2, ,x x and .z

1 2 1 2 3721

1 1 3 33521

2 2 3 36

200 5014 4 3 33

1 0 0 0 300R R

0 1 0 0 300R R0 0 4 8 3 0 600

0 0 0 1 255,000R R

x x s s s zé ù- ê úê úê ú- ê úê ú-ê úê úê ú ê úë û

The optimal solution is $255,000 when 300 Flexscan and 300 Panoramic I sets are produced. (This agrees with the graphical solution found in Exercise 10 of Section 3.3.)

(b) Since 1 2 15 7 3600,x x s+ + = let 1 300x = and 2 300x = and solve for 1.s

1

1

5(300) 7(300) 36000

ss

+ + =

=

Similarly, find s2 and s3.

1 2 2

2

2

2 900300 2(300) 900

0

x x sss

+ + =

+ + =

=

and

1 2 3

3

3

4 4 26004(300) 4(300) 2600

200.

x x sss

+ + =

+ + =

=

There are 200 leftover hours in the testing and packing department.

25. (a) Let x1 be the number of Royal Flush poker sets, x2 be the number of Deluxe Diamond sets, and 3x be the number of Full House sets. The problem can be stated as follows.

Maximize 1 2 338 22 12z x x x= + +

subject to:

1 2 3

1 2 3

1 2 3

1 2 3

1000 600 300 2,800,0004 2 2 10,000

10 5 5 25,0002 6000

x x xx x xx x xx x x

+ + £

+ + £

+ + £

+ + £

Section 4.2 233


with 1 2 30, 0, 0.x x x³ ³ ³

Since there are four constraints, introduce slack variables, 1 2 3, , ,s s s and 4s and set up the initial simplex tableau.

1 2 3 1 2 3 4

1000 600 300 1 0 0 0 0 2,800,0004 2 2 0 1 0 0 0 10,000

10 5 5 0 0 1 0 0 25,0002 1 1 0 0 0 1 0 6000

38 22 12 0 0 0 0 1 0

x x x s s s s zé ùê úê úê úê úê úê úê úê ú- - -ê úë û

Using a graphing calculator or computer program, the maximum profit is $104,000 and is obtained when 1000 Royal Flush poker sets, 3000 Deluxe Diamond poker sets, and no Full House poker sets are assembled.

(b) According to the poker chip constraint:

2

1

1000(1000) 600(3000) 300(0)2,800,000 0.

ss

+ + +

= =

So all of the poker chips are used. Checking the card constraint:

2

2

4(1000) 2(3000) 2(0) 10,0000.

ss

+ + + =

=

So all of the cards are used. Checking the dice constraint:

3

3

10(1000) 5(3000) 5(0) 25,0000.

ss

+ + + =

=

So all of the dice are used. Finally, checking the dealer button constraint:

4

4

2(1000) 3000 0 60001000.

ss

+ + + =

=

This means there are 1000 unused dealer buttons.

26. (a) Let x1 be the number of loaves of raisin bread and x2 be the number of raisin cakes.

Then 1 2

1 2

5 1502 90

x xx x

+ £

+ £

and 1 22 150.x x+ £

To maximize 1 21.75 4 ,z x x= + add s1, s2, and s3 as slack variables. The initial tableau will be as follows.

1 2 1 2 3

74

1 5 1 0 0 0 1501 2 0 1 0 0 902 1 0 0 1 0 150

4 0 0 0 1 0

x x s s s zé ùê úê úê úê úê úê úê ú- -ê úë û


1 2 1 2 3

1 2 2

1 3 319

1 4 4 4

1 5 1 0 0 0 150

3 0 2 5 0 0 1502R 5R R9 0 1 0 5 0 600R 5R R

0 4 0 0 5 6004R 5R R

x x s s s zé ùê úê ú-- + ê úê ú-ê ú- + ê úê ú-+ ê úë û


1 2 1 2 3

2 1 1

2 3 35 95 502519

2 4 4 2 4 24

0 15 5 5 0 0 300R 3R R3 0 2 5 0 0 150

3R R R 0 0 5 15 5 0 150

0 0 0 15R 3R R

x x s s s zé ù-- + ê úê ú-ê úê ú- + -ê úê ú

+ ê úê úë û

Create a 1 in the columns corresponding to x1, x2, s3, and z.

1 2 1 2 31 1 11 115 3 31 522 23 3 31

3 3519 33511

4 4 6 12 215

R R 0 1 0 0 20

R R 1 0 0 0 50

0 0 1 3 1 0 30R R0 0 0 1R R

x x s s s zé ù -ê úê úê ú -ê úê ú- ê úê úê ú ê úë û

The optimal solution occurs when 1 50x = and 2 20;x = that is, when 50 loaves of raisin bread and 20 raisin cakes are baked.

(b) 3352 167.5;= the maximum gross income is

$167.50.

(c) When 1 50x = and 2 20,x = the number of units used are as follows.

Flour: 50 + 5(20) = 150

This is the total amount of available flour.

Sugar: 50 + 2(20) = 90

This is the total amount of available sugar.

Raisins: 2(50) + 20 = 120

This leaves 150 - 120, or 30 units, of raisins.

234 CHAPTER 4 LINEAR PROGRAMMING: THE SIMPLEX METHOD


Since

1 2 3

3

3

2 150,2(50) 20 150

30.

x x sss

+ + =

+ + =

=

27. (a) Let 1x represent the number of racing bicycles, 2x the number of touring bicycles, and 3x the

number of mountain bicycles.

From Exercise 28 in Section 4.1, the initial simplex tableau is as follows.

1 2 3 1 2

17 27 34 1 0 0 91,80012 21 15 0 1 0 42,000

8 12 22 0 0 1 0



1 2 3 1 2

1 2 2

1 3 3

17 27 34 1 0 0 91,80015R 34R R 153 309 0 15 34 0 51, 000

11R 17R R 51 93 0 11 0 17 1, 009,800

x x x s s z

- + -

+


This is a final tableau, since all of the indicators are nonnegative. Create a 1 in the columns corresponding to 3,x 2,s and z.

1 2 3 1 2

271 111 1 2 34 3434

9 309 1512 234 2 34 34

1 93 113 317 17 17

1 0 0 2700R R

R R 0 1 0 1500

R R 3 0 0 1 59, 400

x x x s s zé ù

ê úê úê ú -ê úê úê ú ê úë û

From the tableau, 1 0,x = 2 0,x = and 3x = 2700. The company should make no racing or touring bicycles and 2700 mountain bicycles.

(b) From the third row of the final tableau, the maximum profit is $59,400.

(c) When 1 0,x = 2 0,x = and 3 2700,x = the number of units of steel used is

17(0) 27(0) 34(2700) 91,800+ + =

which is all the steel available. The number of units of aluminum used is

12(0) 21(0) 15(2700) 40,500+ + =

which leaves 42,000 40,500 1500- = units of aluminum unused.

Checking the second constraint:

1 2 3 2

2

2

12 21 15 42,00012(0) 21(0) 15(2700) 42,000

1500.

x x x sss

+ + + =

+ + + =

=

28. (a) The tableau and set up were explained in Exercise 30 of Section 4.1.

1 2 3 1 2 3

2 2 3 1 0 0 0 8001 1 1 0 1 0 0 400

0 1 1 0 0 1 0 20030 40 60 0 0 0 1 0



1 2 3 1 2 3

3 1 1

3 2 2

3 4 4

3R R R 2 1 0 1 0 3 0 2001 0 0 0 1 1 0 200R R R0 1 1 0 0 1 0 200

60R R R 30 20 0 0 0 60 1 12,000

x x x s s s zé ù- + - -ê úê ú-- + ê úê úê úê ú

+ -ê úë û


1 2 3 1 2 3

1 2 2

1 4 4

2 1 0 1 0 3 0 2000 1 0 1 2 1 0 200R 2R R0 1 1 0 0 1 0 2000 5 0 15 0 15 1 15,00015R R R

x x x s s s zé ù- -ê úê ú-- + ê úê úê úê ú+ ê úë û

Create a 1 in the column corresponding to x1.

1 2 3 1 2 331 11

1 1 2 2 221 0 0 0 100R R0 1 0 1 2 1 0 2000 1 1 0 0 1 0 2000 5 0 15 0 15 1 15,000

x x x s s s zé ù- - ê úê ú-ê úê úê úê úê úë û

The maximum profit is $15,000 when 1 100,x =

2 0,x = and 3 200,x = that is, when 100 basic sets, no regular sets, and 200 deluxe sets are made.

(b) Even though regular sets make a larger profit, there are only 200 slicers available. Since slicers are used in regular and deluxe sets, and deluxe sets account for $20 more profit, slicers should be used in deluxe sets (as many as possible) with any leftovers used in regular sets.

Section 4.2 235


29. (a) Let 1x be the number of newspaper ads, 2x be the number of Internet banner ads, and 3x be the number of TV ads. Here is the initial tableau:

1 2 3 1 2 3 4

400 20 2000 1 0 0 0 0 80001 0 0 0 1 0 0 0 300 1 0 0 0 1 0 0 600 0 1 0 0 0 1 0 10

4000 3000 10,000 0 0 0 0 1 0

x x x s s s s z

é ùê úê úê úê úê úê úê úê úê ú- - -ë û


1 2 3 1 2 3 4

1 4 4

1 5 5

400 20 2000 1 0 0 0 0 80001 0 0 0 1 0 0 0 30

0 1 0 0 0 1 0 0 60R 2000R R 400 20 0 1 0 0 2000 0 12,000

5R R R 2000 2900 0 5 0 0 0 1 40,000

x x x s s s s zé ùê úê úê úê úê úê ú

- + - - -ê úê úê ú+ - -ë û


1 2 3 1 2 3 4

3 1 1

3 4 4

3 5 5

20R R R 400 20 2000 1 0 20 0 0 68001 0 0 0 1 0 0 0 300 1 0 0 0 1 0 0 60

20R R R 400 0 0 1 0 20 2000 0 13, 2002900R R R 2000 0 0 5 0 2900 0 1 214,000

x x x s s s s zé ù- + -ê úê úê úê úê úê ú

+ - -ê úê úê ú+ -ë û


1 2 3 1 2 3 4

1 2 2

1 4 4

1 5 5

400 0 2000 1 0 20 0 0 68000 0 2000 1 400 20 0 0 5200R 400R R0 1 0 0 0 1 0 0 600 0 2000 0 0 0 2000 0 20,000R R R0 0 10,000 10 0 2800 0 1 248,0005R R R

x x x s s s s zé ù-ê úê ú- -- + ê úê úê úê ú+ ê úê ú+ ê úë û

Create a 1 in the columns corresponding to 1 2, ,x s and 4.s

1 2 3 1 2 3 41 1 11 1400 400 201 1 12 2400 400 20

14 42000

R R 1 0 5 0 0 0 17

R R 0 0 5 1 0 0 13

0 1 0 0 0 1 0 0 600 0 1 0 0 0 1 0 10R R0 0 10,000 10 0 2800 0 1 248,000

x x x s s s s zé ù -ê úê úê ú - -ê úê úê úê úê úê úê úë û



This is the final tableau. The maximum exposure is 248,000 women when 17 newspaper ads, 60 Internet banner ads, and no TV ads are used.

30. (a) Let 1x represent the number of toy trucks and 2x the number of toy fire engines.

Maximize 1 28.50 12.10z x x= +

subject to: 1 2

1

2

1 2

2 3 24,0006600550010,000

x xxx

x x

+ £

£

£

+ £

The initial tableau:

1 2 1 2 3 4

2 3 1 0 0 0 0 24,0001 1 0 1 0 0 0 10,0001 0 0 0 1 0 0 6600

0 1 0 0 0 1 0 5500850 1210 0 0 0 0 100 0

x x s s s s zé ùê úê úê úê úê úê úê úê úê ú- -ë û

We pivot on the indicated 1 in column 2.

1 2 1 2 3 4

4 1 1

4 2 2

4 5 5

3R R R 2 0 1 0 0 3 0 7500R R R 1 0 0 1 0 1 0 4500

1 0 0 0 1 0 0 66000 1 0 0 0 1 0 5500

1210R R R 850 0 0 0 0 1210 100 6,655,000

x x s s s s zé ù- + -ê úê ú- + -ê úê úê úê úê úê úê ú+ -ë û

Next pivot on the 2 in column 1, row 1.

1 2 1 2 3 4

1 2 2

1 3 3

1 5 5

2 0 1 0 0 3 0 7500

R 2R R 0 0 1 2 0 1 0 1500R 2R R 0 0 1 0 2 3 0 5700

0 1 0 0 0 1 0 5500425R R R 0 0 425 0 0 65 100 9,842,500

x x s s s s zé ù-ê úê ú- + -ê úê ú- + -ê úê úê úê úê ú+ -ë û

Finally, pivot on the 1 in column 6, row 2.

1 2 1 2 3 4

2 1 1

2 3 3

2 4 4

2 5 5

2 0 2 6 0 0 0 12,0003R R R0 0 1 2 0 1 0 15000 0 2 6 2 0 0 12003R R R0 1 1 2 0 0 0 4000R 1R R0 0 360 130 0 0 100 9,940,00065R R R

x x s s s s zé ù-+ ê úê ú-ê úê ú-- + ê úê ú-- + ê úê ú+ ê úë û

From this tableau we can read the solution:

1 212,000 9,940,0006000, 4000, 99,400

2 100x x z= = = = =

Section 4.2 237


Produce 6000 trucks and 4000 fire engines for a maximum profit of $99,400.

x1 6600

x2 5500

2 x1 3 x2 24,000x1 x2 10,000

=

=

==

++x1

x2

(6000, 4000)(6600, 3400)

slope = −1

slope = −2/3

(3750, 5500)

The slopes of the two diagonal constraint lines are 231 and .- -

(b) As we decrease the profit p for five engines from $12.10, the slope of the objective function line

1 28.50z x px= + moves from 8.5012.10- toward 1.- When 8.50,p = the slope reaches 1- and now the

corner point at (6600, 3400) yields a maximum profit. So, for a fire-engine profit of $8.50 produce 6600 trucks and 3400 fire engines.

(c) As we increase the profit p for fire engines from $12.10, the slope of the objective function line

1 28.50z x px= + moves from 8.5012.10- toward 2

3 .- When 12.75,p = the slope is 8.50 212.75 3- = - and now

the corner point at (3750, 5500) yields a maximum profit. So, for a fire-engine profit of $12.75 produce 3750 trucks and 5500 fire engines.

31. (a) The coefficients of the objective function are the profit coefficients from the table: 5, 4, and 3; choice (3) is correct.

(b) The constraints are the available man-hours for the 2 departments, 400 and 600; choice (4) is correct.

(c) 1 2 32 3 1 400X X X+ + £ is the constraint on department 1; choice (3) is correct.

32. (a) Look at the first table, which has to do with the profits. The profit-maximization formula is

$2A $5B $4C ,X+ + =

so the answer is choice (1).

(b) Look at the “Painting” row of the second chart. The “Painting” constraint is

1A 2B 2C 38,000,+ + £

so the answer is choice (3).

33. Maximize 100 200z x y= +

subject to: 2 2 163 12

x yx y+ £+ £

with 0, 0.x y³ ³

Using Excel, we enter the variables x and y in cells Al and Bl, respectively. Enter the x- and y-coordinates of the initial corner point of the feasible region, (0, 0), in cells A2 and B2, respectively, and NAME these cells x and y, respectively. In cells C2, C4, C5, C6, and C7, enter the formula for the function to maximize and each of the constraints: 100 200 ,x y+ 2 2 ,x y+ 3 ,x y+ x, and y. Since x and y have been set to 0, all the cells containing formulas should also show the value 0, as below.



A B C 1 x y 2 0 0 0 3 4 0 5 0 6 0 7 0

Using the SOLVER, ask Excel to maximize the value in cell C2 subject to the constraints C4 16,£ C5 12,£ C6 0,³ C7 0.³ Make sure you have checked off the box Assume Linear Model in SOLVER OPTIONS.

Excel returns the following values and allows you to choose a report.

A B C 1 x y 2 6 2 10003 4 165 126 67 2

Select the sensitivity report. The report will appear on a new sheet of the spread sheet.

Adjustable Cells

Cell Name Final Value

Reduced Cost

Objective Coefficient

Allowable Increase

Allowable Decrease

$A$2 x 6 0 100 100 33.33333333 $B$2 y 2 0 200 100 100

Constraints


Shadow Price

Constraint R.H. Side

Allowable Increase

Allowable Decrease

$C$4 16 25 16 8 8$C$5 12 50 12 12 4$C$6 6 0 0 6 1E 30+$C$7 2 0 0 2 1E 30+

The church group’s allowable increase is $100 and the allowable decrease is $33.33. So their contribution can be as high as $100 $100 $200+ = or as low as $100 $33.33 $66.67- = and the original solution is still optimal. The unions’ allowable increase is $100 and the allowable decrease is $100. So their contribution can be as high as $200 $100 $300+ = or as low as $200 $100 $100- = and the original solution is still optimal.

34. Maximize 350 500z x y= +

subject to: 5 7 36002 900

4 4 2600

x yx yx y

+ £+ £+ £

with 0, 0.x y³ ³

Using Excel, we enter the variables x and y in cells A1 and B1, respectively. Enter the x- and y-coordinates of the initial corner point of the feasible region, (0, 0), in cells A2 and B2, respectively, and NAME these cells x and y, respectively. In cells C2, C4, C5, C6, C7, and C8, enter the formula for the function to maximize and each of the constraints:

Section 4.2 239


350 500 ,x y+ 5 7 ,x y+ 2 ,4 4 , ,x y x y x+ + and y. Since x and y have been set to 0, all the cells containing formulas should also show the value 0, as below.

A B C 1 x y 2 0 0 03 4 05 06 07 08 0

Using the SOLVER, ask Excel to maximize the value in cell C2 subject to the constraints C4 3600,£ C5 900,£ C6 3600,£ C7 0,³ C8 0.³ Make sure you have checked off the box Assume Linear Model in SOLVER OPTIONS.

Excel returns the following values and allows you to choose a report.

A B C 1 x y 2 300 300 2550003 4 36005 9006 24007 3008 300

Select the sensitivity report. The report will appear on a new sheet of the spread sheet.

Adjustable Cells


Reduced Cost

Objective Coefficient

Allowable Increase

Allowable Decrease

$A$2 x 300 0 350 7.142857143 100 $B$2 y 300 0 500 200 10

Constraints


Shadow Price

Constraint R.H. Side

Allowable Increase

Allowable Decrease

$C$8 3600 66.66666667 3600 150 450$C$4 900 16.66666667 900 128.5714286 75$C$7 2400 0 2600 1E 30+ 200$C$5 300 0 0 300 1E 30+$C$6 300 0 0 300 1E 30+

For the Flexscan sets, the allowable increase is $7.14 and the allowable decrease is $100. So the profit from the bargain sets can be as high as $350 $7.14 $357.14+ = or as low as $350 $100 $250- = and the original solution is still optimal. For the Panoramic I sets, the allowable increase is $200 and the allowable decrease is $10. So the profit from the Panoramic I sets can be as high as $500 $200 $700+ = or as low as $500 $10 $490- = and the original solution is still optimal.



35. Let 1x = number of hours running, x2 be the number of hours biking, and x3 be the number hours walking. The problem can be stated as follows.

Maximize 1 2 3531 472 354z x x x= + +

subject to: 1 2 3

1

2 3

153

2 0

x x xx

x x

+ + £

£

- £

with 1 2 30, 0, 0.x x x³ ³ ³

We need three slack variables, s1, s2, and s3. The initial simplex tableau as follows.

1 2 3 1 2 3

1 1 1 1 0 0 0 15

1 0 0 0 1 0 0 30 2 1 0 0 1 0 0

531 472 354 0 0 0 1 0

x x x s s s z

é ùê úê úê úê ú-ê úê ú- - -ê úë û


1 2 3 1 2 3

2 1 1

2 4 4

0 1 1 1 1 0 0 12R R R1 0 0 0 1 0 0 3

0 2 1 0 0 1 0 00 472 354 0 351 0 1 1593531R R R

x x x s s s zé ù-- + ê úê úê úê ú-ê úê ú

- -+ ê úë û


1 2 3 1 2 3

3 1 1

3 4 4

R 2R R 0 0 3 2 2 1 0 241 0 0 0 1 0 0 30 2 1 0 0 1 0 0

236R R R 0 0 590 0 531 236 1 1593

x x x s s s zé ù- + - -ê úê úê úê ú-ê úê ú

+ -ê úë û

Finally pivot on the 3 in row 1, column 3.

1 2 3 1 2 3

1 3 3

1 4 4

0 1 3 2 2 1 0 241 0 0 0 1 0 0 30 6 0 2 2 2 0 24R 3R R

590R 3R R 0 0 0 1180 413 118 3 18,939

x x x s s s zé ù- -ê úê úê úê ú-+ ê úê ú+ ê úë û

Create a 1 in the columns corresponding to x2, x3, and z.

1 2 3 1 2 32 2 11

1 1 3 3 33

1 1 1 13 36 3 3 3

1 1180 413 1184 43 3 3 3

0 0 1 0 8R R1 0 0 0 1 0 0 3

R R 0 1 0 0 4

R R 0 0 0 1 6313

x x x s s s zé ù- - ê úê úê úê úê ú -ê úê úê ú ê úë û

Lauren should run 3 hours, bike 4 hours, and walk 8 hours for a maximum calorie expenditure of 6313 calories.

Section 4.2 241


36. (a) Let 1x represent the number of hours doing calisthenics, 2x be the number of hours swimming, and 3x be the number of hours playing the drums. The problem can be stated as follows.

Maximize 1 2 3388 518 345z x x x= + +

subject to: 1 2 3

1 2 3

3

102 0

4

x x xx x x

x

+ + £

- + - £

£

with 1 2 30, 0, 0.x x x³ ³ ³

We need three slack variables. The initial simplex tableau is as follows.

1 2 3 1 2 3

1 1 1 1 0 0 0 10

1 2 1 0 1 0 0 00 0 1 0 0 1 0 4

388 518 345 0 0 0 1 0

x x x s s s zé ùê úê ú- -ê úê úê úê ú- - -ê úë û


1 2 3 1 2 3

2 1 1

2 4 4

3 0 3 2 1 0 0 20R 2 R R1 2 1 0 1 0 0 00 0 1 0 0 1 0 4

647 0 604 0 259 0 1 0259R R R

x x x s s s zé ù-- + ê úê ú- -ê úê úê úê ú- -+ ê úë û


1 2 3 1 2 3

1 2 2

1 4 4

3 0 3 2 1 0 0 20R 3R R 0 6 0 2 2 0 0 20

0 0 1 0 0 1 0 4647 R 3R R 0 0 129 1294 130 0 3 12,940

x x x s s s zé ù-ê úê ú+ ê úê úê úê ú+ ê úë û

Create a 1 in the columns corresponding to 1 2, ,x x and z.

1 2 3 1 2 3202 11

1 1 3 3 33101 11

2 2 3 3 36

12,9401294 13014 4 3 3 33

1 0 1 0 0R R

0 1 0 0 0R R0 0 1 0 0 1 0 4

0 0 43 0 1R R

x x x s s s zé ù- ê úê úê ú ê úê úê úê úê ú ê úë û

Joe should do 203 hours of calisthenics, 10

3 hours of swimming, and 0 hours of playing the drums for a

maximum calorie expenditure of 12,9403 or 1

34313 calories.

37. (a) Let x1 represent the number of species A, x2 represent the number of species B, and x3 represent the number of species C.

Maximize 1 2 31.62 2.14 3.01z x x x= + +



subject to: 1 2 3

1 2 3

1 2 3

1.32 2.1 0.86 4902.9 0.95 1.52 8971.75 0.6 2.01 653

x x xx x xx x x

+ + £

+ + £

+ + £

with 1 2 30, 0, 0.x x x³ ³ ³

Use a graphing calculator or computer to solve this problem and find that the answer is to stock none of species A, 114 of species B, and 291 of species C for a maximum combined weight of 1119.72 kg.

(b) When 1 0,x = 2 114,x = and 3 291,x = the number of units used are as follows.

Food I: 1.32(0) 2.1(114) 0.86(291) 489.66+ + =

or 490 units, which is the total amount available of Food I.

Food II: 2.9(0) 0.95(114) 1.52(291) 550.62+ + =

or 551 units, which leaves 897 551,- or 346 units of Food II available.

Food III: 1.75(0) 0.6(114) 2.01(291) 653.31+ + =

or 653 units, which is the total amount available of Food III.

(c) Many answers are possible. The idea is to choose average weights for species B and C that are considerably smaller than the average weight chosen for species A, so that species A dominates the objective function.

(d) Many answers are possible. The idea is to choose average weights for species A and B that are considerably smaller than the average weight chosen for species C.

38. (a) Let 1x = amount of P, 2x = amount of Q, 3x = amount of R, and 4x = amount of S (all in kilograms).

We desire to maximize

1 2 3 490 70 60 50z x x x x= + + +

subject to:

3 4

2 3 4

1 2 3

0.375 0.625 5000.75 0.5 0.375 6000.25 0.125 300

x xx x x

x x x

+ £

+ + £

+ + £

with 1 2 3 40, 0, 0, 0.x x x x³ ³ ³ ³

If we rewrite the constraints as

3 4

2 3 4

1 2 3

3 5 5008 8

3 1 3 6004 2 81 1 300,4 8

x x

x x x

x x x

+ £

+ + £

+ + £

and then multiply each inequality by the least common denominator, 8, we get a set of constraints without fractions.

3 4

2 3 4

1 2 3

3 5 40006 4 3 4800

8 2 2400

x xx x x

x x x

+ £

+ + £

+ + £

Section 4.2 243



1 2 3 4 1 2 3

0 0 3 5 1 0 0 0 40000 6 4 3 0 1 0 0 4800

8 2 1 0 0 0 1 0 240090 70 60 50 0 0 0 1 0


The first pivot is the 8 in row 3, column 1.

1 2 3 4 1 2 3

3 4 4

0 0 3 5 1 0 0 0 40000 6 4 3 0 1 0 0 48008 2 1 0 0 0 1 0 24000 190 195 200 0 0 45 4 108,00045R 4 R R

x x x x s s s zé ùê úê úê úê úê úê ú

- - -+ ê úë û


1 2 3 4 1 2 3

1 2 2

1 4 4

0 0 3 5 1 0 0 0 4000

0 30 11 0 3 5 0 0 12,0003R 5R R8 2 1 0 0 0 1 0 24000 190 75 0 40 0 45 4 268,00040 R R R

x x x x s s s z

é ùê úê ú-- + ê úê úê úê ú

- -+ ê úë û


1 2 3 4 1 2 3

2 3 3

2 4 4

0 0 3 5 1 0 0 0 4000

0 30 11 0 3 5 0 0 12,000120 0 4 0 3 5 15 0 24,000R 15R R

0 0 16 0 63 95 135 12 1,032,00019 R 3R R

x x x x s s s z

é ùê úê ú-ê úê ú-- + ê úê ú

-+ ê úë û


1 2 3 4 1 2 3

2 1 1

2 3 3

2 4 4

0 90 0 55 20 15 0 0 80003R 11R R0 30 11 0 3 5 0 0 12,000

1320 120 0 0 45 75 165 0 216,0004 R 11R R0 480 0 0 645 1125 1485 132 11,544,00016 R 11R R

x x x x s s s zé ù- -- + ê úê ú-ê úê ú- -- + ê úê ú+ ê úë û

1 2 3 4 1 2 318 4 3 16001

1 1 11 11 11 115530 3 5 12,0001

2 2 11 11 11 11111 3 15 1 18001

3 3 11 88 264 8 11132040 215 1125 45 962,0001

4 4 11 44 132 4 11132

0 0 1 0 0R R

0 1 0 0 0R R

1 0 0 0R R

0 0 0 1R R

x x x x s s s zé ù- -ê úê úê úê ú- ê úê úê ú- - ê úê úê ú

ê úê úë û



This final tableau gives the solution

1 2

3

4

1800 163.6, 0,1112,000 1090.9,111600 145.5,11

x x

x

x

= » =

= »

= »

and 962,000 87, 454.5.11z = »

Produce 163.6 kg of food P, none of food Q, 1090.9 kg of R, and 145.5 kg of S.

(b) The maximum total growth value is read from the bottom row of the final tableau: 962,00011

87,454.5.»

(c) When ,11800

11x = 2 0,x = ,312,000

11x = and ,41600

11x = the number of units of nutrient A used is

12,000 16000.375 0.625 50011 11

æ ö æ ö÷ ÷ç ç+ =÷ ÷ç ç÷ ÷÷ ÷ç çè ø è ø

which is the total amount of nutrient A available. The number of units of nutrient B used is 12,000 16000.75(0) 0.5 0.375 600

11 11æ ö æ ö÷ ÷ç ç+ + =÷ ÷ç ç÷ ÷÷ ÷ç çè ø è ø

which is all the units of nutrient B. The amount of nutrient C used is 1800 12,0000.25(0) 0.125 300

11 11æ ö æ ö÷ ÷ç ç+ + =÷ ÷ç ç÷ ÷÷ ÷ç çè ø è ø

which is all of the nutrient C. So none of the nutrients are left over.

39. Let x1 represent the number of minutes for the senator, x2 the number of minutes for the congresswoman, and x3 the number of minutes for the governor.

Of the half-hour show’s time, at most only 30 3 27- = min are available to be allotted to the politicians. The given information leads to the inequality

1 2 3 27x x x+ + £

and the inequalities

1 32x x³ and 1 3 22 ,x x x+ ³

and we are to maximize the objective function

1 2 335 40 45 .z x x x= + +

Rewrite the equation as

3 1 227x x x£ - -

and the inequalities as

1 32 0x x- + £ and 1 2 32 0.x x x- + - £

Substitute 1 227 x x- - for x3 in the objective function and the inequalities, and the problem is as follows.

Maximize 1 2 335 40 45z x x x= + +

subject to: 1 3

1 2 3

1 2 3

2 02 0

27

x xx x xx x x

- + £

- + - £

+ + £

Section 4.2 245


with 1 2 30, 0, 0.x x x³ ³ ³


1 2 3 1 2 3

1 0 2 1 0 0 0 0

1 2 1 0 1 0 0 01 1 1 0 0 1 0 27

35 40 45 0 0 0 1 0

x x x s s s zé ù-ê úê ú- -ê úê úê úê ú- - -ê úë û


1 2 3 1 2 3

2 3 3

2 4 4

1 0 2 1 0 0 0 01 2 1 0 1 0 0 03 0 3 0 1 2 0 54R 2R R

20R R R 55 0 65 0 20 0 1 0

x x x s s s zé ù-ê úê ú- -ê úê ú-- + ê úê ú+ - -ê úë û


1 2 3 1 2 3

1 2 2

1 3 3

1 4 4

1 0 2 1 0 0 0 03 4 0 1 2 0 0 0R 2R R

3R 2R R 9 0 0 3 2 4 0 10865R 2R R 175 0 0 65 40 0 2 0

x x x s s s zé ù-ê úê ú-+ ê úê ú- + - -ê úê ú+ -ê úë û


1 2 3 1 2 3

3 1 1

3 2 2

3 4 4

R 9R R 0 0 18 6 2 4 0 108R 3R R 0 12 0 0 4 4 0 108

9 0 0 3 2 4 0 108175R 9R R 0 0 0 60 10 700 18 18,900

x x x s s s z+ é ù-ê ú

ê ú+ ê úê ú- -ê úê ú+ ê úë û

Create a 1 in the columns corresponding to x1, x2, x3, and z.

13

1 2 3 1 2 31 1 1 21 118 3 9 9

1 1 12 212 3 31 2 43 39 9 9

1 10 5 3504 418 3 9 9

0 0 1 0 6R R

0 1 0 0 0 9R R

1 0 0 0 12R R

0 0 0 1 1050R R

x x x s s s zé ù -ê úê úê ú ê úê úê ú ê ú- -ê úê ú

ê úê úë û

The maximum value of z is 1050 when 1 12,x = 2 9,x = and 3 6.x = That is, for a maximum of 1,050,000 viewers, the time allotments should be 12 minutes for the senator, 9 minutes for the congresswoman, and 6 minutes for the governor.



40. (a) Let 1x = number of large fund-raising parties, 2x = number of letters requesting funds, and 3x = number of dinner parties.

Maximize 1 2 3200,000 100,000 600,000z x x x= + +

subject to: 1 2 3

1 2 3

253000 1000 12,000 102,000

x x xx x x

+ + £

+ + £

with 1 2 30, 0, 0.x x x³ ³ ³

We need two slack variables. The initial simplex tableau is as follows.

1 2 3 1 2

1 1 1 1 0 0 253000 1000 0 1 0 102,00012,000

200,000 100,000 600,000 0 0 1 0

x x x s s zé ùê úê úê úê úê ú- - -ê úë û

Pivot on the 12,000 in row 2, column 3.

1 2 3 1 2

2 1 1

2 3 3

R 12,000R R 9000 11,000 0 12,000 1 0 198,0003000 1000 12,000 0 1 0 102,000

50R R R 50,000 50,000 0 0 50 1 5,100,000

x x x s s zé ù- + -ê úê úê úê ú+ - -ê úë û

Pivot on 9000 in row 1 column 1.

1 2 3 1 2

1 2 2

1 3 3

9000 11,000 0 12,000 1 0 198,000R 3R R 0 8000 36,000 12,000 4 0 108,000

50R 9R R 0 100,000 0 600,000 400 9 55,800,000

x x x s s zé ù-ê úê ú- + - -ê úê ú+ ê úë û


1 2 3 1 2

1 11 4 11 19000 9 3 90001 2 1 12 236,000 9 3 9000

1 100,000 200,000 4003 39 9 3 9

R R 1 0 0 22

R R 0 1 0 3

0 0 1 6,200,000R R

x x x s s zé ùê ú ê ú-ê úê ú ê ú- -ê úê úê úê úë û

The maximum amount of money is $6,200,000 when 1 22,x = 2 0,x = and 3 3,x = that is, when 22 fund-raising parties, no mailings, and 3 dinner parties are planned.

Section 4.3 247


4.3 Minimization Problems; Duality

Your Turn 1

Write the augmented matrix.

3 3 4 245 1 3 27

25 12 27 0


Transpose to get the matrix for the dual problem. 3 5 253 1 124 3 27

24 27 0


Write the dual problem. Maximize 1 224 27z x x= +

subject to: 1 2

1 2

1 2

3 5 253 12

4 3 27

x xx x

x x

+ £

+ £

+ £

with 1 20, 0x x³ ³

Your Turn 2


3 5 203 1 18

15 12 0


Transpose to get the matrix for the dual problem.

3 3 155 1 12

20 18 0


Write the dual problem.


subject to: 1 2

1 2

3 3 155 12x xx x+ £

+ £

with 1 20, 0.x x³ ³

The initial tableau for this problem is

1 2 1 2

3 3 1 0 0 15

5 1 0 1 0 1220 18 0 0 1 0



1 2 1 2

2 1 1

2 3 3

3R 5R R 0 12 5 3 0 395 1 0 1 0 12

4R R R 0 14 0 4 1 48


Now pivot around the indicated 12:

1 2 1 2

1 2 2

1 3 3

0 12 5 3 0 39R 12R R 60 0 5 15 0 1057R 6R R 0 0 35 3 6 561


Finally divide the last row by 6 to produce a 1 in the z column:

1 2 1 2

35 18713 3 6 2 2

0 12 5 3 0 3960 0 5 15 0 105

R /6 R 0 0 1

x x s s zé ùê ú-ê úê ú-ê úê ú ê úë û

In the original problem, w has a minimum of 1872 when

351 6y = and 1

2 2 .y =

4.3 Warmup Exercises

W1. Maximize 1 25 3z x x= +

subject to:

1 2

1 2

1

2

5 2 203 6 18

00

x xx x

xx

+ ≤+ ≤

≥≥


1 2 1

1 2 2

5 2 203 6 18

x x sx x s

+ + =+ + =


1 25 3 0z x x− − =

The intial simplex tableau is

1 2 1 2

5 2 1 0 0 203 6 0 1 0 185 3 0 0 1 0

x x s s z

− −

The most negative element in the last rwo is in column 1, and the corresponding quotients are



20 45

= and 18 63

=

Since 4 6,< we pivot on the 5 in the upper left corner.

1 2 1 2

1 2 2

1 3 3

5 2 1 0 0 200 24 3 5 0 303R 5R R0 1 1 0 1 20R R R

x x s s z

−− + → −+ →

A negative element remains in column 2 of the last row. The corresponding quotients are

20 102

= and 30 1.2524

=

Since 1.25 10,< we now pivot on the 24 in column 2.

1 2 1 2

2 1 1

2 3 3

( 1)R 12R R 60 0 15 5 0 2100 24 3 5 0 300 0 21 5 24 510R 24R R

x x s s z− + → −

−

+ →

The last row has no negative elements, so we can now read the solution.

1 2210 30 5103.5 1.25 21.2560 24 24

x x z= = = = = =

A maximum of 21.25 is obtained at 1 3.5x = and 2 1.25x = .

W2. Maximize 1 23 2z x x= +

subject to:

1 2

1 2

1

2

2 3 184 3 24

00

x xx x

xx

+ ≤+ ≤

≥≥


1 2 1

1 2 2

2 3 184 3 24

x x sx x s

+ + =+ + =


1 23 2 0z x x− − =


1 2 1 2

2 3 1 0 0 184 3 0 1 0 243 2 0 0 1 0

x x s s z

− −

The most negative element in the last row is in column 1, and the corresponding quotients are

18 92

= and 24 64

=

Since 6 9,< we pivot on the 4 in column 1.

1 2 1 2

2 1 1

2 3 3

( 1)R 2R R 0 3 2 1 0 124 3 0 1 0 240 1 0 3 4 723R 4R R

x x s s z

− + → − + →

The last row has no negative elements, do we can now read the solution.

1 224 726 0 184 4

x x z= = = = =

A maximum of 18 is obtained at 1 6x = and 2 0.x =

4.3 Exercises

1. To form the transpose of a matrix, the rows of the original matrix are written as the columns of the transpose. The transpose of

1 2 33 2 11 10 0

é ùê úê úê úê úë û

is 1 3 12 2 10 .3 1 0


2. The transpose of a matrix is found by exchanging the rows and columns. The transpose of

3 4 2 0 12 0 11 5 7é ù-ê úê úë û

is 3 24 0

.2 110 51 7

é ùê úê úê úê ú-ê úê úê úê úê úë û

3. The transpose of

4 5 3 157 14 20 85 0 2 23

é ù-ê úê ú-ê úê ú-ë û

is

Section 4.3 249


4 7 55 14 0

.3 20 2

15 8 23

é ùê úê úê úê ú- -ê úê ú-ê úë û

4. The transpose of

1 11 150 10 64 12 21 1 132 25 1

é ùê úê ú-ê úê ú-ê úê ú-ê úê ú-ê úë û

is 1 0 4 1 2

11 10 12 1 25 .15 6 2 13 1

é ùê úê ú-ê úê ú- - -ë û

5. Maximize 1 2 34 3 2z x x x= + +

subject to: 1 2 3

1 2

1 2 3

54

2 3 15

x x xx xx x x

+ + £

+ £

+ + £

with 1 2 30, 0, 0.x x x³ ³ ³

To form the dual, first write the augmented matrix for the given problem.

1 1 1 51 1 0 42 1 3 154 3 2 0


Then form the transpose of this matrix. 1 1 2 41 1 1 31 0 3 25 4 15 0


The dual problem is stated from this second matrix (using y instead of x).

Minimize 1 2 35 4 15w y y y= + +

subject to: 1 2 3

1 2 3

1 3

2 43

3 2

y y yy y yy y

+ + ³

+ + ³

+ ³

with 1 2 30, 0, 0.y y y³ ³ ³

6. Maximize 1 2 32 7 4z x x x= + +

subject to: 1 2 3

1 2 3

4 2 267 8 33

x x xx x x

+ + £

+ + £

with 1 2 30, 0, 0.x x x³ ³ ³

To find the dual, first write the augmented matrix for the problem.

4 2 1 261 7 8 332 7 4 0


Then form the transpose of this matrix. 4 1 22 7 71 8 4

26 33 0


The dual problem is:

Minimize 1 226 33w y y= +

subject to: 1 2

1 2

1 2

4 22 7 7

8 4

y yy yy y

+ ³

+ ³

+ ³

with 1 20, 0.y y³ ³

7. Minimize 1 2 3 43 6 4w y y y y= + + +

subject to: 1 2 3 4

1 2 3 4

1502 2 3 4 275

y y y yy y y y

+ + + ³

+ + + ³

with 1 2 3 40, 0, 0, 0.y y y y³ ³ ³ ³

To find the dual problem, first write the augmented matrix for the problem.

1 1 1 1 1502 2 3 4 2753 6 4 1 0


Then form the transpose of this matrix.

1 2 31 2 61 3 41 4 1

150 275 0

é ùê úê úê úê úê úê úê úê úê úë û

The dual problem is




subject to: 1 2

1 2

1 2

1 2

2 32 63 44 1

x xx xx xx x

+ £

+ £

+ £

+ £

with 1 20, 0.x x³ ³

8. Minimize 1 2 34w y y y= + +

subject to: 1 2 3

1 2 3

1 3

2 3 1152 8 200

50

y y yy y yy y

+ + ³

+ + ³

+ ³

with 1 2 30, 0, 0.y y y³ ³ ³

Write the augmented matrix for the problem.

1 2 3 1152 1 8 2001 0 1 501 1 4 0


Form the transpose of this matrix.

1 2 1 12 1 0 13 8 1 4

115 200 50 0


The dual problem is:

Maximize 1 2 3115 200 50z x x x= + +

subject to: 1 2 3

1 2

1 2 3

2 12 13 8 4

x x xx xx x x

+ + £

+ £

+ + £

with 1 2 30, 0, 0.x x x³ ³ ³

9. Find 1 0y ³ and 2 0y ³ such that

1 2

1 2

2 3 62 7

y yy y

+ ³

+ ³

and 1 25 2w y y= + is minimized.

Write the augmented matrix for this problem.

2 3 62 1 75 2 0



2 2 53 1 26 7 0


Use this matrix to write the dual problem.

Find 1 0x ³ and 2 0x ³ such that

1 2

1 2

2 2 53 2

x xx x

+ £

+ £


Introduce slack variables s1 and s2. The initial tableau is as follows.

1 2 1 2

2 2 1 0 0 5

3 1 0 1 0 26 7 0 0 1 0


Pivot on the 1 in row 2, column 2, since that column has the most negative indicator and that row has the smallest nonnegative quotient.

1 2 1 2

2 1 1

2 3 3

2R R R 4 0 1 2 0 13 1 0 1 0 2

7R R R 15 0 0 7 1 14

x x s s zé ù- + - -ê úê úê úê ú+ ê úë û

The minimum value of w is the same as the maximum value of z. The minimum value of w is 14 when 1 0y = and 2 7.y = (Note that the values of y1 and y2 are given by the entries in the bottom row of the columns corresponding to the slack variables in the final tableau.)


1 2

1 2

2 3 155 6 35

y yy y

+ ³

+ ³



2 3 155 6 352 3 0


Form the transpose to get the matrix for the dual problem.

Section 4.3 251


2 5 23 6 3

15 35 0


Use this matrix to write the dual problem:


1 2

1 2

2 5 23 6 3

x xx x

+ £

+ £


Introduce slack variables and write the initial tableau.

1 2 1 2

2 1 0 0 253 6 0 1 0 3

15 35 0 0 1 0



1 2 1 2

1 2 2

1 3 3

2 5 1 0 0 2

6R 5R R 3 0 6 5 0 37R R R 1 0 7 0 1 14

x x s s zé ùê úê ú- + -ê úê ú+ -ê úë û

Because the quotients in the pivot column are the same, we have a choice for the second pivot. Choose the 3 in row 2, column 1, as the second pivot.

1 2 1 2

2 1 1

2 3 3

2R 3R R 0 15 15 10 0 03 0 6 5 0 3

R 3R R 0 0 15 5 3 45

x x s s zé ù- + -ê úê ú-ê úê ú+ ê úë û


1 2 1 2

1 21 115 3

1 52 23 3

1 53 33 3

R R 0 1 1 0 0

R R 1 0 2 0 1

R R 0 0 5 1 15

x x s s zé ùê ú -ê úê úê ú -ê úê úê ú

ê úê úë û

The minimum is 15 when y1 = 5 and 52 3

.y =

For the second pivot, if the 2 in row 1, column 1, was chosen instead, the minimum would still be 15 but would occur when 15

21y = and y2 = 0. So,

any point on the line segment between ( )53

5, and

( )152

, 0 is a solution.


1 2

1 2

10 5 10020 10 150

y yy y

+ ³

+ ³



10 5 10020 10 1504 5 0



10 20 45 10 5

100 150 0


Write the dual problem from this matrix.


1 2

1 2

10 20 45 10 5x xx x

+ £

+ £



1 2 1 2

10 20 1 0 0 45 10 0 1 0 5

100 150 0 0 1 0



1 2 1 2

1 1 2

1 3 3

10 20 1 0 0 4R 2R R 0 0 1 2 0 6

15R 2R R 50 0 15 0 2 60





1 2 1 2

1 3 3

10 20 1 0 0 40 0 1 2 0 6

5R R R 0 100 20 0 2 80

x x s s zé ùê úê ú-ê úê ú+ ê úë û

Create a 1 in the columns corresponding to x1, s2, and z.

1 2 1 21 1 2

1 110 10 51 1

2 22 21

3 32

R R 1 2 0 0

R R 0 0 1 0 3

0 50 10 0 1 40R R

x x s s zé ù ê úê úê ú

ê - úê úê úê ú ê úë û

The minimum value of w is 40 when 1 10y = and

2 0.y = (These values of y1 and y2 are read from the last row of the columns corresponding to s1 and s2 in the final tableau.)

12. Minimize 1 229 10w y y= +

subject to: 1 2

1 2

3 2 25 3

y yy y

+ ³

+ ³

with 1 20, 0.y y³ ³


3 2 25 1 3

29 10 0


From the transpose to get the matrix for the dual problem.

3 5 292 1 102 3 0


Write the dual problem from this matrix:


subject to: 1 2

1 2

3 5 292 10x xx x

+ £

+ £

with 1 20, 0.x x³ ³

Write the initial tableau.

1 2 1 2

3 5 1 0 0 292 1 0 1 0 102 3 0 0 1 0



1 2 1 2

1 2 2

1 3 3

3 5 1 0 0 29

R 5R R 7 0 1 5 0 213R 5R R 1 0 3 0 5 87



1 2 1 2

2 1 1

2 3 3

3R 7R R 0 35 10 15 0 1407 0 1 5 0 21

R 7R R 0 0 20 5 35 630

x x s s zé ù- + -ê úê ú-ê úê ú+ ê úë û


1 2 1 2

1 2 31 135 7 7

1 1 52 27 7 7

1 4 13 335 7 7

R R 0 1 0 4

R R 1 0 0 3

R R 0 0 1 18

x x s s zé ù

ê - úê úê úê ú -ê úê úê ú ê úë û

The minimum is 18 when 41 7

y = and 12 7

.y =

13. Minimize 1 26 10w y y= +

subject to: 1 2

1 2

3 5 154 7 20

y yy y

+ ³

+ ³

with 1 20, 0.y y³ ³ Write the augmented matrix.

3 5 154 7 206 10 0



3 4 65 7 10

15 20 0




Section 4.3 253


subject to: 1 2

1 2

3 4 65 7 10

x xx x

+ £

+ £

with 1 20, 0.x x³ ³ Write the initial tableau for this problem.

1 2 1 2

3 4 1 0 0 65 7 0 1 0 10

15 20 0 0 1 0


Pivot around the indicated 3 to obtain this final tableau:

1 2 1 2

1 2 2

1 3 3

3 4 1 0 0 65R 3R R 0 1 5 3 0 0

5R R R 0 0 5 0 1 30

x x s s zé ùê úê ú- + -ê úê ú+ ê úë û

Instead we could pivot around the 5 in the first row, second column. This produces the following final tableau:

1 2 1 2

2 1 1

2 3 3

3R 5R R 0 1 5 3 0 65 7 0 1 0 10

3R R R 0 1 0 3 1 30

x x s s zé ù- + - -ê úê úê úê ú+ ê úë û

In the original problem, w has a minimum of 30 when 1 5y = and 2 0y = (reading from the first final tableau) or when 1 0y = and 2 3y = (reading from the second final tableau). Any point on the line segment between (5, 0) and (0, 3) also gives the minimum of 30.

14. Minimize 1 23 2w y y= +

subject to: 1 2

1 2

1 2

2 108

2 12

y yy yy y

+ ³

+ ³

+ ³

with 1 20, 0.y y³ ³


1 2 101 1 82 1 123 2 0



1 1 2 32 1 1 2

10 8 12 0


Write the dual problem:

Maximize 1 2 310 8 12z x x x= + +

subject to: 1 2 3

1 2 3

2 32 2x x xx x x

+ + £

+ + £

with 1 2 30, 0, 0.x x x³ ³ ³


1 2 3 1 2

1 1 2 1 0 0 32 1 1 0 1 0 2

10 8 12 0 0 1 0



1 2 3 1 2

1 2 2

3 1 3

1 1 2 1 0 0 3

R 2R R 3 1 0 1 2 0 1R 6R R 4 2 0 6 0 1 18

x x x s s zé ùê úê ú- + -ê úê ú+ - -ê úë û


1 2 3 1 2

2 1 1

2 3 3

R 3R R 0 2 6 4 2 0 83 1 0 1 2 0 1

4R 3R R 0 2 0 14 8 3 58

x x x s s zé ù- + -ê úê ú-ê úê ú+ -ê úë û


1 2 3 1 2

2 1 1

3 2 3

2R R R 6 0 6 6 6 0 63 1 0 1 2 0 1

R 2R R 6 0 0 12 12 3 60

x x x s s zé ù- + - -ê úê ú-ê úê ú+ ê úë û


1 2 3 1 2

11 16

13 33

R R 1 0 1 1 1 0 13 1 0 1 2 0 12 0 0 4 4 1 20R R

x x x s s z

é ù - -ê úê ú-ê úê ú ê úë û

This solution is optimal. The minimum is 20 when 1 4y = and 2 4.y =



15. Minimize 1 2 32 3w y y y= + +

subject to: 1 2 3

1 2

1002 50

y y yy y

+ + ³

+ ³

with 1 2 30, 0, 0.y y y³ ³ ³


1 1 1 1002 1 0 502 1 3 0



1 2 21 1 11 0 3

100 50 0


The dual problem is as follows.


subject to: 1 2

1 2

1

2 213

x xx xx

+ £

+ £

£

with 1 20, 0.x x³ ³


1 2 1 2 3

1 2 1 0 0 0 2

1 1 0 1 0 0 11 0 0 0 1 0 3

100 50 0 0 0 1 0



1 2 1 2 3

2 1 1

2 3 3

2 4 4

R R R 0 1 1 1 0 0 11 1 0 1 0 0 1

R R R 0 1 0 1 1 0 2100R R R 0 50 0 100 0 1 100

x x s s s zé ù- + -ê úê úê úê ú- + - -ê úê ú+ ê úë û

The minimum value of w is 100 when 1 0,y = 2 100,y = and 3 0.y =

16. Minimize 1 2 34 7 9w y y y= + +

subject to: 1 2 3

1 2 3

2 3 4 455 2 40

y y yy y y

+ + ³

+ + ³

with 1 2 30, 0, 0.y y y³ ³ ³ Write the augmented matrix for this problem.

2 3 4 451 5 2 404 7 9 0


Form the transpose of this matrix for the dual problem.

2 1 43 5 74 2 9

45 40 0


This corresponds to the following dual problem.


subject to: 1 2

1 2

1 2

2 43 5 74 2 9

x xx xx x

+ £

+ £

+ £

with 1 20, 0.x x³ ³


1 2 1 2 3

2 1 1 0 0 0 43 5 0 1 0 0 74 2 0 0 1 0 9

45 40 0 0 0 1 0



1 2 1 2 3

1 2 2

1 3 3

1 4 4

2 1 1 0 0 0 4

0 7 3 2 0 0 23R 2R R0 0 2 0 1 0 12R R R0 35 45 0 0 2 18045R 2R R

x x s s s zé ùê úê ú-- + ê úê ú-- + ê úê ú

-+ ê úë û

Now pivot around the indicated 7.

1 2 1 2 3

2 1 1

2 4 4

R 7R R 14 0 10 2 0 0 260 7 3 2 0 0 20 0 2 0 1 0 1

5R R R 0 0 30 10 0 2 190

x x s s s zé ù- + -ê úê ú-ê úê ú-ê úê ú+ ê úë û

Now divide the last row by 2.

Section 4.3 255


1 2 1 2 3

4 4

14 0 10 2 0 0 260 7 3 2 0 0 20 0 2 0 1 0 1

R /2 R 0 0 15 5 0 1 95

x x s s s zé ù-ê úê ú-ê úê ú-ê úê ú ê úë û

The minimum is 95 when 1 15,y = 2 5,y = and

3 0.y =

17. Minimize 1 22z x x= +

subject to: 1 2

1 2

2 12 1

x xx x

- + ³

- ³

with 1 20, 0.x x³ ³

A quick sketch of the constraints 1 22 1x x- + ³ and 1 22 1x x- ³ will verify that the two corresponding half planes do not overlap in the first quadrant of the x1x2-plane. Therefore, this problem (P) has no feasible solution. The dual of the given problem is as follows:

Maximize 1 2w y y= +

subject to: 1 2

1 2

2 12 2

y yy y

- + £

- £

with 1 20, 0.y y³ ³

A quick sketch here will verify that there is a feasible region in the y1y2-plane, and it is unbounded. Therefore, there is no maximum value of w in this problem (D).

(P) has no feasible solution and the objective function of (D) is unbounded; this is choice (a).

18. Since the constraints in Example 4 allow arbitrarily large 1y and 2,y the objective function can be made negative and as large in absolute value as desired by choosing a large enough positive value of 1.y

19. (a) Let y1 = the number of units of regular beer

and y2 = the number of units of light beer.

Minimize 1 232,000 50,000w y y= +

subject to: 1

2

1 2

1 2

1 2

101545

120,000 300,000 9,000,00020

yy

y yy yy y

³

³

+ ³

+ ³

+ ³

with 1 20, 0.y y³ ³

Write the augmented matrix for this problem. 1 0 100 1 151 1 45

120,000 300,000 9,000,0001 1 20

32,000 50,000 0

é ùê úê úê úê úê úê úê úê úê úê úê úë û


1 0 1 120,000 1 32,0000 1 1 300,000 1 50,000

10 15 45 9,000,000 20 0




The dual problem is

Maximize 1 2 3 4 510 15 45 9,000,000 20z x x x x x= + + + +

subject to: 1 3 4 5

2 3 4 5

120,000 32,000300,000 50,000

x x x xx x x x

+ + + £

+ + + £

with 1 2 3 4 50, 0, 0, 0, 0.x x x x x³ ³ ³ ³ ³

Write the initial simplex tableau.

1 2 3 4 5 1 2

1 0 1 120,000 1 1 0 0 32,000

0 1 1 300,000 1 0 1 0 50,00010 15 45 9,000,000 20 0 0 1 0

x x x x x s s zé ùê úê úê úê ú- - - - -ê úë û

Pivot on the 300,000 in row 2, column 3.

1 2 3 4 5 1 2

2 1 1

2 3 3

2R 5R R 5 2 3 0 3 5 2 0 60,0000 1 1 300,000 1 0 1 0 50,000

30R R R 10 15 15 0 10 0 30 1 1,500,000

x x x x x s s zé ù- + - -ê úê úê úê ú+ - -ê úë û


1 2 3 4 5 1 2

1 2 2

1 3 3

5 2 3 0 3 5 2 0 60,000R 3R R 5 5 0 900,000 0 5 5 0 90,0005R R R 15 5 0 0 25 25 20 1 1,800,000

x x x x x s s zé ù- -ê úê ú- + - -ê úê ú+ ê úë û

Create a 1 in the columns corresponding to x3 and x4.

1 2 3 4 5 1 2

1 1

3 3

1 5 2 5 2R R 1 0 1 0 20,0003 3 3 3 31 1 1 1 1 1R R 0 1 0 0900,000 180,000 180,000 180,000 180,000 10

15 5 0 0 25 25 20 1 1,800,000

x x x x x s s zé ù - -ê úê úê úê ú - -ê úê úê úê úë û

The minimum value of w is 1,800,000 when 1 25y = and 2 20.y =

Therefore, 25 units of regular beer and 20 units of light beer should be made for a minimum cost of $1,800,000.

(b) The shadow cost for revenue is 110 dollar or $0.10. An increase in $500,000 in revenue will increase costs to

$1,800,000 $0.10(500,000) $1,850,000.+ =

Section 4.3 257


20. (a) Let y1 = the number of small test tubes and y2 = the number of large test tubes.


Subject to: 1

2

1 2

1 2

90060027002

yy

y yy y

³

³

+ ³

³

with 1 20, 0.y y³ ³

The last constraint can be written as

1 22 0.y y- ³


1 0 9000 1 6001 1 27001 2 0

18 15 0

é ùê úê úê úê úê úê ú-ê úê úê úë û


1 0 1 1 180 1 1 2 15

900 600 2700 0 0

é ùê úê ú-ê úê úê úë û


Maximize 1 2 3900 600 2700z x x x= + +

Subject to: 1 3 4

2 3 4

182 15

x x xx x x

+ + £

+ - £

with 1 2 3 40, 0, 0, 0.x x x x³ ³ ³ ³


1 2 3 4 1 2

1 0 1 1 1 0 0 18

0 1 1 2 0 1 0 15900 600 2700 0 0 0 1 0

x x x x s s zé ùê úê ú-ê úê ú- - -ê úë û


1 2 3 4 1 2

2 1 1

2 3 3

R R R 1 1 0 3 1 1 0 3

0 1 1 2 0 1 0 15

2700 R R R 900 2100 0 5400 0 2700 1 40,500

x x x x s s z

- + - -

-

+ - -



1 2 3 4 1 2

1 2 2

1 3 3

1 1 0 3 1 1 0 32R 3R R 2 1 3 0 2 1 0 51

1800R R R 900 300 0 0 1800 900 1 45,900

x x x x s s zé ù- -ê úê ú+ ê úê ú+ ê úë û

Create a 1 in the columns corresponding to x3 and x4.

1 2 3 4 1 21 1 1 1 1

1 13 3 3 3 31 2 1 2 1

2 23 3 3 3 3

R R 0 1 0 1

R R 1 0 0 17

900 300 0 0 1800 900 1 45,900

x x x x s s zé ù - -ê úê úê ú

ê úê úê úê úê úë û

The minimum cost is 45,900¢, or $459, when 1800 small test tubes and 900 test tubes are ordered.

(b) The shadow cost for the test tubes is $0.17. An increase in the minimum number of test tubes by (3000 2700) 300- = will increase the cost to

$459 $0.17(300) $510.+ =

21. (a) The initial matrix for the original problem is

1 1 1 100400 160 280 20,000 .120 40 60 0


The transposed matrix, for the dual problem, is

1 400 1201 160 40

.1 280 60100 20,000 0


Minimize 1 2100 20,000w y y= +

subject to: 1 2

1 2

1 2

400 120160 40280 60

y yy yy y

+ ³

+ ³

+ ³

with 1 20, 0.y y³ ³

(b) We apply the simplex algorithm to the original maximization problem. The initial tableau is

1 2 3 1 2

1 1 1 1 0 0 100

400 160 280 0 1 0 20,000120 40 60 0 0 1 0





1 2 3 1 2

2 1 1

32 3 310

R R R400 0 240 120 400 1 0 20,000400 160 280 0 1 0 20,000

R R R 0 8 24 0 0.3 1 6000

x x x s s zé ù- + -ê úê úê úê ú+ ê úë û

Create a 1 in the columns corresponding to x1 and s1.

1 2 3 1 21 1

1 1400 4001 1

2 2400 400

R R 0 0.6 0.3 1 0 50

R R 1 0.4 0.7 0 0 50

0 8 24 0 0.3 1 6000

x x x s s z

-é ù ê úê úê ú ê úê úê úê úê úë û

This solution is optimal. A maximum profit of $6000 is achieved by planting 50 acres of potatoes, 0 acres of corn, and 0 acres of cabbage.

From the dual solution, the shadow cost of acreage is 0 and of capital is 3

10.

3New profit 6000 0( 10) 100010

$6300

æ ö÷ç= + - + ÷ç ÷÷çè ø

=

Now calculate the number of acres of each:

Profit 120 P 40C 60 B6300 120 P 40(0) 60(0)

P 52.5.

= + +

= + +=

The farmer will make a profit of $6300 by planting 52.5 acres of potatoes and no corn or cabbage.

(c) 3New profit 6000 0(10) ( 1000)10

$5700

æ ö÷ç= + + -÷ç ÷÷çè ø

=

Calculate the number of acres of each:

Profit 120 P 40C 60 B5700 120 P 40(0) 60(0)

P 47.5.

= + +

= + +=

The farmer will make a profit of $5700 by planting 47.5 acres of potatoes and no corn or cabbage.

22. (a) Maximize 1 21.5x x z+ =

subject to: 1 2

1 2

2

2 2004 3 600

0 90

x xx x

x

+ £

+ £

£ £

with 1 0.x ³

(b) Write the initial tableau.

1 2 1 2 3

1 2 1 0 0 0 2004 3 0 1 0 0 600

0 1 0 0 1 0 901 1.5 0 0 0 1 0



1 2 1 2 3

3 1 1

3 2 2

3 4 4

2 R R R 1 0 1 0 2 0 203R R R 4 0 0 1 3 0 330

0 1 0 0 1 0 901.5R R R 1 0 0 0 1.5 1 135

x x s s s zé ù- + -ê úê ú- + -ê úê úê úê ú+ -ê úë û


1 2 1 2 3

1 2 2

1 4 4

1 0 1 0 2 0 20

0 0 4 1 5 0 2504 R R R0 1 0 0 1 0 900 0 1 0 0.5 1 155R R R

x x s s s zé ù-ê úê ú-- + ê úê úê úê ú

-+ ê úë û


1 2 1 2 32 3 2

2 1 15 5 51 4 1

2 25 5 51 4 1

2 3 35 5 51

2 4 410

R R R 1 0 0 0 120

R R 0 0 1 0 50

R R R 0 1 0 0 40

0 0 0.6 0.1 0 1 180R R R

x x s s s z

é ù+ -ê úê úê ú ê - úê úê ú

- + ê ú-ê úê úê ú+ ë û

The maximum profit is $180 when 1 120x = and 2 40,x = that is, when 120 bears and 40 monkeys are produced.

Section 4.3 259


(c) The corresponding dual problem is as follows:

Minimize 1 2 3200 600 90w y y y= + +

subject to: 1 2

1 2 3

4 12 3 1.5

y yy y y

+ ³

+ + ³

with 1 2 30, 0, 0.y y y³ ³ ³

(d) From the given final tableau, the optimal solution to the dual problem is 1 0.6,y =

2 0.1,y = 3 0,y = and 180.w =

(e) The shadow value for felt is 0.6; an increase in supply of 10 units of felt will increase profit to

$180 0.6(10) $186.+ =

(f) The shadow values are 0.1 for stuffing and 0 for trim. If stuffing and trim are each decreased by 10 units, the profit will be

$180 0.1(10) 0(10) $179.- - =

23. Let y1 = the number of political interviews conducted

and y2 = the number of market interviews conducted.

The problem is:


subject to: 1 2

1 2

1 2

88 10 606 5 40

y yy yy y

+ ³

+ ³

+ ³

with 1 20, 0.y y³ ³


1 1 88 10 606 5 40

45 55 0



1 8 6 451 10 5 558 60 40 0


Write the dual problem:

Maximize 1 2 38 60 40z x x x= + +

subject to: 1 2 3

1 2 3

8 6 4510 5 55

x x xx x x

+ + £

+ + £

with 1 2 30, 0, 0.x x x³ ³ ³


1 2 3 1 2

1 8 6 1 0 0 45

1 10 5 0 1 0 558 60 40 0 0 1 0



1 2 3 1 2

2 1 1

2 3 3

4R 5R R 1 0 10 5 4 0 51 10 5 0 1 0 55

6R R R 2 0 10 0 6 1 330

x x x s s zé ù- + -ê úê úê úê ú+ - -ê úë û


1 2 3 1 2

1 2 2

1 3 3

1 0 10 5 4 0 5R 2 R R 1 20 0 5 6 0 105

R R R 1 0 0 5 2 1 335

x x x s s zé ù-ê úê ú- + -ê úê ú+ -ê úë û


1 2 3 1 2

1 2 2

1 3 3

1 0 10 5 4 0 5

R R R 0 20 10 10 10 0 100R R R 0 0 10 10 2 1 340

x x x s s zé ù-ê úê ú- + - -ê úê ú+ -ê úë û


1 2 3 1 2

2 1 1

2 3 3

2R 5R R 5 40 30 5 0 0 2250 20 10 10 10 0 100

R 5R R 0 20 40 40 0 5 1800

x x x s s zé ù+ ê úê ú- -ê úê ú+ ê úë û

Create a 1 in the columns corresponding to x1, s2, and z.

1 2 3 1 2

11 15

12 210

13 35

R R 1 8 6 1 0 0 45R R 0 2 1 1 1 0 10

0 4 8 8 0 1 360R R

x x x s s z

é ùê úê ú - -ê úê úê ú ë û

The minimum time spent is 360 min when 1 8y = and 2 0,y = that is, when 8 political interviews and no market interviews are done.



24. (a) Let y1 = the number of grams of soybean meal,

y2 = the number of grams of meat byproducts

and y3 = the number of grams of grain.

Minimize 1 2 38 9 10w y y y= + +

subject to: 1 2 3

1 2 3

2.5 4.5 5 545 3 10 60

y y yy y y

+ + ³

+ + ³

with 1 2 30, 0, 0.y y y³ ³ ³


2.5 4.5 5 545 3 10 608 9 10 0



2.5 5 84.5 3 95 10 1054 60 0




subject to: 1 2

1 2

1 2

2.5 5 84.5 3 9

5 10 10

x xx xx x

+ £

+ £

+ £

with 1 20, 0.x x³ ³


1 2 1 2 3

2.5 5 1 0 0 0 84.5 3 0 1 0 0 9

5 10 0 0 1 0 1054 60 0 0 0 1 0


To eliminate the decimal entries, multiply rows 1 and 2 by 2.

1 2 1 2 3

5 10 2 0 0 0 169 6 0 2 0 0 18

5 10 0 0 1 0 1054 60 0 0 0 1 0



1 2 1 2 3

3 1 1

3 2 2

3 4 4

0 0 2 0 1 0 6R R R3R 5R R 30 0 0 10 3 0 60

5 10 0 0 1 0 106 R R R 24 0 0 0 6 1 60

x x s s s zé ù-- + ê úê ú- + -ê úê úê úê ú

+ -ê úë û


1 2 1 2 3

2 3 3

2 4 4

0 0 2 0 1 0 630 0 0 10 3 0 600 60 0 10 9 0 0R 6 R R0 0 0 40 18 5 5404 R 5R R

x x s s s zé ù-ê úê ú-ê úê ú-- + ê úê ú+ ê úë û

Create a 1 in the columns representing x1, x2, s1, and z.

1 2 1 2 31 1

1 12 21 1 1

2 230 3 101 1 3

3 360 6 201

4 45

R R 0 0 1 0 0 3

R R 1 0 0 0 2

R R 0 1 0 0 0

0 0 0 8 3.6 1 108R R

x x s s s z

é ù -ê úê úê ú -ê úê úê ú ê ú-ê úê úê ú ë û

The minimum cost is obtained when 0 g of soybean meal, 8g of meat by products, and 3.6 g of grain are used, or 0 g of soybean meal, 0 g of meat by products and 10.8 g of grain are used.

(b) The minimum cost is $1.08.

(c) After the initial pivot, the tableau is

1 2 1 2 3

0 0 2 0 1 0 630 0 0 10 3 0 60

5 10 0 0 1 0 1024 0 0 0 6 1 60

x x s s s zé ù-ê úê ú-ê úê úê úê ú-ê úë û

Now pivot on the 5 in row 3, column 1.

1 2 1 2 3

3 2 2

3 4 4

0 0 2 0 1 0 66 R R R 0 60 0 10 9 0 0

5 10 0 0 1 0 1024 R 5R R 0 240 0 0 54 5 540

x x s s s zé ù-ê úê ú- + - -ê úê úê úê ú+ ê úë û

Section 4.3 261


1 2 1 2 3

14 45

0 0 2 0 1 0 60 60 0 10 9 0 05 10 0 0 1 0 10

R R 0 48 0 0 10.8 1 108

x x s s s z

é ù-ê úê ú- -ê úê úê úê ú ê úë û

The minimum cost is $108 when 1 0,y =

2 8y = and 3 10.8,y = that is, when 0 grams of soybean meal, 0 grams of meat by-products, and 10.8 grams of grain are mixed.

25. Organize the information in a table.

Units of Nutrient

A (per bag)

Units of Nutrient

B (per bag)

Cost (per bag)

Feed 1 1 2 $3 Feed 2 3 1 $2 Minimum 7 4

Let y1 = the number of bags of feed 1

and y2 = the number of bags of feed 2.

(a) We want the cost to equal $7 for 7 units of A and 4 units of B exactly. Therefore, use a system of equations rather than a system of inequalities.

1 2

1 2

1 2

3 2 73 7

2 4

y yy yy y

+ =

+ =

+ =

Use Gauss-Jordan elimination to solve this system of equations.

3 2 71 3 72 1 4


1 2 2

1 3 3

3 2 7R 3R R 0 7 14

2R 3R R 0 1 2

é ùê úê ú- + ê úê ú- + - -ë û

2 1 1

2 3 3

2R 7 R R 21 0 210 7 14

R 7 R R 0 0 0

é ù- + ê úê úê úê ú+ ë û

11 121

12 27

R R 1 0 10 1 2R R0 0 0

é ùê úê ú ê úê úë û

Thus, 1 1y = and 2 2,y = so use 1 bag of feed 1 and 2 bags of feed 2. The cost will be 3(1) 2+ (2) $7= as desired. The number of units of A is 1(1) 3(2) 7,+ = and the number of units of B is 2(1) 1(2) 4.+ =

(b)

Units of Nutrient

A (per bag)

Units of Nutrient

B (per bag)

Cost (per bag)

Feed 1 1 2 $3 Feed 2 3 1 $2 Minimum 5 4

The problem is:


subject to: 1 2

1 2

3 52 4

y yy y

+ ³

+ ³

with 1 20, 0.y y³ ³



subject to: 1 2

1 2

2 33 2x xx x

+ £

+ £

with 1 20, 0.x x³ ³


1 2 1 2

1 2 1 0 0 3

3 1 0 1 0 25 4 0 0 1 0


Pivot as indicated.

1 2 1 2

2 1 1

2 3 3

R 3R R 0 5 3 1 0 73 1 0 1 0 2

5R 3R R 0 7 0 5 3 10


1 2 1 2

1 2 2

1 3 3

0 5 3 1 0 7R 5R R 15 0 3 6 0 3

7 R 5R R 0 0 21 18 15 99





1 2 1 2

1 3 1 71 15 5 5 5

1 1 2 12 215 5 5 5

1 7 6 333 315 5 5 5

R R 0 1 0

R R 1 0 0

R R 0 0 1

x x s s zé ù

ê - úê úê úê ú -ê úê úê ú ê úë û

Reading from the final column of the final tableau, 2 $1.40x = is the cost of nutrient B and 1x =

$0.20 is the cost of nutrient A. With 5 units of A and 4 units of B, this gives a minimum cost of

5($0.20) 4($1.40) $6.60+ =

as given in the lower right corner. 1.4 ( )75or bags

of feed 1 and 1.2 ( )65or bags of feed 2 should be

used.

26. Let y1 = the number of large bowls.

y2 = the number of small bowls.

y3 = the number of pots for plants.

Minimize 1 2 35 6 4w y y y= + +

subject to: 1 2 3

1 2 3

3 2 4 726 6 2 108

y y yy y y

+ + ³

+ + ³

with 1 2 30, 0, 0.y y y³ ³ ³


3 2 4 726 6 2 1085 6 4 0


Transpose to get the matrix for the dual problem. 3 6 52 6 64 2 4

72 108 0




subject to: 1 2

1 2

1 2

3 6 52 6 64 2 4

x xx xx x

+ £

+ £

+ £

with 1 20, 0.x x³ ³


1 2 1 2 3

3 6 1 0 0 0 52 6 0 1 0 0 64 2 0 0 1 0 4

72 108 0 0 0 1 0



1 2 1 2 31 1 51

1 1 2 6 66

1 2 21 71

1 3 3 3 33

1 4 4

1 0 0 0R R

1 0 1 1 0 0 1R R R

3 0 0 1 0R R R

18 0 18 0 0 1 9018R R R

x x s s s zé ù

ê úê úê ú- -- + ê úê úê ú-- + ê úê ú-+ ê úë û


1 2 1 2 32 1 41

3 1 1 9 6 9610 1 161

3 2 2 9 3 931 1 71

3 3 9 3 93

3 4 4

0 1 0 0R R R

0 0 1 0R R R

1 0 0 0R R

6 R R R 0 0 16 0 6 1 104

x x s s s zé ù-- + ê úê úê úê - ú+ ê úê úê ú- ê úê ú+ ê úë û

The minimum time is 104 hours when 1 16,y =

2 0,y = and 3 6,y = that is, when 16 large bowls, 0 small bowls, and 6 pots for flowers are made.

27. Let y1 = the number of minutes spent walking,

y2 = the number of minutes spent cycling,

and y3 = the number of minutes spent swimming.

Minimize 1 2 3w y y y= + +

subject to: 1 2 3

1 2 3

1

3.5 4 8 1500330

y y yy y yy

+ + ³

+ ³

³

with 1 2 30, 0, 0.y y y³ ³ ³

The second constraint can be written as

1 2 33 0.y y y+ - ³


3.5 4 8 15001 1 3 01 0 0 301 1 1 0

é ùê úê ú-ê úê úê úê úê úë û

Section 4.3 263



3.5 1 1 14 1 0 18 3 0 1

1500 0 30 0

é ùê úê úê úê ú-ê úê úê úë û



subject to: 1 2 3

1 2

1 2

3.5 14 18 3 1

x x xx xx x

+ + £

+ £

- £

with 1 2 30, 0, 0.x s x³ ³ ³


1 2 3 4 1 2

3.5 1 1 1 0 0 0 14 1 0 0 1 0 0 18 3 0 0 0 1 0 1

1500 0 30 0 0 0 1 0

x x x x s s zé ùê úê úê úê ú-ê úê ú- -ê úë û

Using a graphing calculator or computer program, such as Solver in Microsoft Excel, we obtain the optimal answer: 30 minutes walking, 197.25 minutes cycling, and 75.75 minutes swimming for a total minimum time of 303 minutes per week.

28. Let y1 = the number of #1 pills

and y2 = the number of #2 pills.

Organize the given information in a table.

Vitamin

A Vitamin

B1 Vitamin

C Cost

#1 1600 1 20 $0.10 #2 400 1 70 $0.20

Total Needed

3200 5 200

The problem is:

Minimize 1 20.1 0.2w y y= +

subject to: 1 2

1 2

1 2

1600 400 32005

20 70 200

y yy yy y

+ ³

+ ³

+ ³

with 1 20, 0.y y³ ³

Write the augmented matrix for the given problem.

1600 400 32001 1 520 70 2000.1 0.2 0



1600 1 20 0.1400 1 70 0.2

3200 5 200 0


This corresponds to the following dual problem.

Maximize 1 2 33200 5 200z x x x= + +

subject to: 1 2 3

1 2 3

1600 20 0.1400 70 0.2

x x xx x x

+ + £

+ + £

with 1 2 30, 0, 0.x x x³ ³ ³


1 2 3 1 2

1600 1 20 1 0 0 0.1400 1 70 0 1 0 0.2

3200 5 200 0 0 1 0


Pivot on 1600.

1 2 3 1 2

1 2 2

1 3 3

1600 1 20 1 0 0 0.1

R 4R R 0 3 260 1 4 0 0.72R R R 0 3 160 2 0 1 0.2

x x x s s zé ùê úê ú- + -ê úê ú+ - -ê úë û

Pivot on 260.

1 2 3 1 2

2 1 1

2 3 3

R 13R R 20,800 10 0 14 4 0 0.60 3 260 1 4 0 0.7

8R 13R R 0 15 0 18 32 13 8.2

x x x s s z

- + -

-

+ -


Pivot on 10.

1 2 3 1 2

1 2 2

1 3 3

20,800 10 0 14 4 0 0.63R 10R R 62,400 0 2600 52 52 0 5.23R 2R R 62,400 0 0 78 52 26 18.2

x x x s s zé ù-ê úê ú- + - -ê úê ú+ ê úë û



Now divide the last row by 26.

1 2 3 1 2

2 3

20,800 10 0 14 4 0 0.662, 400 0 2600 52 52 0 5.2

R /26 R 2400 0 0 3 2 1 0.7

x x x s s zé ù-ê úê ú- -ê úê ú ê úë û

Greg should buy 3 of pill #1 and 2 of pill #2 for a minimum cost of $0.70.

29. Let y1 = the number of units of ingredient I;

y2 = the number of units of ingredient II;

and y3 = the number of units of ingredient III.

The problem is:

Minimize 1 2 34 7 5w y y y= + +

subject to: 1 2 3

1 2 3

2 3

4 10 103 2 12

4 5 20

y y yy y y

y y

+ + ³

+ + ³

+ ³

with 1 2 30, 0, 0.y y y³ ³ ³


Maximize 1 2 310 12 20z x x x= + +

subject to: 1 2

1 2 3

1 2 3

4 3 42 4 7

10 5 5

x xx x xx x x

+ £

+ + £

+ + £

with 1 2 30, 0, 0.x x x³ ³ ³


1 2 3 1 2 3

4 3 0 1 0 0 0 41 2 4 0 1 0 0 7

10 1 5 0 0 1 0 510 12 20 0 0 0 1 0


Pivot as indicated.

1 2 3 1 2 3

3 2 2

3 4 4

4 3 0 1 0 0 0 44R 5R R 35 6 0 0 5 4 0 15

10 1 5 0 0 1 0 54R R R 30 8 0 0 0 4 1 20

x x x s s s zé ùê úê ú- + - -ê úê úê úê ú

+ -ê úë û

1 2 3 1 2 3

1 2 2

1 3 3

1 4 4

4 3 0 1 0 0 0 443 0 0 2 5 4 0 72R R R26 0 15 1 0 3 0 11R 3R R

122 0 0 8 0 12 3 928R 3R R

x x x s s s z

- - -- +

-- +

+



1 2 3 1 2 31 4 1 4

1 13 3 3 3

1 26 1 1 113 3 15 15 5 515

122 8 9214 4 3 3 33

R R 1 0 0 0 0

43 0 0 2 5 4 0 7

0 1 0 0R R

0 0 0 4 1R R

x x x s s s zé ù ê úê úê ú- - -ê úê ú

- ê úê úê úê ú ê úë û

From the last row, the minimum value is 923

when 8

1 3,y = 2 0,y = and 3 4.y = The biologist can

meet his needs at a minimum cost of $30.67 by using 83

units of ingredient I and 4 units of ingredient III.

(Ingredient II should not be used at all.)

4.4 Nonstandard Problems

Your Turn 1


subject to: 1 2

1 2

3 4 109 7 18

y yy y

+ ³

+ £

with 1 20, 0.y y³ ³ Instead we maximize 1 26 4z w y y= - = - - subject to the same constraints. Inserting slack and surplus variables produces the following initial tableau.

1 2 1 2

3 4 1 0 0 10

9 7 0 1 0 186 4 0 0 1 0

y y s s zé ù-ê úê úê úê úê úë û

Because 1s is negative, we choose the positive entry farthest to the left in row 1, which is the 3 in column 1. The entry 9 in this column gives the smallest quotient so we choose 9 as the pivot.

Section 4.4 265


1 2 1 2

2 1 1

2 3 3

R 3R R 0 5 3 1 0 129 7 0 1 0 18

2R 3R R 0 2 0 2 3 36

y y s s zé ù- + - -ê úê úê úê ú- + - - -ê úë û

2s is still negative, so we pivot on the 5 in column 2.

1 2 1 2

1 2 2

1 3 3

0 5 3 1 0 12

7R 5R R 45 0 21 12 0 62R 5R R 0 0 6 12 15 156

y y s s zé ù- -ê úê ú- + ê úê ú+ - - -ê úë û

Now we work on the largest negative indicator and pivot on the 12 in column 4.

1 2 1 2

2 1 1

2 3 3

R 12R R 45 60 15 0 0 15045 0 21 12 0 6

R R R 45 0 15 0 15 150

y y s s zé ù+ -ê úê úê úê ú+ -ê úë û

From this we can read the solution: The minimum is

( )15015 10-- = when 1 0y = and 2

150 5 .60 2

y = =

Your Turn 2

We start with this tableau.

1 2 3 4 1 2 3 4

0 1 0 1 0 0 0 1 0 160 0 0 0 1 1 1 1 0 01 0 0 1 1 0 0 1 0 12

0 0 1 1 1 0 1 1 0 80 0 0 300 180 0 400 480 1 10,640

y y y y s s s s zé ù-ê úê úê úê ú-ê úê ú

- - -ê úê úê ú- -ë û

We pivot on the 1 in row 4 of column 4.

1 2 3 4 1 2 3 4

4 1 1

4 3 3

4 5 5

0 1 1 0 1 0 1 0 0 8R R R0 0 0 0 1 1 1 1 0 0

R R R 1 0 1 0 0 0 1 0 0 200 0 1 1 1 0 1 1 0 8

300R R R 0 0 300 0 120 0 100 180 1 8240

y y y y s s s s zé ù-- + ê úê úê úê ú+ -ê úê úê - - - úê úê ú+ - -ê úë û

Finally we pivot on the 1 in row 2 of column 5.

1 2 3 4 1 2 3 4

2 1 1

2 4 4

5 52

0 1 1 0 0 1 0 1 0 8R R R0 0 0 0 1 1 1 1 0 01 0 1 0 0 0 1 0 0 200 0 1 1 0 1 0 0 0 8R R R0 0 300 0 0 120 220 300 1 8240120R R R

y y y y s s s s zé ùê úê úê úê úê úê úê úê úê úê úë û

- - -- +

-+

-+

Since there are now no negative indicators this tableau gives the solution:

1 2 3 420, 8, 0, 8,y y y y= = = = with a minimum cost of $8240.

4.4 Exercises 1. 1 2

1 2

2 3 84 7

x xx x

+ £

+ ³

Introduce the slack variable s1 and the surplus variable s2 to obtain the following equations:

1 2 1

1 2 2

2 3 84 7.

x x sx x s

+ + =

+ - =

2. 1 2

1 2

3 7 94 5 11

x xx x

+ £

+ ³

We need one slack variable, s1, and one surplus variable, s2. The system becomes

1 2 1

1 2 2

3 7 94 5 11.x x sx x s

+ + =

+ - =

3. 1 2 3

1 2 3

1 2

2 2 503 352 15

x x xx x xx x

+ + £

+ + ³

+ ³

Introduce the slack variable s1 and the surplus variables s2 and s3 to obtain the following equations:

1 2 3 1

1 2 3 2

1 2 3

2 2 503 352 15.

x x x sx x x sx x s

+ + + =

+ + - =

+ - =

4. 1 3

1 2

1 3

2 401820

x xx xx x

+ £

+ ³

+ ³

We need one slack variable, s1, and two surplus variables, s2 and s3.

The system becomes

1 3 1

1 2 2

1 3 3

2 401820.

x x sx x sx x s

+ + =

+ - =

+ - =

5. Minimize 1 2 33 4 5w y y y= + +

subject to: 1 2 3

2 3

1 2 3

2 3 92 8

2 2 6

y y yy y

y y y

+ + ³

+ ³

+ + ³

with 1 2 30, 0, 0.y y y³ ³ ³



Change this to a maximization problem by letting .z w= - The problem can now be stated

equivalently as follows:

Maximize 1 2 33 4 5z y y y= - - -

subject to: 1 2 3

2 3

1 2 3

2 3 92 8

2 2 6

y y yy y

y y y

+ + ³

+ ³

+ + ³

with 1 2 30, 0, 0.y y y³ ³ ³

6. Minimize 1 2 38 3w y y y= + +

subject to: 1 2 3

1 2 3

7 6 8 184 5 10 20

y y yy y y

+ + ³

+ + ³

with 1 2 30, 0, 0.y y y³ ³ ³

To minimize 1 2 38 3 ,w y y y= + +

we maximize 1 2 38 3 .z w y y y= - = - - -

The constraints are not changed.

7. Minimize 1 2 3 42 5w y y y y= + + +

subject to: 1 2 3 4

1 2 3 4

503 2 100y y y yy y y y

+ + + ³

+ + + ³

with 1 2 3 40, 0, 0, 0.y y y y³ ³ ³ ³

Change this to a maximization problem by letting .z w= - The problem can now be stated

equivalently as follows:

Maximize 1 2 3 42 5z y y y y= - - - -

subject to: 1 2 3 4

1 2 3 4

503 2 100

y y y yy y y y

+ + + ³

+ + + ³

with 1 2 3 40, 0, 0, 0.y y y y³ ³ ³ ³

8. Minimize 1 2 37w y y y= + +

subject to: 1 2 3

1 2 3

1 2

5 2 1254 6 756 8 84

y y yy y yy y

+ + ³

+ + £

+ ³

with 1 2 30, 0, 0.y y y³ ³ ³

To minimize 1 2 3 7 ,w y y y= + +

we maximize 1 2 37 .z w y y y= - = - - -



1 2

1 2

2 2440

x xx x

+ ³

+ £


Subtracting the surplus variable s1 and adding the slack variable s2 leads to the equations

1 2 1

1 2 2

2 2440.

x x sx x s

+ - =

+ + =


1 2 1 2

1 2 1 0 0 241 1 0 1 0 40

12 10 0 0 1 0

x x s s zé ù-ê úê úê úê ú- -ê úë û

The initial basic solution is not feasible since 1 24s = - is negative, so row transformations

must be used. Pivot on the 1 in row 1, column 1, since it is the positive entry that is farthest to the left in the first row (the row containing the 1)-

and since, in the first column, 241

24= is a smaller

quotient than 401

40.= After row transformations,

we obtain the following tableau.

1 2 1 2

1 2 2

1 3 3

1 2 1 0 0 24

R R R 0 1 1 1 0 1612R R R 0 14 12 0 1 288

x x s s zé ù-ê úê ú- + -ê úê ú+ -ê úë û

The basic solution is now feasible, but the problem is not yet finished since there is a negative indicator. Continue in the usual way. The 1 in column 3 is the next pivot. After row transformations, we get the following tableau.

1 2 1 2

1 2 1

2 3 3

R R R 1 1 0 1 0 400 1 1 1 0 16

12R R R 0 2 0 12 1 480

x x s s zé ù+ ê úê ú-ê úê ú+ ê úë û

This is a final tableau since the entries in the last row are all nonnegative. The maximum value is 480 when 1 40x = and 2 0.x =


1 2

1 2

2 22 5 80

x xx x

+ ³

+ £


Section 4.4 267


Introducing one surplus variable and one slack variable, the system becomes

1 2 1

1 2 2

2 22 5 80.

x x sx x s

+ - =

+ + =


1 2 1 2

2 1 1 0 0 22 5 0 1 0 80 .6 2 0 0 1 0

x x s s zé ù-ê úê úê úê ú- -ê úë û

The initial basic solution is not feasible since 1 2.s = - Pivot on the 2 in row 1, column 1.

1 2 1 2

1 2 2

1 3 3

2 1 1 0 0 2

R R R 0 4 1 1 0 783R R R 0 1 3 0 1 6

x x s s zé ù-ê úê ú- + ê úê ú+ -ê úë û


1 2 1 2

2 1 1

2 3 3

R R R 2 5 0 1 0 800 4 1 1 0 78

3R R R 0 13 0 3 1 240

x x s s zé ù+ ê úê úê úê ú+ ê úë û

Create a 1 in the column corresponding to x1.

1 2 1 21 5 1

1 12 2 2R R 1 0 0 40

0 4 1 1 0 780 13 0 3 1 240

x x s s zé ù ê úê úê úê úê úê úê úë û

The maximum is 240 when 1 40x = and 2 0.x =

11. Find 1 2 30, 0, and 0x x x³ ³ ³ such that

1 2 3

1 2 3

150100

x x xx x x

+ + £

+ + ³

and 1 2 3 2 5 3z x x x= + + is maximized.


1 2 3 1 2

1 1 1 1 0 0 150

1 1 1 0 1 0 1002 5 3 0 0 1 0

x x x s s zé ùê úê ú-ê úê ú- - -ê úë û

Note that s1 is a slack variable, while s2 is a surplus variable. The initial basic solution is not feasible,

since 2 100s = - is negative. Pivot on the 1 in row 2, column 1.

1 2 3 1 2

2 1 1

2 3 3

0 0 0 1 1 0 50R R R1 1 1 0 1 0 100

2R R R 0 3 1 0 2 1 200

x x x s s zé ù- + ê úê ú-ê úê ú+ - - -ê úë û


1 2 3 1 2

2 3 3

0 0 0 1 1 0 501 1 1 0 1 0 100

3R R R 3 0 2 0 5 1 500

x x x s s zé ùê úê ú-ê úê ú+ -ê úë û


1 2 3 1 2

1 2 2

1 3 3

0 0 0 1 1 0 50R R R 1 1 1 1 0 0 150

5R R R 3 0 2 5 0 1 750

x x x s s zé ùê úê ú+ ê úê ú+ ê úë û

This is a final tableau. The maximum value is 750 when 1 2 30, 150, and 0.x x x= = =

12. Find 1 2 30, 0, and 0x x x³ ³ ³ such that

1 2 3

1 2 3

154 4 2 48

x x xx x x

+ + £

+ + ³

and 1 2 32 3z x x x= + + is maximized.


1 2 3 1 2

1 1 1 1 0 0 154 4 2 0 1 0 48 .2 1 3 0 0 1 0

x x x s s zé ùê úê ú-ê úê ú- - -ê úë û


1 2 3 1 2

1 2 2

1 3 3

0 1 1 1 0 0 15

2R R R 2 2 0 2 1 0 183R R R 1 2 0 3 0 1 45

x x x s s zé ùê úê ú- + - -ê úê ú+ ê úë û

The initial basic solution is still not feasible since 2 18.s = - To choose a pivot, locate the positive

entry farthest to the left in row 2. The 2 in row 2, column 1, determines the pivot column and is also the pivot element, since it forms the smaller quotient.



1 2 3 1 2

2 1 1

2 3 3

R 2R R 0 0 2 4 1 0 122 2 0 2 1 0 18

R 2R R 0 2 0 8 1 2 72

x x x s s zé ù- + ê úê ú- -ê úê ú- + ê úë û

The solution is feasible since all variables and indicators are nonnegative. Therefore, create a 1 in the columns corresponding to x1, x3, and z.

1 2 3 1 2

1 11 12 2

1 12 22 2

1 13 32 2

R R 0 0 1 2 0 6

R R 1 1 0 1 0 9

R R 0 1 0 4 1 36

x x x s s zé ù

ê úê úê úê ú - -ê úê úê ú ê úë û

The maximum is 36 when 1 29, 0,x x= = and

3 6.x =

13. Find 1 20 and 0x x³ ³ such that

1 2

1 2

1 2

1002 3 75

4 50

x xx xx x

+ £

+ £

+ ³

and 1 25 3z x x= - is maximized.


1 2 1 2 3

1 1 1 0 0 0 100

2 3 0 1 0 0 75.

1 4 0 0 1 0 505 3 0 0 0 1 0

x x s s s zé ùê úê úê úê ú-ê úê ú-ê úë û


1 2 3 1 2

2 1 1

2 3 3

2 4 4

0 1 2 1 0 0 125R 2R R2 3 0 1 0 0 75

.0 5 0 1 2 0 25R 2R R0 21 0 5 0 2 3755R 2R R

x x x s s zé ù- -- + ê úê úê úê ú- -- + ê úê ú

+ ê úë û

This solution is still not feasible since 253 2

.s = -


1 2 3 1 2

3 1 1

3 2 2

3 4 4

0 0 10 6 2 0 650R 5R R10 0 0 8 6 0 3003R 5R R

0 5 0 1 2 0 2521R 5R R 0 0 0 46 42 10 1350

x x x s s zé ù- -+ ê úê ú- + ê úê ú- -ê úê ú

- + ê úë û


1 2 3 1 23 11

1 1 5 5104 31

2 2 5 5101 21

3 3 5 5523 211

4 4 5 510

0 0 1 0 65R R

1 0 0 0 30R R.

0 1 0 0 5R R

0 0 0 1 135R R

x x x s s zé ù- - ê úê úê úê ú ê úê úê ú- - ê úê úê ú ê úë û

This is a final tableau. The maximum is 135 when 1 230, 5.x x= =


1 2

1 2

1 2

2 183 12

2 2 24

x xx xx x

+ £

+ ³

+ £

and 1 25 10z x x= - is maximized.

Introduce slack and surplus variables to get the system

1 2 1

1 2 2

1 2 3

2 183 12

2 2 24.

x x sx x sx x s

+ + =

+ - =

+ + =

The initial tableau is

1 2 1 2 3

1 2 1 0 0 0 183 0 1 0 0 121

.2 2 0 0 1 0 245 10 0 0 0 1 0

x x s s s zé ùê úê ú-ê úê úê úê ú-ê úë û

2 12s = - is not a feasible solution. Pivot on the 1 in row 1, column 1.

1 2 1 2 3

2 1 1

2 3 3

2 4 4

0 1 1 1 0 0 6R R R1 3 0 1 0 0 120 4 0 1 0 02R R R 2

5R R R 0 25 0 5 0 1 60

x x s s s zé ù-- + ê úê ú-ê úê ú-- + ê úê ú+ -ê úë û

Section 4.4 269



1 2 1 2 3

3 1 1

3 2 2

3 4 4

0 6 2 0 1 0 12R 2R R2 2 0 0 1 0 24R 2R R0 4 0 2 1 0 0

5R 2R R 0 30 0 0 5 2 120

x x s s s zé ù-- + ê úê ú+ ê úê ú-ê úê ú+ ê úë û

1 2 1 2 311

1 1 2211

2 2 2211

3 3 2251

4 4 22

0 3 1 0 0 6R R

1 1 0 0 0 12R R

0 2 0 1 0 0R R

0 15 0 0 1 60R R

x x s s s zé ù- ê úê úê úê ú ê úê úê ú- ê úê úê ú ê úë û


15. Find 1 2 30, 0, and 0y y y³ ³ ³ such that

1 2 3

1 2 3

5 3 2 1505 10 3 90

y y yy y y

+ + £

+ + ³

and 1 2 310 12 10w y y y= + + is minimized.

Let 2 310 12 10 .z w y y y= - = - - - Maximize z.


1 2 3 1 2

5 3 2 1 0 0 150

5 10 3 0 1 0 9010 12 10 0 0 1 0

y y y s s zé ùê úê ú-ê úê úê úë û


1 2 3 1 2

2 1 1

2 3 3

R R R 0 7 1 1 1 0 60

5 10 3 0 1 0 902R R R 0 8 4 0 2 1 180

y y y s s zé ù- + - -ê úê ú-ê úê ú- + - -ê úë û


2 1 1

2 3 3

1 2 3 1 2

7R 10R R 35 0 11 10 3 0 12305 10 3 0 1 0 90

8R 10R R 40 0 64 0 12 10 1080

y y s s zy

+

-

+ -


Create a 1 in the columns corresponding to y2, s1, and z.

1 2 3 1 2

1 7 11 31 110 2 10 10

1 1 3 12 210 2 10 10

1 32 63 3 5 5

R R 0 1 0 123

R R 1 0 0 9

R R 4 0 0 1 10810

y y y s s zé ù

ê úê úê úê ú -ê úê úê ú -ê úë û

This is a final tableau. The minimum is 108 when 1 2 30, 9, and 0.y y y= = =

16. Minimize 1 2 33 2 3w y y y= + +

subject to: 1 2 3

1 2 3

2 3 6 604 5 40

y y yy y y

+ + £

+ + ³

with 1 2 30, 0, 0y y y³ ³ ³

Let 1 2 33 2 3 .z w y y y= - = - - - Maximize z.


1 2 3 1 2

2 3 6 1 0 0 601 4 5 0 1 0 403 2 3 0 0 1 0

y y y s s zé ùê úê ú-ê úê úê úë û


1 2 3 1 2

1 2 2

1 3 3

2 3 6 1 0 0 60

R 2R R 0 5 4 1 2 0 203R 2R R 0 5 12 3 0 2 180

y y y s s zé ùê úê ú- + - -ê úê ú- + - - - -ê úë û

There are negative indicators, so now pivot on the 4 in row 2, column 3.

1 2 3 1 2

2 1 1

2 3 3

3R 2R R 4 9 0 5 6 0 600 5 4 1 2 0 20

3R R R 0 10 0 6 6 2 120

y y y s s zé ù- + -ê úê ú- -ê úê ú+ - - -ê úë û


1 2 3 1 2

1 2 2

1 3 3

4 9 0 5 6 0 60

R 5R R 4 16 20 0 4 0 1606R 5R R 24 4 0 0 6 10 240

y y y s s zé ù-ê úê ú+ -ê úê ú+ - -ê úë û


1 2 3 1 2

2 1 1

2 3 3

9R 16R R 100 0 180 80 60 0 24004 16 20 0 4 0 160

R 4R R 100 0 20 0 20 40 800

y y y s s zé ù+ ê úê ú-ê úê ú+ -ê úë û



1 2 3 1 2

1 5 9 31 180 4 4 4

1 1 5 12 216 4 4 4

5 1 113 3 2 2 240

R R 0 1 0 30

R R 1 0 0 10

0 0 1 20R R

y y y s s zé ù

ê úê úê úê ú -ê úê úê ú- ê úë û

The minimum is 20w = when 1 0,y = 2 10,y = and 3 0.y =


subject to: 1 2

1 2

1 2

504 2 1205 2 200

x xx xx x

+ =

+ ³

+ £

with 1 20, 0.x x³ ³

The artificial variable a1 is used to rewrite 1 2 50x x+ = as 1 2 1 50;x x a+ + = note

that a1 must equal 0 for this equation to be a true statement. Also the surplus variable s1 and the slack variable s2 are needed. The initial tableau is as follows.

1 2 1 1 2

1 1 1 0 0 0 504 2 0 1 0 0 1205 2 0 0 1 0 2003 2 0 0 0 1 0

x x a s s zé ùê úê ú-ê úê úê úê ú- -ê úë û

The initial basic solution is not feasible. Pivot on the 4 in row 2, column 1.

1 2 1 1 2

2 1 1

2 3 3

2 4 4

0 2 4 1 0 0 80R 4R R4 2 0 1 0 0 1200 2 0 4 0 2005R 4R R 5

3R 4R R 0 2 0 3 0 4 360

x x a s s zé ù- + ê úê ú-ê úê ú-- + ê úê ú+ - -ê úë û

The basic solution is now feasible, but there are negative indicators. Pivot on the 5 in row 3, column 4 (which is the column with the most negative indicator and the row with the smallest nonnegative quotient).

1 2 1 1 2

3 1 1

3 2 2

3 4 4

0 20 0 4 0 200R 5R R 1220 8 0 0 4 0 800R 5R R0 2 0 5 4 0 2000 16 0 0 12 20 24003R 5R R

x x a s s zé ù-- + ê úê ú+ ê úê ú-ê úê ú

-+ ê úë û


1 2 1 1 2

1 2 2

1 3 3

1 4 4

0 12 20 0 4 0 20060 0 40 0 20 0 20002R 3R R0 0 20 30 20 0 1400 R 6R R0 0 80 0 20 60 8000 4R 3R R

x x a s s zé ù-ê úê ú-- + ê úê ú+ ê úê ú+ ê úë û

We now have 1 0,a = so drop the a1 column.

1 2 1 2

0 12 0 4 0 20060 0 0 20 0 20000 0 30 20 0 14000 0 0 20 60 8000

x x s s zé ù-ê úê úê úê úê úê úê úë û

We are finished pivoting. Create a 1 in the columns corresponding to x1, x2, s1, and z.

1 2 1 21 501

1 1 3 3121 1001

2 2 3 3602 1401

3 3 3 3301 4001

4 4 3 360

0 1 0 0R R

1 0 0 0R R

0 0 1 0R R

0 0 0 1R R

x x s s zé ù- ê úê úê úê ú ê úê úê ú ê úê úê ú ê úë û

The maximum value is 4003

when 1001 3

x = and 50

2 3.x =


subject to: 1 2

1 2

1 2

152 4 303 5 10

x xx xx x

+ =

+ ³

+ ³

with 1 20, 0.x x³ ³

With artificial, slack, and surplus variables, we have

1 2

1 2 1

1 2 2

152 4 303 5 10.

x x ax x sx x s

+ + =

+ - =

+ - =

The initial tableau is

[ ]1 2 1 2

1 1 0 0 1 0 152 4 1 0 0 0 30

.3 5 0 1 0 0 105 7 0 0 0 1 0

x x s s a zé ùê úê ú-ê úê ú-ê úê ú- -ê úë û

First, eliminate the artificial variable a. Pivot on the 1 in row 1, column 1.

Section 4.4 271


1 2 1 2

1 2 2

1 3 3

1 4 4

1 1 0 0 1 0 150 2 1 0 2 0 02R R R0 2 0 1 3 0 353R R R0 2 0 0 5 1 755R R R

x x s s a zé ùê úê ú- -- + ê úê ú- -- + ê úê ú-+ ê úë û

Now 0,a = so we can drop the a column.

1 2 1 2

1 1 0 0 0 15

0 2 1 0 0 00 2 0 1 0 350 2 0 0 1 75

x x s s zé ùê úê ú-ê úê ú-ê úê ú

-ê úë û

Because 2 –35,s = we choose the 2 in row 2, column 2, as the next pivot. (Note that 1 0.)s =

1 2 1 2

2 1 1

2 3 3

2 4 4

2 0 1 0 0 30R 2R R0 2 1 0 0 00 0 1 1 0 35R R R0 0 1 0 1 75R R R

x x s s zé ù- + ê úê ú-ê úê ú-- ê úê ú

-+ ê úë û


1 2 1 2

1 2 2

1 3 3

1 4 4

2 0 1 0 0 30R R R 2 2 0 0 0 30R R R 2 0 0 1 0 65R R R 2 0 0 0 1 105

x x s s zé ùê úê ú+ ê úê ú+ ê úê ú+ ê úë û

Create a 1 in the column for x2.

1 2 1 2

12 22

2 0 1 0 0 301 1 0 0 0 15R R2 0 0 1 0 652 0 0 0 1 105

x x s s zé ùê úê ú ê úê úê úê úê úë û


19. Minimize 1 2 332 40 48w y y y= + +

subject to: 1 2 3

1 2 3

1 2 3

20 10 5 20025 40 50 50018 24 12 300

y y yy y yy y y

+ + =

+ + £

+ + ³

with 1 2 30, 0, 0y y y³ ³ ³

With artificial, slack, and surplus variables, this problem becomes

Maximize 1 2 332 40 48z y y y= - - -

subject to:

1 2 3 1

1 2 3 1

1 2 3 2

20 10 5 20025 40 50 50018 24 12 300.

y y y ay y y sy y y s

+ + + =

+ + + =

+ + - =


1 2 3 1 1 2

20 10 5 1 0 0 0 20025 40 50 0 1 0 0 50018 24 12 0 0 1 0 30032 40 48 0 0 0 1 0

y y y a s s zé ùê úê úê úê ú-ê úê úê úë û

The initial basic tableau is not feasible. Pivot on the 20 in row 1, column 1.

1 2 3 1 1 2

1 2 2

1 3 3

1 4 4

20 10 5 1 0 0 0 200

0 110 175 5 4 0 0 10005R 4R R0 150 75 9 0 10 0 12009R 10R R0 120 200 8 0 0 5 16008R 5R R

y y y a s s z

-- +

- -- +

- -- +


Eliminate the a1 column.

1 2 3 1 2

20 10 5 0 0 0 2000 110 175 4 0 0 1000

0 150 75 0 10 0 12000 120 200 0 0 5 1600

y y y s s zé ùê úê úê úê ú-ê úê ú

-ê úë û


1 2 3 1 2

3 1 1

3 2 2

3 4 4

300 0 0 0 10 0 1800R 15R R0 0 1800 60 110 0 180011R 15R R0 150 75 0 10 0 1200

4R 5R R 0 0 700 0 40 25 12,800

y y y s s zé ù- + ê úê ú- + ê úê ú-ê úê ú- + -ê úë û

Create ones in the columns corresponding to y1, y2, s1, and z.

1 2 3 1 211

1 1 30300111

2 2 6601 11

3 3 2 1515081

4 4 525

1 0 0 0 0 6R R

0 0 30 1 0 30R R

0 1 0 0 8R R

0 0 28 0 1 512R R

y y y s s zé ù

ê úê úê úê ú ê úê úê ú- ê úê úê ú- ê úë û

This is a final tableau. The minimum value is 512 when 1 2 36, 8, and 0.y y y= = =



20. Minimize 1 2 315 12 18w y y y= + +

subject to: 1 2 3

1 2 3

1 2 3

2 3 123 3 18

10

y y yy y yy y y

+ + £

+ + ³

+ + =

with 1 2 30, 0, 0y y y³ ³ ³

Let 1 2 315 12 18z w y y y= - = - - - and maximize z. Introduce the slack variable s1, the surplus variable s2, and the artificial variable a1. The initial tableau is as follows.

1 2 3 1 2 1

1 2 3 1 0 0 0 123 1 3 0 1 0 0 18

1 1 1 0 0 1 0 1015 12 18 0 0 0 1 0

y y y s s a zé ùê úê ú-ê úê úê úê úê úë û

First, eliminate the artificial variable a1. Pivot on the 1 in row 3, column 1.

1 2 3 1 2 1

3 1 1

3 2 2

3 4 4

0 1 2 1 0 1 0 2R R R0 2 0 0 1 3 0 123R R R1 1 1 0 0 1 0 100 3 3 0 0 15 1 15015R R R

y y y s s a zé ù-- + ê úê ú- - - -- + ê úê úê úê ú- - -- + ê úë û

Now 1 0,a = so we can drop the a1 column.

1 2 3 1 2

0 1 2 1 0 0 20 2 0 0 1 0 121 1 1 0 0 0 100 3 3 0 0 1 150

y y y s s zé ùê úê ú- - -ê úê úê úê ú

- -ê úë û


1 2 3 1 2

1 2 2

1 3 3

1 4 4

0 1 2 1 0 0 20 0 4 2 1 0 82 R R R1 0 1 1 0 0 8R R R

3R R R 0 0 9 3 0 1 144

y y y s s zé ùê úê ú- -+ ê úê ú- -- + ê úê ú+ -ê úë û

The maximum value of –z w= is –144. Therefore, the minimum value of w is 144 when 1 8,y = 2 2,y = and 3 0.y =

23. (a) Let y1 = amount shipped from S1 to D1,

y2 = amount shipped from S1 to D2,

y3 = amount shipped from S2 to D1,

and y4 = amount shipped from S2 to D2.

Minimize 1 2 3 430 20 25 22w y y y y= + + +

Section 4.4 273


subject to: 1 3

2 4

1 2

3 4

1 2 3 4

3000500050005000

2 6 5 4 40,000

y yy yy yy y

y y y y

+ ³

+ ³

+ £

+ £

+ + + £

with 1 2 3 40, 0, 0, 0.y y y y³ ³ ³ ³

Maximize 1 2 3 430 20 25 22 .z w y y y y= - = - - - -

1 2 3 4 1 2 3 4 5

1 0 1 0 1 0 0 0 0 0 30000 1 0 1 0 1 0 0 0 0 50001 1 0 0 0 0 1 0 0 0 50000 0 1 1 0 0 0 1 0 0 50002 6 5 4 0 0 0 0 1 0 40,000

30 20 25 22 0 0 0 0 0 1 0

y y y y s s s s s zé ù-ê úê ú-ê úê úê úê úê úê úê úê úê úê úë û

Pivot on the 1 in row 1, column 1 since the feasible solution has a negative value, 1 3000.s = -

1 2 3 4 1 2 3 4 5

1 3 3

1 5 5

1 6 6

1 0 1 0 1 0 0 0 0 0 30000 1 0 1 0 1 0 0 0 0 5000

R R R 0 1 1 0 1 0 1 0 0 0 20000 0 1 1 0 0 0 1 0 0 5000

2R R R 0 6 3 4 2 0 0 0 1 0 34,00030R R R 0 20 5 22 30 0 0 0 0 1 90,000

y y y y s s s s s zé ù-ê úê ú-ê úê ú- + -ê úê úê úê úê ú- + ê úê ú- + - -ê úë û

Since the feasible solution has a negative value 2( 5000),s = - pivot on the 1 in row 3, column 2.

1 2 3 4 1 2 3 4 5

3 2 2

3 5 5

3 6 6

1 0 1 0 1 0 0 0 0 0 30000 0 1 1 1 1 1 0 0 0 3000R R R0 1 1 0 1 0 1 0 0 0 20000 0 1 1 0 0 0 1 0 0 5000

6R R R 0 0 9 4 4 0 6 0 1 0 22,00020R R R 0 0 15 22 10 0 20 0 0 1 130,000

y y y y s s s s s zé ù-ê úê ú- - -- + ê úê ú-ê úê úê úê úê ú- + - -ê úê ú- + - -ê úë û

Since the feasible solution has a negative value ( 2 3000s = - ), pivot on the 9 in row 5, column 3.

1 2 3 4 1 2 3 4 5

5 1 1

5 2 2

5 3 3

5 4 4

5 6 6

9 0 0 4 5 0 6 0 1 0 5000R 9R RR 9R R 0 0 0 5 5 9 3 0 1 0 5000R 9R R 0 9 0 4 5 0 3 0 1 0 40,000R 9R R 0 0 0 5 4 0 6 9 1 0 23,000

0 0 9 4 4 0 6 0 1 0 22,0005R 3R R 0 0 0 46 50 0 30 0 5 3 500,000

y y y y s s s s s zé ù- - -- + ê úê ú- + - - - -ê úê ú+ ê úê ú

- + -êêê - -êê- + - - -êë û

úúúúúú




1 2 3 4 1 2 3 4 5

2 1 1

2 3 3

2 4 4

2 5 5

2 6 6

45 0 0 0 45 36 18 0 9 0 45,0004R 5R R0 0 0 5 5 9 3 0 1 0 50000 45 0 0 45 36 27 0 9 0 180,0004R 5R R

R R R 0 0 0 0 9 9 9 9 0 0 18,0004R 5R R 0 0 45 0 0 36 18 0 9 0 90,000

46R 5R R 0 0 0 0 480 414 12 0 21 15 2,730,000

y y y y s s s s s zé - - -+

- - - -

- + - +

- + -- + - -ë

ùê úê úê úê úê úê úê úê úê úê úê úê úû


1 2 3 4 1 2 3 4 5

4 1 1

4 2 2

4 3 3

4 5 5

4 6 6

45 0 0 0 63 54 0 18 9 0 90002R R R0 0 0 15 6 18 0 9 3 0 33,000R 3R R0 45 0 0 18 9 0 27 9 0 126,0003R R R0 0 0 0 9 9 9 9 0 0 18,0000 0 45 0 18 54 0 18 9 0 126,0002R R R0 0 0 0 1476 450 0 36 63 45 8,118,0004R 3R R

y y y y s s s s s zé - - - -- + êê - - -+

-- +

+ -+ ë

ùúú

ê úê úê úê úê úê úê úê úê úû

Create a 1 in the columns corresponding to y1, y2, y3, y4, and z.

1 2 3 4 1 2 3 4 51 7 6 2 11 145 5 5 5 5

1 6 32 12 215 5 5 5 532 1 11

3 3 5 5 5 545

1 62 2 15 545 5 5 5 5

1 164 746 645 5 5 5

R R 1 0 0 0 0 0 200R R 0 0 0 1 0 0 2200

0 1 0 0 0 0 2800R R0 0 0 0 9 9 9 9 0 0 18,000

R R 0 0 1 0 0 0 2800R R 0 0 0 0 10 0 1 180,400

y y y y s s s s s zé ù - - - -ê úê ú - - -ê úê úê ú- êêêê êêê -êë û

úúúúúúúú

Here, 1 200,y = 2 2800,y = 3 2800,y = 4 2200,y = and 180, 400.z w- = = So, ship 200 barrels of oil from supplier 1S to distributor 1D . Ship 2800 barrels of oil from supplier 1S to distributor 2D . Ship 2800 barrels of oil from supplier 2S to distributor 1D . Ship 2200 barrels of oil from supplier 2S to distributor 2D . The minimum cost is $180,400.

(b) From the final tableau, 39 18,000,s = so 3 2000.s = Therefore, 1S could furnish 2000 more barrels of oil.

24. Let y1 = amount shipped from S1 to D1, y2 = amount shipped from S1 to D2, y3 = amount shipped from S2 to D1, and y4 = amount shipped from S2 to D2. Maximize 1 2 3 430 20 25 22w y y y y= + + +

subject to: 1 3

2 4

1 2

3 4

1 2 3 4

3000500050005000

2 6 5 4 40,000

y yy y

y yy y

y y y y

+ ³

+ ³

+ =

+ =

+ + + £

with 1 2 3 40, 0, 0, 0.y y y y³ ³ ³ ³

Section 4.4 275


Maximize 1 2 3 430 20 25 22 .z w y y y y= - = - - - -

1 2 3 4 1 2 3 1 2

1 0 1 0 1 0 0 0 0 0 30000 1 0 1 0 1 0 0 0 0 5000

1 1 0 0 0 0 0 1 0 0 50000 0 1 1 0 0 0 0 1 0 50002 6 5 4 0 0 1 0 0 0 40,000

30 20 25 22 0 0 0 0 0 1 0

y y y y s s s a a zé ù-ê úê ú-ê úê úê úê úê úê úê úê úê úê úë û

Pivot on the 1 in row 3, column 1 to remove the a columns.

1 2 3 4 1 2 3 1 2

1 3 1

3 5 5

3 6 6

R R R 0 1 1 0 1 0 0 1 0 0 20000 1 0 1 0 1 0 0 0 0 50001 1 0 0 0 0 0 1 0 0 50000 0 1 1 0 0 0 0 1 0 5000

2R R R 0 4 5 4 0 0 1 2 0 0 30,00030R R R 0 10 25 22 0 0 0 30 0 1 150,000

y y y y s s s a a zé ù- + -ê úê ú-ê úê úê úê úê úê ú- + -ê úê ú- + - - -ê úë û

Column a1 can be removed since 1 0.a =

1 2 3 4 1 2 3 2

0 1 1 0 1 0 0 0 0 20000 1 0 1 0 1 0 0 0 50001 1 0 0 0 0 0 0 0 5000

0 0 1 1 0 0 0 1 0 50000 4 5 4 0 0 1 0 0 30,0000 10 25 22 0 0 0 0 1 150,000

y y y y s s s a zé ù-ê úê ú-ê úê úê úê úê úê úê úê úê ú- -ê úë û


1 2 3 4 1 2 3 2

4 1 1

4 5 5

4 6 6

0 1 0 1 1 0 0 1 0 7000R R R0 1 0 1 0 1 0 0 0 50001 1 0 0 0 0 0 0 0 50000 0 1 1 0 0 0 1 0 5000

5R R R 0 4 0 1 0 0 1 5 0 500025R R R 0 10 0 3 0 0 0 25 1 275,000

y y y y s s s a zé ù+ ê úê ú-ê úê úê úê úê úê úê ú- + - -ê úê ú- + - - - -ê úë û

Column 2a can be removed since 2 0.a = Pivot on the 1 in row 2, column 2 since the basic solution is not yet feasible 2( 5000).s = -



1 2 3 4 1 2 3

2 1 l

2 3 3

2 5 5

2 6 6

0 0 0 0 1 1 0 0 2000R R R0 1 0 1 0 1 0 0 50001 0 0 1 0 1 0 0 0R R R0 0 1 1 0 0 0 0 5000

4R R R 0 0 0 5 0 4 1 0 15,00010R R R 0 0 0 7 0 10 0 1 225,000

y y y y s s s zé ù- + ê úê ú-ê úê ú-- + ê úê úê úê úê ú- + - -ê úê ú+ - -ê úë û

Pivot on the –5 in row 5, column 4.

1 2 3 4 1 2 3

5 2 2

5 3 3

5 4 4

5 6 6

0 0 0 0 1 1 0 0 2000R 5R R 0 5 0 0 0 1 1 0 10,000R 5R R 5 0 0 0 0 1 1 0 15,000R 5R R 0 0 5 0 0 4 1 0 10,000

0 0 0 5 0 4 1 0 15,0007R 5R R 0 0 0 0 0 22 7 5 1, 230,000

y y y y s s s zé ùê úê ú+ -ê úê ú- - - -ê úê ú

+ ê úê úê ú- -ê úê ú+ - -ê úë û


1 2 3 4 1 2 3

1 2 2

1 3 3

1 4 4

1 5 5

1 6 6

0 0 0 0 1 1 0 0 20000 5 0 0 1 0 1 0 12,000R R R5 0 0 0 1 0 1 0 13,000R R R0 0 5 0 4 0 1 0 20004 R R R0 0 0 5 4 0 1 0 23,0004 R R R0 0 0 0 22 0 7 5 1,186,00022R R R

y y y y s s s zé ùê úê ú+ ê úê ú- -+ ê úê ú-- + ê úê ú- - -- + ê úê ú

-ê ú+ ë û

1 2 3 4 1 2 3

1 112 2 5 55

1 113 3 5 55

4 114 4 5 55

4 115 5 5 55

72216 6 5 55

0 0 0 0 1 1 0 0 2000

0 1 0 0 0 0 2400R R

1 0 0 0 0 0 2600R R

0 0 1 0 0 0 400R R

0 0 0 1 0 0 4600R R

0 0 0 0 0 1 237,200R R

y y y y s s s zé ùê úê ú ê úê úê ú- -- ê úê úê ú- ê úê úê ú-- ê úê ú-ê ú ê úë û

Here, 1 2600,y = 2 2400,y = 3 400,y = 4 4600,y = and 237,200z w= - = Therefore, ship 2600 barrels from 1S to 1D , 2400 barrels from 1S to D2, 400 barrels from 2S to 1D , and 4600 barrels from 2S to 2D for a minimum cost of $237,200.

Section 4.4 277


25. (a) Let 1x = the number of computers shipped from W1 to D1,

2x = the number of computers shipped from W1 to D2,

3x = the number of computers shipped from W2 to D1,

and 4x = the number of computers shipped from W2 to D2.

Minimize 1 2 3 414 12 12 10w x x x x= + + +

subject to: 1 3

2 4

1 2

3 4

32202530

x xx xx xx x

+ ³

+ ³

+ £+ £

with 1 2 3 40, 0, 0, 0.x x x x³ ³ ³ ³

Maximize 1 2 3 414 12 12 10 .z w x x x x= - = - - - -

The initial tableau looks like the following.

1 2 3 4 1 2 3 4

1 0 1 0 1 0 0 0 0 320 1 0 1 0 1 0 0 0 20

1 0 0 0 0 1 0 0 250 0 1 1 0 0 0 1 0 30

14 12 12 10 0 0 0 0 1 0

1

x x x x s s s s zé ù-ê úê ú-ê úê úê úê úê úê úê úë û

The variable 1s is negative; we pivot on the 1 in row 3 of column 1.

1 2 3 4 1 2 3 4

3 1 1

3 5 5

R R R 0 1 1 0 1 0 1 0 0 70 1 0 1 0 1 0 0 0 201 1 0 0 0 0 1 0 0 250 0 1 1 0 0 0 1 0 30

14R R R 0 2 12 10 0 0 14 0 1 350

x x x x s s s s zé ù- + - - -ê úê ú-ê úê úê úê úê úê úê ú- + - - -ê úë û

The variable 1s is still negative; we pivot on the 1 in row 1 of column 3.

1 3 4 1 2 3 42

1 4 45 51

0 1 1 0 1 0 1 0 0 7

0 1 0 1 0 1 0 0 0 201 1 0 0 0 0 1 0 0 25

R R R 0 1 0 1 1 0 1 1 0 2312R R R 0 10 0 10 12 0 2 0 1 434

x x x x s s s s zé ù- - -ê úê ú-ê úê úê úê ú

- + ê úê úê ú- + - -ë û

The variable s2 is still negative; we pivot on the 1 in row 2 of column 2.

1 2 3 4 1 2 3 4

2 1 1

2 3 3

2 4 4

2 5 5

0 0 1 1 1 1 1 0 0 27R R R0 1 0 1 0 1 0 0 0 201 0 0 1 0 1 1 0 0 5R R R

R R R 0 0 0 0 1 1 1 1 0 310R R R 0 0 0 0 12 10 2 0 1 634

x x x x s s s s zé ù- - -+ ê úê ú-ê úê ú-- + ê úê ú

- + ê úê úê ú- + - -ë û

Now we eliminate the only negative indicator by pivoting on the 1 in row 4 of column 7.

1 2 3 4 1 2 3 4

4 1 1

4 3 3

4 5 5

R R R 0 0 1 1 0 0 0 1 0 300 1 0 1 0 0 0 0 0 20

R R R 1 0 0 1 1 0 0 1 0 20 0 0 0 1 1 1 1 0 3

2R R R 0 0 0 0 14 12 0 2 1 628

x x x x s s s s z

+

- + - - -

+ -


From this we can read the solution: Ship 2 computers from W1 to D1, ship 20 computers from W1 to D2, ship 30 computers from W2 to D1, and 0 computers from W2 to D2. The resulting minimum cost is $628.

(b) From the final tableau, 3 3.s = Therefore, warehouse W1 has three more computers that it could ship.

26. Let x1 = the amount invested in government

securities,

x2 = the amount invested in municipal bonds,

and x3 = the amount invested in mutual funds.

Maximize 1 2 30.07 0.06 0.10z x x x= + +

subject to:

1 2 3

1

2 3

1 2 3

100,00040,00050,000

0.02 0.01 0.03 2400

x x xx

x xx x x

+ + =

³

+ ³

+ + £

or 1 2 32 3 240,000x x x+ + £

with 1 2 30, 0, 0.x x x³ ³ ³

1 2 3 1 2 3 4

1 1 1 1 0 0 0 0 100,000

1 0 0 0 1 0 0 0 40,0000 1 1 0 0 1 0 0 50,0002 1 3 0 0 0 1 0 240,000

0.07 0.06 0.10 0 0 0 0 1 0

x x x s s s s zé ùê úê ú-ê úê ú-ê úê úê úê úê ú- - -ë û

Since 2s is negative we pivot on the 1 in row 2, column 1.



1 2 3 1 2 3 4

2 1 1

2 4 4

2 5 5

0 1 1 1 1 0 0 0 60,000R R R1 0 0 0 1 0 0 0 40,000

0 1 1 0 0 1 0 0 50,0002R R R 0 1 3 0 2 0 1 0 160,000

7R 100R R 0 6 10 0 7 0 0 100 280,000

x x x s s s s z

- +

-

-

- +

+ - - -


2s is no longer basic, but 3s is negative, so we pivot on the 1 in row 3 of column 2.

1 2 3 1 2 3 4

3 1 1

3 4 4

3 5 5

R R R 0 0 0 1 1 1 0 0 10, 0001 0 0 0 1 0 0 0 40, 0000 1 1 0 0 1 0 0 50, 000

R R R 0 0 2 0 2 1 1 0 110, 0006R R R 0 0 4 0 7 6 0 100 580, 000

x x x s s s s z

- +

-

-

- +

+ - - -


Now all the basic variables are nonnegative, so we work on the most negative indicator in the last row, which is 7.- We pivot on the 1 in the first row of column 5.

1 2 3 1 2 3 4

1 2 2

1 4 4

1 5 5

0 0 0 1 1 1 0 0 10,0001 0 0 1 0 1 0 0 50,000R R R0 1 1 0 0 1 0 0 50,000

2R R R 0 0 2 2 0 1 1 0 90,0007R R R 0 0 4 7 0 1 0 100 650,000

x x x s s s s zé ùê úê ú+ ê úê ú-ê úê ú- + - -ê úê ú

+ ê ú-ë û

Now we have just a single negative indicator, so we pivot on the 2 in row 4 of column 3.

1 2 3 1 2 3 4

4 3 3

4 5 5

0 0 0 1 1 1 0 0 10,0001 0 0 1 0 1 0 0 50,000

R 2R R 0 2 0 2 0 1 1 0 10,0000 0 2 2 0 1 1 0 90,000

2R R R 0 0 0 3 0 1 2 100 830,000

x x x s s s s zé ùê úê úê úê ú- + - -ê úê ú

- -ê úê ú

+ ê ú-ë û

There is still a negative indicator, so we pivot on the 1 in row 1 of column 6.

1 2 3 1 2 3 4

1 2 2

1 3 3

1 4 4

1 5 5

0 0 0 1 1 1 0 0 10,0001 0 0 0 1 0 0 0 40,000R R R0 2 0 3 1 0 1 0 20,000R R R

R R R 0 0 2 1 1 0 1 0 100,000R R R 0 0 0 4 1 0 2 100 840,000

x x x s s s s zé ùê úê ú-- + ê úê ú-+ ê úê ú+ -ê úê ú+ ê úë û

We now read the solution.

1

2

3

40,000,20,000 10,000,

2100,000 50,000,

2

x

x

x

=

= =

= =

Invest $40,000 in government securities, $10,000 in municipal bonds, and $50,000 in mutual funds; the maximum interest is $8400.

27. Let x1 = the number of million dollars for home

loans and x2 = the number of million dollars for

commercial loans. Maximize 1 20.12 0.10z x x= +

subject to: 1 2 1 2

1 2

1 2

1 2

4 or 4 010

3 2 7225

x x x xx xx xx x

³ - ³

+ ³

+ £

+ £

with 1 20, 0.x x³ ³

1 2 1 2 3 4

1 4 1 0 0 0 0 01 1 0 1 0 0 0 103 2 0 0 1 0 0 721 1 0 0 0 1 0 25

0.12 0.10 0 0 0 0 1 0

x x s s s s zé ù- -ê úê ú-ê úê úê úê úê úê ú- -ê úë û

Eliminate the decimals in the last row by multiplying by 100

1 2 1 2 3 4

1 4 1 0 0 0 0 0

1 1 0 1 0 0 0 103 2 0 0 1 0 0 721 1 0 0 0 1 0 25

12 10 0 0 0 0 100 0

x x s s s s zé ù- -ê úê ú-ê úê úê úê úê úê úê ú- -ë û


1 2 1 2 3 4

2 1 1

2 3 3

2 4 4

2 5 5

0 5 1 1 0 0 0 10R R R1 1 0 1 0 0 0 10

3R R R 0 1 0 3 1 0 0 42R R R 0 0 0 1 0 1 0 15

12R R R 0 2 0 12 0 0 100 120

x x s s s s zé ù- - -- + ê úê ú-ê úê ú- + -ê úê ú

- + ê úê úê ú+ -ë û


Section 4.4 279


1 2 1 2 3 4

3 1 1

3 2 2

3 4 4

3 5 5

0 14 3 0 1 0 0 72R 3R R3 2 0 0 2 0 0 72R 3R R0 1 0 3 1 0 0 42

R 3R R 0 1 0 0 1 3 0 34R R R 0 2 0 0 4 0 100 288

x x s s s s zé ù- - - -- + ê úê ú-+ ê úê ú-ê úê ú

- + -ê úê úê ú+ -ë û


1 2 1 2 3 4

4 1 1

4 2 2

4 3 3

4 5 5

14R R R 0 0 3 0 15 42 0 302R R R 3 0 0 0 0 6 0 66

R R R 0 0 0 3 0 3 0 450 1 0 0 1 3 0 3

2R R R 0 0 0 0 2 6 100 294

x x s s s s zé ù+ - - -ê úê ú- + -ê úê ú+ ê úê ú-ê úê ú+ ê úë û

Create a 1 in the columns corresponding to 1x and .z

1 2 1 2 3 4

12 23

15 5100

0 0 3 0 15 42 0 30R R 1 0 0 0 0 2 0 22

0 0 0 3 0 3 0 450 1 0 0 1 3 0 30 0 0 0 0.02 0.06 1 2.94R R

x x s s s s z

é ù- - -ê úê ú -ê úê úê úê ú-ê úê ú ê úë û

Here, 1 22,x = 2 3,x = and 2.94.z = Make $22 million ($22,000,000) in home loans and $3 million ($3,000,000) in commercial loans for a maximum return of $2.94 million, or $2,940,000.

28. Let x1 = the number of pounds of bluegrass seed,

x2 = the number of pounds of rye seed,

and x3 = the number of pounds of Bermuda seed.

If each batch must contain at least 25% bluegrass seed, then

1 1 2 3

1 2 3

0.25( )0.75 0.25 0.25 0.

y y y yy y y

³ + +

- - ³

And if the amount of Bermuda must be no more than 2

3 the amount of rye, then

3 2

2 3

23

2 3 0.

y y

y y

£

- + =

Using these forms for our constraints, we can now state the problem as follows.

Minimize 1 2 316 14 12w y y y= + +

subject to: 1 2 3

2 3

1 2 3

0.75 0.25 0.25 02 3 0

6000

y y yy y

y y y

- - ³

- + £

+ + ³

with 1 2 30, 0, 0.y y y³ ³ ³


1 2 3 1 2 3

0.75 0.25 0.25 1 0 0 0 00 2 3 0 1 0 0 0

1 1 1 0 0 1 0 600016 14 12 0 0 0 1 0

y y y s s s zé ù- - -ê úê ú-ê úê ú-ê úê úê úë û

Since 3 6000,s = - the basic solution is not feasible. So pivot on the 1 in row 3, column 1.

1 2 3 1 2 3

3 1 1

3 4 4

0 1 1 1 0 0.75 0 45000.75R R R0 2 3 0 1 0 0 01 1 1 0 0 1 0 6000

16 R R R 0 2 4 0 0 16 1 96, 000

y y y s s s z

--

-

-

- + - - -


All of the variables are now nonnegative so choose the pivot by locating the most negative number in the bottom row and forming the quotients. Pivot on the 3 in row 2, column 3.

1 2 3 1 2 3

2 1 1

2 3 3

2 4 4

R 3R R 0 5 0 3 1 2.25 0 13,5000 2 3 0 1 0 0 0

R 3R R 3 5 0 0 1 3 0 18,0004R 3R R 0 14 0 0 4 48 3 288,000

y y y s s s zé ù- + - -ê úê ú-ê úê ú- + - -ê úê ú+ - -ê úë û


1 2 3 1 2 3

1 2 2

1 3 3

1 4 4

0 5 0 3 1 2.25 0 13,5002R 5R R 0 0 15 6 3 4.5 0 27,000

R R R 3 0 0 3 0 0.75 0 450014R 5R R 0 0 0 42 6 208.5 15 1,251,000

y y y s s s zé ù- -ê úê ú+ -ê úê ú- + - -ê úê ú+ -ê úë û

Create a 1 in the columns corresponding to y1, y2, y3, and z.

1 2 3 1 2 31

1 151

2 2151

3 331

4 415

R R 0 1 0 0.6 0.2 0.45 0 2700R R 0 0 1 0.4 0.2 0.3 0 1800

1 0 0 1 0 0.25 0 1500R R0 0 0 2.8 0.4 13.9 1 83,400R R

y y y s s s z

é ù- -ê úê ú -ê úê ú- - ê úê ú-ê ú ë û

Here, 1 1500,y = 2 2700,y = 3 1800,y = and 83,400.z w= - = Therefore, use 1500 lb of



bluegrass, 2700 lb of rye, and 1800 lb of Bermuda for a minimum cost of $834.

29. Let x1 = the number of pounds of bluegrass seed,

x2 = the number of pounds of rye seed,

and x3 = the number of pounds of Bermuda seed. If each batch must contain at least 25% bluegrass

seed, then 1 1 2 3

1 2 3

0.25( )0.75 0.25 0.25 0.

y y y yy y y

³ + +

- - ³

And if the amount of Bermuda must be no more than 2

3 the amount of rye, then

23 23

2 32 3 0.

y y

y y

£

- + =

Using these forms for our constraints, we can now state the problem as follows.

Minimize 1 2 316 14 12w y y y= + +

subject to: 1 2 3

2 3

1 2 3

0.75 0.25 0.25 02 3 0

6000

y y yy y

y y y

- - ³

- + £

+ + =

with 1 2 30, 0, 0.y y y³ ³ ³


1 2 3 1 2

0.75 0.25 0.25 1 0 0 0 00 2 3 0 1 0 0 0

1 1 1 0 0 1 0 600016 14 12 0 0 0 1 0

y y y s s a zé ù- - -ê úê ú-ê úê úê úê úê úë û

First eliminate the artificial variable a. Pivot on the 1 in row 3, column 1.

1 2 3 1 2

3 1 1

3 4 4

0.75R R R 0 1 1 1 0 0.75 0 45000 2 3 0 1 0 0 01 1 1 0 0 1 0 6000

16R R R 0 2 4 0 0 16 1 96,000

y y y s s a zé ù- ê úê ú-ê úê úê úê ú- + - - - -ê úë û

Since 0,a = we can drop the a column.

1 2 3 1 2

0 1 1 1 0 0 4500

0 2 3 0 1 0 01 1 1 0 0 0 60000 2 4 0 0 1 96,000

y y y s s zé ùê úê ú-ê úê úê úê ú

- - -ê úë û


1 2 3 1 2

2 1 1

2 3 3

2 4 4

0 5 0 3 1 0 13,500R 3R R0 2 3 0 1 0 03 5 0 0 1 0 18, 000R 3R R0 14 0 0 4 3 288, 0004R 3R R

y y y s s z

-- +

-

-- +

- -+



1 2 3 1 2

1 2 2

1 3 3

1 4 4

0 5 0 3 1 0 13,5002R 5R R 0 0 15 6 3 0 27,000

R R R 3 0 0 3 0 0 450014R 5R R 0 0 0 42 6 15 1, 251,000

y y y s s zé ù-ê úê ú+ ê úê ú- + -ê úê ú+ -ê úë û


1 2 3 1 21

1 151

2 2151

3 331

4 415

R R 0 1 0 0.6 0.2 0 2700R R 0 0 1 0.4 0.2 0 1800

1 0 0 1 0 0 1500R R0 0 0 2.8 0.4 1 83,400R R

y y y s s z

é ù-ê úê úê úê ú- ê úê ú- ê úë û

Here, 1 1500,y = 2 2700,y = 3 1800,y = and 83,400.z w= - = Therefore, use 1500 lb of

bluegrass, 2700 lb of rye, and 1800 lb of Bermuda for a minimum cost of $834.

30. Let 1y = the number of gallons of ingredient 1, 2y = the number of gallons of ingredient 2, 3y = the number of gallons of ingredient 3, 4y = the number of gallons of ingredient 4, 5y = the number of gallons of ingredient 5, and 6y = the number of gallons of water.

Note that 10%(15,000) = 1500, and 0.01(15,000) = 150.

The problem becomes:

Minimize

1 2 3 4

5 6

0.48 0.32 0.53 0.280.43 0.04

w y y y yy y

= + + +

+ +

subject to:

1 2 3 4 5

3 4

2 5

1 4

1 2 3 4 5 6

0.28 0.19 0.43 0.57 0.22 1500150

150150

15,000

y y y y yy y

y yy yy y y y y y

+ + + + £

+ ³

³++ ³

+ + + + + =

Section 4.4 281


with 1 2 3 4 50, 0, 0, 0, 0,y y y y y³ ³ ³ ³ ³ 6 0.y ³

This exercise should be solved by graphing calculator or computer methods. The answer is to use 0 gal of ingredient 1, 150 gal of 2, 0 gal of 3, 150 gal of 4, 0 gal of 5, and 14,700 gal of water for a minimum cost of $678.

31. Let 1x = the amount of chemical I,

2x = the amount of chemical II,

and 3x = the amount of chemical III.

Minimize 1 2 31.09 0.87 0.65w x x x= + +

subject to: 1 2 3

1 2 3

2 3

7500.09 0.04 0.03 30

3 4

x x xx x x

x x

+ + ³

+ + ³

=

with 1 2 30, 0, 0.x x x³ ³ ³

We follow the suggestion in the note in the text to reduce the number of variables by using the fact that 3 20.75x x= to express our constraints as follows:

Minimize 1 21.09 1.3575w x x= +

subject to 1 2

1 2

1.75 7500.09 0.0625 30

x xx x

+ ³

+ ³

We maximize z w= - and after multiplying the second constraint through by 100, our initial tableau is the following:

1 2 1 2

1 1.75 1 0 0 750

9 6.25 0 100 0 30001.09 1.3575 0 0 1 0

x x s s zé ù-ê úê ú-ê úê úê úë û

Since 1s is negative, we look for a pivot in the first column, and choose 9 because it has the smallest ratio with the corresponding entry in the last column.

1 2 1 2

2 1 1

2 3 3

R 9R R 0 9.5 9 100 0 37509 6.25 0 100 0 3000

1.09R 9R R 0 5.405 0 109 9 3270

x x s s zé ù- + -ê úê ú-ê úê ú- + -ê úë û

1s is still negative so we pivot on the 9.5 in column 2.

1 2 1 2

1 2 2

1 3 3

0 9.5 9 100 0 37506.25R 9.5R R 85.5 0 56.25 1575 0 5062.55.405R 9.5R R 0 0 48.645 495 85.5 51,333.75

x x s s zé ù-ê úê ú- + -ê úê ú- + -ê úë û

This tableau yields the following solution.

1 2

3

5062.5 375059.21, 394.74,85.5 9.5

3750 3 296.059.5 4

x x

x

= = = =

= ⋅ =

Minimum = 51,333.75 600.3985.5

æ ö- ÷ç- =÷ç ÷÷çè ø

So use 59.21 kg of chemical I, 394.74 kg of chemical II, and 296.05 kg of chemical III, for a minimum cost of $600.39.

32. Let 1y = the number of ounces of ingredient I, 2y = the number of ounces of ingredient II, and 3y = the number of ounces of ingredient III.

Expressing the problem in cents, the problem is:

Minimize 1 2 330 9 27w y y y= + +

subject to 1 2

1 2 3

1

3

14

3

2

1

1015

y y yy y y

y y

y y

+ + ³

+ + £

³

³

with 1 2 30, 0, 0.y y y³ ³ ³

Rewrite the last two inequalities so that the problem becomes:

Minimize 1 2 330 9 27w y y y= + +

subject to: 1 2

1 2 3

1

1

3

2

3

1015

4 00

y y yy y yyy y

y

+ + ³

+ + £

- £+- £

with 1 2 30, 0, 0.y y y³ ³ ³

We maximize z w=- and have the following initial tableau.



1 2 3 1 2 3 41 1 1 1 0 0 0 0 101 1 1 0 1 0 0 0 154 1 0 0 0 1 0 0 0

0 1 0 0 0 1 0 030 9 27 0 0 0 0 1 01

y y y s s s s z-

-

-


Because the solution is not feasible 1( 10),s = - pivot on the 1 in row 4, column 1.

4 1 1

4 2 2

4 3 3

4 5 5

1 2 3 1 2 3 40 1 2 1 0 0 1 0 10R R R0 1 2 0 1 0 1 0 15R R R

4R R R 0 4 0 0 1 4 0 01 0 1 0 0 0 1 0 0

30R R R 0 9 57 0 0 0 30 1 0

1

y y y s s s s z- -- +

-- +

+ -

-

- + -


Because the solution is still not feasible 1( 10),s = - pivot on the 1 in row 3, column 2.

3 1 1

3 2 2

3 5 5

1 2 3 1 2 3 4

0 1 6 1 0 1 5 0 10R R R0 1 6 0 1 1 5 0 15R R R0 1 4 0 0 1 4 0 01 0 1 0 0 0 1 0 00 0 93 0 0 9 66 1 09R R R

y y y s s s s z- - -- +

- -- +

-

-

- -- +


Because the solution is still not feasible 1( 10),s =- pivot on the 6 in row 1, column 3.

1 2 3 1 2 3 4

1 2 2

1 3 3

1 4 4

3 5 5

0 0 6 1 0 1 5 0 100 0 0 1 1 0 0 0 5R R R0 3 0 2 0 1 2 0 202R 3R R6 0 0 1 0 1 1 0 10R 6R R0 0 0 31 0 13 23 2 31031R 2R R

y y y s s s s z

- - -

- +

-+

- -+

-- +



1 2 3 1 2 3 4

2 2

3 3

4 4

5 5

1 1 5 56 6 6 3

16

2 1 2 2013 3 3 331 1 1 516 6 6 36

31 13 2312 2 22

0 0 1 0 0

R R 0 0 0 1 1 0 0 0 5

0 1 0 0 0R R

1 0 0 0 0R R

0 0 0 0 1 155R R

y y y s s s s z

- - -

-

- -

-

é ùê úê úê úê úê úê úê úê úê úê úê úê úë û

Here 1 2 35 20 5

, , ,3 3 3

y y y= = = and 155.w z=- =

Therefore, the additive should consist of 53

oz of

ingredient I, 203

oz of ingredient II, and 53

oz of

ingredient III, for a minimum cost of 155¢/gal, or $1.55/gal. The amount of additive that should be

used per gallon of gasoline is 5

3

20 510

3 3+ + = oz.

33. (a) Let 1x = the number of hours spent doing calisthenics,

2x = the number of hours spent swimming,

and 3x = the number of hours spent playing the drums.

The problem can be stated as follows.

Maximize 1 2 3388 518 345z x x x= + +

subject to: 1 2 3

1 2 3

3

3

102 0

41.

x x xx x x

xx

£+ +- + ³

£³

with 1 2 30, 0, 0.x x x³ ³ ³


1 2 3 1 2 3 4

1 1 1 1 0 0 0 0 10

1 2 1 0 1 0 0 0 00 0 1 0 0 1 0 0 40 0 1 0 0 0 1 0 1

388 518 345 0 0 0 0 1 0

x x x s s s s zé ùê úê ú- -ê úê úê úê ú

-ê úê úê ú- - -ë û

Since 2s is negative, we pivot on the 1 in row 2 of column 1.

1 2 3 1 2 3 4

2 1 1

2 5 5

0 3 0 1 1 0 0 0 10R R R1 2 1 0 1 0 0 0 00 0 1 0 0 1 0 0 4

0 0 1 0 0 0 1 0 1388R R R 0 1294 43 0 388 0 0 1 0

x x x s s s s zé ù- + ê úê ú- -ê úê úê úê ú

-ê úê ú

+ ê ú- -ë û

Since 4s is negative, we pivot on the 1 in row 4 of column 3.

Chapter 4 Review 283


1 2 3 1 2 3 4

4 2 2

4 3 3

4 5 5

0 3 0 1 1 0 0 0 10R R R 1 2 0 0 1 0 1 0 1R R R 0 0 0 0 0 1 1 0 3

0 0 1 0 0 0 1 0 143R R R 0 1294 0 0 388 0 43 1 43

x x x s s s s zé ùê úê ú- + - - -ê úê ú- + ê úê ú

-ê úê úê ú- + - - -ë û

Now we work on the column with the most negative indicator and pivot on 3 in column 2.

1 2 3 1 2 3 4

1 2 2

1 5 5

0 3 0 1 1 0 0 0 103 0 0 2 1 0 3 0 172R 3R R0 0 0 0 0 1 1 0 30 0 1 0 0 0 1 0 1

1294R 3R R 0 0 0 1294 130 0 129 3 12,811

x x x s s s s zé ùê úê ú-+ ê úê úê úê ú-ê úê ú+ ê úë û

This tableau gives the solution: Joe should do 17/3 hours of calisthenics, 10/3 hours of swimming, and 1 hour of playing the drums, for a maximum calorie expenditure of 12,811/3 or 1

34270 calories.

Chapter 4 Review Exercises

1. True

2. False

3. True

4. False

5. False

6. True

7. True

8. False

9. False

10. True

11. False

12. True

13. False

14. True

15. The simplex method should be used for problems with more than two variables or problems with two variables and many constants.

16. If a surplus variable cannot be made nonnegative, then the inequality which represents one of the constraints can never exist. This means that no solution is possible.

17. (a) Maximize 1 22 7z x x= +

subject to: 1 2

1 2

1 2

1 2

4 6 603 182 5 20

15

x xx xx xx x

+ £

+ £

+ £

+ £

with 1 20, 0.x x³ ³

Adding slack variables s1, s2, s3, and s4, we obtain the following equations.

1 2 1

1 2 2

1 2 3

1 2 4

4 6 603 182 5 20

15.

x x sx x sx x sx x s

+ + =

+ + =

+ + =

+ + =

(b) The initial simplex tableau is as follows.

1 2 1 2 3 44 6 1 0 0 0 0 603 1 0 1 0 0 0 182 5 0 0 1 0 0 20 .1 1 0 0 0 1 0 152 7 0 0 0 0 1 0

x x s s s s zé ùê úê úê úê úê úê úê úê ú- -ê úë û

18. Maximize 1 225 30z x x= +

subject to: 1 2

1 2

1 2

1 2

3 5 47

25

5 2 35

2 30

x x

x x

x x

x x

+ £

+ £

+ £

+ £

with 1 20, 0.x x³ ³

(a) Add 1 2 3, , ,s s s and 4s as slack variables to obtain

1 2 1

1 2 2

1 2 3

1 2 4

3 5 4725

5 2 352 30.

x x sx x sx x sx x s

+ + =

+ + =

+ + =

+ + =

(b) The initial tableau is



1 2 1 2 3 43 5 1 0 0 0 0 471 1 0 1 0 0 0 255 2 0 0 1 0 0 35 .2 1 0 0 0 1 0 30

25 30 0 0 0 0 1 0

x x s s s s zé ùê úê úê úê úê úê úê úê ú- -ê úë û

19. Maximize 1 2 35 8 6z x x x= + +

subject to: 1 2 3

1 2 3

1 2

902 5 120

3 80

x x xx x xx x

+ + £

+ + £

+ ³

with 1 2 30, 0, 0.x x x³ ³ ³

(a) Adding the slack variables 1s and 2s and subtracting the surplus variable 3,s we obtain the following equations:

1 2 3 1

1 2 3 2

1 2 3

902 5 120

3 80.

x x xx x x sx x s

s+ + =

+ + + =

+ - =

+

(b) The initial tableau is

1 2 3 1 2 31 1 1 1 0 0 0 902 5 1 0 1 0 0 120

.1 3 0 0 0 1 0 805 8 6 0 0 0 1 0

x x x s s s zé ùê úê úê úê ú-ê úê ú- - -ê úë û

20. Maximize 1 2 34 6 8z x x x= + +

subject to: 1 2 3

1 3

1 2 3

2 2008 6 4003 5 300

x x xx xx x x

+ + ³

+ £

+ £+

with 1 2 30, 0, 0.x x x³ ³ ³

(a) Introduce 1s as a surplus variable and 2s and

3s as slack variables to obtain the following equations.

1 2 3

1 3 2

1 2 3

1

3

2 2008 6 4003 5 300.

x x xx x sx x x s

s+ + =

+ + =

+ + + =

-

(b) The initial tableau is as follows.

1 2 3 1 2 31 1 2 1 0 0 0 2008 0 6 0 1 0 0 400

.3 5 1 0 0 1 0 3004 6 8 0 0 0 1 0

x x x s s s zé ù-ê úê úê úê úê úê ú- - -ê úë û

21. 1 2 3 1 24 5 2 1 0 0 18

2 8 0 1 0 245 3 6 0 0 1 0

6 .

x x x s s z

- - -


The most negative entry in the last row is 6,- and the smaller of the two quotients is 24

6 4.= Hence, the 6 in row 2, column 3, is the first pivot. Performing row transformations leads to the following tableau.

2 1 1

2 3 3

1 2 3 1 2

R 3R R 10 7 0 3 1 0 302 8 6 0 1 0 24

R R R 3 5 0 0 1 1 24.

x x x s s z

- + -

+ -



1 2 2

1 3 3

1 2 3 1 2

10 7 0 3 1 0 30R 5R R 0 33 30 3 6 0 90

3R 10R R 0 71 0 9 7 10 330.

x x x s s z

-

- + -

+



1 1

2

3 3

1 7 3 110 10 10 101 11 1 130 10 10 51 71 9 7

10 10 10 10

1

2

2 3 1 2

R R 1 0 0 3

R R 0 1 0 3

R R 0 0 10 33

.

x x x s s z

-

-

é ùê úê úê úê úê úê úê úê úë û

The maximum value is 33 when 1 23, 0,x x= =

3 13, 0,x s= = and 2 0.s =

22. 1 2 1 2

2 7 1 0 0 142 3 0 1 0 102 4 0 0 1 0




The most negative indicator is in the second

column. The smaller quotient is 14 2.7

= Pivot on

the 7 in row 1, column 2.

1 2 1 2

1 2 2

1 3 3

2 7 1 0 0 14

3R 7R R 8 0 3 7 0 284R 7R R 6 0 4 0 7 56



2 1 1

2 3 3

1 2 1 2

R 4R R 0 28 7 7 0 288 0 3 7 0 28

3R 4R R 0 0 7 21 28 308

x x s s z- + -

-

+



1 2 1 2

1 1

2

3 3

1 1 128 4 41 3 7 78 8 8 21 1 328 4 4

2

1

0

0

R R 0 0 1

R R 1 0

R R 0 1 11

x x s s zé ù

-ê úê úê ú

-ê úê úê ú

ê úê úë û

The maximum value is 11 when

1 272 1, 1, 0,x x s= = = and 2 0.s =

23. 1 2 3 1 2 31 2 2 1 0 0 0 50

3 1 0 0 1 0 0 201 0 2 0 0 1 0 155 3 2 0 0 0 1 0

s s s zx x xé ùê úê úê úê ú-ê úê ú- - -ê úë û

The initial basic solution is not feasible since 3 15.s = - In the third row where the negative

coefficient appears, the nonnegative entry that appears farthest to the left is the 1 in the first column. In the first column, the smallest nonnegative quotient is 20

3. Pivot on the 3 in row 2,

column 1.

2 1 1

2 3 3

2 4 4

1 2 3 1 2 30 5 6 3 1 0 0 130R 3R R3 1 0 0 1 0 0 20

0 1 0 1 3 0 25R 3R R0 4 6 0 5 0 3 1005R 3R R

6

x x x s s s z-- +

- - -- +

- -+


Continue by pivoting on each boxed entry.

3 2 1

3 4 4

1 2 3 1 2 3

R R R 0 0 3 0 3 0 1053 1 0 0 1 0 0 200 1 6 0 1 3 0 25

R R R 0 5 0 0 4 3 3 125

6

x x x s s s z

- +

- - -

+ - -


The basic solution is now feasible, but there are negative indicators.

Continue pivoting.

1 2 2

1 3 3

1 4 4

1 2 3 1 2 3

0 6 0 3 0 3 0 10518 0 0 3 6 3 0 15R 6R R0 0 36 3 0 15 0 255R 6R R0 0 0 15 24 3 18 12755R 6R R

x x x s s s z

- -- +

-+

-+


1 2 2

1 3 3

1 4 4

1 2 3 1 2 30 6 0 3 0 0 105

18 6 0 0 6 0 0 120R R R0 30 36 18 0 0 0 7805R R R0 6 0 18 24 0 18 1380R R R

3x x x s s s z

+

+

+



1 2 3 1 2 31

1 131

2 2181

3 3361

4 418

R R 0 2 0 1 0 1 0 35R R 1 .33 0 0 .33 0 0 6.67

0 .83 1 .5 0 0 0 21.67R R0 .33 0 1 1.33 0 1 76.67

R R

x x x s s s z

é ùê úê úê úê ú

ê úê úê úë û

The maximum value is about 76.67 when 1 6.67,x » 2 0,x = 3 21.67,x » 1 0,s =

2 0,s = and 3 35.s =

24. 1 2 1 2 3

3 6 1 0 0 0 281 1 0 1 0 0 122 1 0 0 1 0 161 2 0 0 0 1 0

x x s s s zé ù-ê úê úê úê úê úê ú- -ê úë û


1 2 1 2 3

1 2 2

1 3 3

1 4 4

3 6 1 0 0 0 28

3 0 1 6 0 0 44R 6R R9 0 1 0 6 0 68R 6R R0 0 1 0 0 3 28R 3R R

x x s s s zé ù-ê úê ú- + ê úê ú- + ê úê ú

-+ ê úë û




1 2 1 2 3

2 1 1

2 3 3

2 4 4

6 6 0 6 0 0 72R R R3 0 1 6 0 0 446 0 0 6 6 0 24R R R3 0 0 6 0 3 72R R R

x x s s s zé ù+ ê úê úê úê ú-- + ê úê ú+ ê úë û

1 2 1 2 31

1 16

13 36

14 43

1 1 0 1 0 0 12R R3 0 1 6 0 0 44

R R 1 0 0 1 1 0 41 0 0 2 0 1 24R R

x x s s s zé ù ê úê úê úê ú -ê úê ú

ê úë û

The maximum is 24 when 1 2 0, 12,x x= =

1 44,s = 2 3 0, and 4.s s= =

25. Minimize 1 210 15w y y= +

subject to: 1 2

1 2

175 8 42

y yy y

+ ³

+ ³

with 1 20, 0.y y³ ³

Using the dual method:

To form the dual, write the augmented matrix for the given problem.

1 1 175 8 42

10 15 0



1 5 101 8 15

17 42 0




subject to: 1 2

1 2

5 108 15

x xx x

+ £

+ £

with 1 20, 0.x x³ ³


1 2 1 2

1 5 1 0 0 10

1 8 0 1 0 1517 42 0 0 1 0


Pivot on the 8 in row 2 column 2.

1 2 1 2

2 1 1

2 3 3

5R 8R R 3 0 8 5 0 51 8 0 1 0 15

21R 4R R 47 0 0 21 4 315



1 2 1 2

1 2 2

1 3 3

3 0 8 5 0 5

R 3R R 0 24 8 8 0 4047 R 3R R 0 0 376 172 12 1180

x x s s zé ù-ê úê ú- + -ê úê ú+ -ê úë û


1 2 1 2

2 1 1

2 3 3

5R 8R R 24 120 24 0 0 2400 24 8 8 0 40

43R 2R R 0 1032 408 0 24 4080

x x s s zé ù+ ê úê ú-ê úê ú+ ê úë û


1 2 1 21

1 1241

2 281

3 324

R R 1 5 1 0 0 10R R 0 3 1 1 0 5

0 43 17 0 1 170R R

x x s s z

é ùê úê ú -ê úê úê ú ë û

The minimum value is 170 when 1 17y = and

2 0.y =

Using the method of 4.4:

Change the objective function to

Maximize 1 210 15 .z w y y= - = - -



1 2 1 2

1 1 1 0 0 175 8 0 1 0 42

10 15 0 0 1 0

y y s s zé ù-ê úê ú-ê úê úê úë û

The solution is not feasible since 1 17s = - and

2 42.s = - Pivot on the 1 in row 1, column 1.

1 2 1 2

1 2 2

1 3 3

1 1 1 0 0 175R 5R R 0 3 5 1 0 4310R R R 0 5 10 0 1 170

y y s s zé ù-ê úê ú- + - -ê úê ú- + -ê úë û



Thus when 17, 170 so the minimum is 170.y z= =-

26. Minimize 1 2 322 44 33w y y y= + +

subject to: 1 2 3

1 3

1 2 3

2 33

3 2 2 8

y y yy yy y y

+ + ³

+ ³

+ + ³

with 1 2 30, 0, 0.y y y³ ³ ³



1 2 1 31 0 1 33 2 2 8

22 44 33 0


Form the transpose of the matrix.

1 1 3 222 0 2 441 1 2 333 3 8 0



Maximize 1 2 33 3 8z x x x= + +

subject to: 1 2 3

1 3

1 2 3

3 222 2 44

2 33

x x xx xx x x

+ + £

+ £

+ + £

with 1 2 30, 0, 0.x x x³ ³ ³


1 2 3 1 2 3

1 1 3 1 0 0 0 222 0 2 0 1 0 0 441 1 2 0 0 1 0 333 3 8 0 0 0 1 0



1 2 3 1 2 3

1 2 2

1 3 3

1 4 4

1 1 3 1 0 0 0 224 2 0 2 3 0 0 882R 3R R1 1 0 2 0 3 0 552R 3R R1 1 0 8 0 0 3 1768R 3R R

x x x s s s zé ùê úê ú- -- + ê úê ú-- + ê úê ú- -+ ê úë û

There is a choice of pivot columns; choose column 2. Pivot on the 1 in row 1, column 2.

1 2 3 1 2 3

1 2 2

1 3 3

1 4 4

1 1 3 1 0 0 0 222R R R 6 0 6 0 3 0 0 132

R R R 0 0 3 3 0 3 0 33R R R 0 0 3 9 0 0 3 198

x x x s s s zé ùê úê ú+ ê úê ú- + - -ê úê ú+ ê úë û

Create a 1 in the columns corresponding to s2, s3, and z.

1 2 3 1 2 3

12 23

13 33

14 43

1 1 3 1 0 0 0 22R R 2 0 2 0 1 0 0 44

0 0 1 1 0 1 0 11R R0 0 1 3 0 0 1 66R R

x x x s s s zé ùê úê úê úê ú- - ê úê ú ê úë û

The minimum value is 66 when 1 23, 0,y y= = and 3 0.y =



Maximize 1 2 322 44 33 .z w y y y= - = - - -

The constraints are not changed. The initial simplex tableau is as follows.

1 2 3 1 2 3

1 2 1 1 0 0 0 31 0 1 0 1 0 0 33 2 2 0 0 1 0 8

22 44 33 0 0 0 1 0

y y y s s s zé ù-ê úê ú-ê úê ú-ê úê úê úë û

The solution is not feasible since 1 3,s = -

2 3,s = - and 3 8.s = - Pivot on the 1 in row 1, column 1.

1 2 3 1 2 3

1 2 2

1 3 3

1 4 4

1 2 1 1 0 0 0 30 2 0 1 1 0 0 0R R R0 4 1 3 0 1 0 13R R R0 0 11 22 0 0 1 6622R R R

y y y s s s zé ù-ê úê ú- -- + ê úê ú-- ê úê ú-- + ê úë û

The solution is feasible and there are no negative indicators so the solution is optimal. Create a 1 in the column corresponding to s2.

1 2 3 1 2 3

2 2

1 2 1 1 0 0 0 30 2 0 1 1 0 0 0R R0 4 1 3 0 1 0 10 0 11 22 0 0 1 66

y y y s s s zé ù-ê úê ú-- ê úê ú-ê úê ú-ê úë û



Since 66,z w= - = - the minimum value is 66 when 1 2 33, 0, and 0.y y y= = =

27. Minimize 1 2 37 2 3w y y y= + +

subject to: 1 2 3

1 2

3

1 3

2 481210

3 30

y y yy y

yy y

+ + ³

+ ³

³

+ ³

with 1 2 30, 0, 0.y y y³ ³ ³



1 1 2 481 1 0 120 0 1 103 0 1 307 2 3 0



1 1 0 3 71 1 0 0 22 0 1 1 3

48 12 10 30 0



Maximize 1 2 3 448 12 10 30z x x x x= + + +

subject to: 1 2 4

1 2

1 3 4

3 72

2 3

x x xx x

x x x

+ + £

+ £

+ + £

with 1 2 3 40, 0, 0, 0.x x x x³ ³ ³ ³


1 2 3 4 1 2 3

1 1 0 3 1 0 0 0 71 1 0 0 0 1 0 0 2

2 0 1 1 0 0 1 0 348 12 10 30 0 0 0 1 0



1 2 3 4 1 2 3

3 1 1

3 2 2

3 4 4

R 2 R R 0 2 1 5 2 0 1 0 11

R 2 R R 0 2 1 1 0 2 1 0 12 0 1 1 0 0 1 0 3

24 R R R 0 12 14 6 0 0 24 1 72

x x x x s s s z

- + - -

- + - - -

+ - -



1 2 3 4 1 2 3

2 1 1

2 4 4

0 0 0 6 2 2 0 0 10R R R0 2 1 1 0 2 1 0 12 0 1 1 0 0 1 0 30 0 8 12 0 12 18 1 786R R R

x x x x s s s zé ù-- + ê úê ú- - -ê úê úê úê ú

-+ ê úë û


1 2 3 4 1 2 3

1 2 2

1 3 3

1 4 4

0 0 0 6 2 2 0 0 100 12 6 0 2 10 6 0 16R 6 R R

12 0 6 0 2 2 6 0 8R 6 R R0 0 8 0 4 8 18 1 982 R R R

x x x x s s s z

-

- -+

-- +

+


Create a 1 in the columns corresponding to x1, x2, and x4.

1 2 3 4 1 2 31 1 51

1 1 3 3 361 1 5 1 41

2 2 2 6 6 2 3121 1 1 1 21

3 3 2 6 6 2 312

0 0 0 1 0 0R R

0 1 0 0R R

1 0 0 0R R

0 0 8 0 4 8 18 1 98

x x x x s s s zé ù- ê úê úê úê - - ú ê úê úê ú- ê úê úê úë û

The minimum value is 98 when 1 24, 8,y y= = and 3 18.y =



Maximize 1 2 37 2 3 .z w y y y= - = - - -





1 2 3 1 2 3 4

1 1 2 1 0 0 0 0 481 1 0 0 1 0 0 0 120 0 1 0 0 1 0 0 10

3 0 1 0 0 0 1 0 307 2 3 0 0 0 0 1 0

y y y s s s s zé ù-ê úê ú-ê úê ú-ê úê ú

-ê úê úê úë û

The solution is not feasible since 1 48,s = -

2 12,s = - 3 10,s = - and 4 30.s = - Pivot on the 3 in row 4, column 1.

1 2 3 1 2 3 4

4 1 1

4 2 2

4 5 5

R 3R R 0 3 5 3 0 0 1 0 114R 3R R 0 3 1 0 3 0 1 0 6

0 0 1 0 0 1 0 0 103 0 1 0 0 0 1 0 30

7R 3R R 0 6 2 0 0 0 7 3 210

y y y s s s s zé ù- + -ê úê ú- + - -ê úê ú-ê úê ú

-ê úê úê ú- + -ë û

The solution is still not feasible since 1 38,s = -

2 2,s = - and 3 10.s = - Pivot on the 3 in row 1, column 2.

1 2 3 1 2 3 4

1 2 2

1 5 5

0 3 5 3 0 0 1 0 1140 0 6 3 3 0 0 0 108R R R0 0 1 0 0 1 0 0 103 0 1 0 0 0 1 0 300 0 8 6 0 0 5 3 4382R R R

y y y s s s s zé ù-ê úê ú-- ê úê ú-ê úê ú

-ê úê úê ú- -- + ë û

Again, the solution is not feasible since 3 10s = - and 4 30.s = - Pivot on the 1 in row 3, column 3.

1 2 3 1 2 3 4

3 1 1

3 2 2

3 4 4

3 5 5

0 3 0 3 0 5 1 0 645R R R6R R R 0 0 0 3 3 6 0 0 48

0 0 1 0 0 1 0 0 10R R R 3 0 0 0 0 1 1 0 20

8R R R 0 0 0 6 0 8 5 3 358

y y y s s s s zé ù-- + ê úê ú- + -ê úê ú-ê úê ú

- + -ê úê úê ú+ - -ë û

The solution is feasible because all variables are nonnegative. But it is still not optimal. Pivot on the 6 in row 2, column 6.

1 2 3 1 2 3 4

2 1 1

2 3 3

2 4 4

2 5 5

0 18 0 3 15 0 6 0 1445R 6R R0 0 0 3 3 6 0 0 480 0 6 3 3 0 0 0 108R 6R R

18 0 0 3 3 0 6 0 72R 6R R4R 3R R 0 0 0 6 12 0 15 9 882

y y y s s s s z

é ù- -- + ê úê ú-ê úê ú-+ ê úê ú- -- + ê úê ú+ -ê úë û

Create a 1 in the columns corresponding to 1,y

2,y 3,y 3s and .z

1 2 3 1 2 3 451 11

1 1 6 6 3181 11

2 2 2 261 11

3 3 2 261 1 11

4 4 6 6 3181 52 4

5 59 3 3 3

0 1 0 0 0 8R R

0 0 0 1 0 0 8R R

0 0 1 0 0 0 18R R

1 0 0 0 0 4R R

R R 0 0 0 0 1 98

y y y s s s s zé ù- - ê úê úê ú- ê úê úê ú- ê úê ú- -ê úê úê ú -ê úë û

Since 98,z w= - = - the minimum value is 98 when 1 4,y = 2 8,y = and 3 18.y =

28. Minimize 1 2 3 43 4 2w y y y y= + + +

subject to: 1 2 3 4

1 2 3 4

4 6 3 8 1913 7 2 6 16

y y y yy y y y

+ + + ³

+ + + ³

with 1 2 3 40, 0, 0, 0.y y y y³ ³ ³ ³



4 6 3 8 1913 7 2 6 16

3 4 1 2 0



4 13 36 7 43 2 18 6 2

19 16 0


Write the dual problem. Maximize 1 219 16z x x= +

subject to: 1 2

1 2

1 2

1 2

4 13 36 7 43 2 18 6 2

x xx xx xx x

+ £

+ £

+ £

+ £

with 1 20, 0.x x³ ³




1 2 1 2 3 4

4 13 1 0 0 0 0 36 7 0 1 0 0 0 43 2 0 0 1 0 0 1

8 6 0 0 0 1 0 219 16 0 0 0 0 1 0

x x s s s s zé ùê úê úê úê úê úê úê úê úê ú- -ë û

Pivot on the 8 in row 4 of column 1.

1 2 1 2 3 4

4 1 1

4 2 2

4 3 3

4 5 5

R 2R R 0 20 2 0 0 1 0 43R 4R R 0 10 0 4 0 3 0 103R 8R R 0 2 0 0 8 3 0 2

8 6 0 0 0 1 0 219R 8R R 0 14 0 0 0 19 8 38

x x s s s s zé ù- + -ê úê ú- + -ê úê ú- + - -ê úê úê úê úê ú+ -ë û


1 2 1 2 3 4

1 2 2

1 3 3

1 4 4

1 5 5

0 20 2 0 0 1 0 40 0 2 8 0 5 0 16R 2R R0 0 2 0 80 31 0 24R 10R R

80 0 6 0 0 13 0 83R 10R R0 0 14 0 0 183 80 4087R 10R R

x x s s s s zé ù-ê úê ú- -- + ê úê ú-+ ê úê ú-- + ê úê ú

+ ê úë û

Get a l in the last row of the z column.

1 2 1 2 3 4

7 183 515 5 40 80 10

0 20 2 0 0 1 0 40 0 2 8 0 5 0 160 0 2 0 80 31 0 24

80 0 6 0 0 13 0 8

0 0 0 0 1R /80 R

x x s s s s zé ù-ê úê ú- -ê úê ú-ê úê ú-ê úê úê ú ê úë û

Now we can read the solution: The minimum is 51/10 when 1 7/40y = and 4 183/80.y =


Change the objective function to Maximize 1 2 3 4.3 4 2z w y y y y= - = - - - -



1 2 3 4 1 2

4 6 3 8 1 0 0 19

13 7 2 6 0 1 0 163 4 1 2 0 0 1 0

y y y y s s zé ù-ê úê ú-ê úê úê úë û


1 2 3 4 1 2

2 1 1

2 3 3

4R 13R R 0 50 31 80 13 4 0 18313 7 2 6 0 1 0 16

3R 13R R 0 31 7 8 0 3 13 48

y y y y s s zé ù- + -ê úê ú-ê úê ú- + -ê úë û

Now pivot on the 80 in row 1 of column 4. (We take advantage of the fact mentioned in Section 4 that any positive entry in the row containing the variable we want to eliminate can be used to identify a pivot column.)

1 2 3 4 1 2

1 2 2

1 3 3

0 50 31 80 13 4 0 1833R 40R R 520 130 13 0 39 52 0 91R 10R R 0 260 39 0 13 26 130 663

y y y y s s zé ù-ê úê ú- + - -ê úê ú

- + -ê úë û

We now find 1 491 7 183,

520 40 80y y= = =

and a

minimum of 663 51.130 10

æ ö- ÷ç- =÷ç ÷÷çè ø

29. 1 2 1 2

5 10 1 0 0 120

10 15 0 1 0 20020 30 0 0 1 0

x x s s zé ùê úê ú-ê úê ú- -ê úë û

The initial tableau is not feasible. Pivot on the 10 in row 2, column 1.

1 2 1 2

2 1 1

2 3 3

R 2R R 0 5 2 1 0 4010 15 0 1 0 200

2R R R 0 0 0 2 1 400

x x s s zé ù- + ê úê ú-ê úê ú+ -ê úë û

The basic solution is feasible, but there are negative indicators. Pivot on the 1 in row 1, column 4.

1 2 1 2

2 1 2

1 3 3

0 5 2 1 0 40R 2R R 10 20 2 0 0 2402R R R 0 10 4 0 1 480

x x s s zé ùê úê ú+ ê úê ú+ ê úë û

Create a one in the column corresponding to x1.

1 2 1 2

1 12 210 5

0 5 2 1 0 40

R R 1 2 0 0 24

0 10 4 0 1 480

x x s s zé ùê úê ú

ê úê úê úê úë û

The maximum value is 480z = when 1 24x = and 2 0.x =



30. Minimize 1 24 2w y y= +

subject to: 1 2

1 2

1 2

3 62 8 21

0, 0

y yy y

y y

+ ³

+ £

³ ³

Let 1 24 2z w y y= - = - - and maximize z.

Introduce the surplus variable s1 and the slack variable s2. The initial tableau is as follows.

1 2 1 2

1 3 1 0 0 62 8 0 1 0 214 2 0 0 1 0

y y s s zé ù-ê úê úê úê úê úë û


1 2 1 2

1 2 2

1 3 3

1 3 1 0 0 62R R R 0 2 2 1 0 94R R R 0 10 4 0 1 24

y y s s z

é ù-ê úê ú- + ê úê ú- + - -ê úë û


1 2 1 2

1 2 2

1 3 3

1 3 1 0 0 62R 3R R 2 0 8 3 0 15

10 0 2 0 3 1210R 3R R

y y s s zé ù-ê úê ú- + -ê úê ú-ê ú+ ë û

Create a 1 in the columns corresponding to y2, s2, and z.

1 2 1 2

1 111 1 3 33

81 22 23 3 3

1 10 23 33 3 3

1 0 0 2R R

R R 0 1 0 5

R R 0 0 1 4

y y s s zé ù- ê úê úê ú -ê úê úê ú -ê úë û

The maximum value of z w= - is 4.- Therefore, the minimum value of w is 4 when 1 0y = and

2 2.y =

31. Maximize 1 210 12z x x= +

subject to: 1 2

1 2

1 2

2 2 172 5 224 3 28

x xx xx x

+ =

+ ³

+ £

with 1 20, 0.x x³ ³

Introduce artificial variable a, surplus variable s1, and slack variable s2. The initial simplex tableau as follows.

1 2 1 2s s

2 2 1 0 0 0 172 5 0 1 0 0 224 3 0 0 1 0 28

10 12 0 0 0 1 0

x x a zé ùê úê ú-ê úê úê úê ú- -ê úë û

First, eliminate the artificial variable a. Pivot on the 2 in row 1, column 1.

1 2 1 2

1 2 2

1 3 3

1 4 4

s s2 2 1 0 0 0 170 3 1 1 0 0 5R R R0 1 2 0 1 0 62R R R0 2 5 0 0 1 855R R R

x x a zé ùê úê ú- -- + ê úê ú-- ê úê ú-+ ê úë û

Now 0,a = so we can drop the a column.

1 2 1 2

2 2 0 0 0 17

0 3 1 0 0 50 1 0 1 0 60 2 0 0 1 85

x x s s zé ùê úê ú-ê úê ú-ê úê ú

-ê úë û

Because 1 5,s = - we choose the 3 in row 2, column 2, as the next pivot.

1 2 1 2

2 1 1

2 3 3

2 4 4

2R 3R R 6 0 2 0 0 410 3 1 0 0 5

0 0 1 3 0 13R 3R R0 0 2 0 3 2652R 3R R

x x s s zé ù- + ê úê ú-ê úê ú-- + ê úê ú

-+ ê úë û

The solution is still not feasible since 132 3

.s = -


1 2 l 2

3 1 1

3 2 2

3 4 4

2R R R 6 0 0 6 0 15R R R 0 3 0 3 0 18

0 0 1 3 0 132R R R 0 0 0 6 3 291

x x s s zé ù- + ê úê ú+ -ê úê ú-ê úê ú

+ -ê úë û

The solution is now feasible but is not yet optimal. Pivot on the 6 in row 1, column 4.



1 2 1 2

1 2 2

1 3 3

1 4 4

6 0 0 6 0 156 6 0 0 0 51R 2R R6 0 2 0 0 41R 2R R6 0 0 0 3 306R R R

x x s s zé ùê úê ú+ ê úê ú+ ê úê ú+ ê úë û

Create a 1 in the columns corresponding to x2, s1, s2, and z.

1 2 1 21 51 16 21 172 26 21 413 32 21

4 43

R R 1 0 0 1 0

R R 1 1 0 0 0

R R 3 0 1 0 0

2 0 0 0 1 102R R

x x s s zé ù ê úê úê úê úê ú

ê úê úê ú ê úë û

The maximum is 102 when 1 0x = and 172 2 .x =

32. Minimize 1 2 324 30 36w y y y= + +

subject to: 1 2 3

1 2 3

1 2 3

5 10 15 120050

0, 0, 0

y y yy y y

y y y

+ + ³

+ + £

³ ³ ³

Let 24z w= - = - 1 30y - 2 336y y- and maximize z. Introduce surplus variable s1 and slack variable s2. The initial tableau is as follows.

1 2 3 1 2

5 10 15 1 0 0 1200

1 1 1 0 1 0 5024 30 36 0 0 1 0

y y y s s zé ù-ê úê úê úê úê úë û

The initial basic solution is not feasible since 1 1200.s = -


1 2 3 1 2

2 1 1

2 3 3

0 5 10 1 5 0 9505R R R1 1 1 0 1 0 50

24R R R 0 6 12 0 24 1 1200

y y y s s zé ù- -- + ê úê úê úê ú- + - -ê úë û


1 2 3 1 2

2 1 1

2 3 3

5 0 5 1 10 0 7005R R R1 1 1 0 1 0 50

6R R R 6 0 6 0 30 1 1500

y y y s s zé ù- - -- + ê úê úê úê ú- + - - -ê úë û


1 2 3 1 3

2 1 1

2 3 3

5R R R 10 5 0 1 15 0 4501 1 1 0 1 0 50

6R R R 12 6 0 0 36 1 1800

y y y s s zé ù- + - - - -ê úê úê úê ú- + - - - -ê úë û

Now 1 450s = - is not a feasible solution, but it is not possible to choose a pivot point. Therefore there is no solution.

33. Any maximizing or minimizing problems can be solved using slack, surplus, and artificial variables. Slack variables are used in problems involving “£ ”constraints. Surplus variables are used in problems involving “³ ” constraints. Artificial variables are used in problems involving “= ” constraints.

34. A dual can be used to solve any standard minimization problem.

35. 4 2 3 1 0 0 95 4 1 0 1 0 106 7 5 0 0 1 0

é ùê úê úê úê ú- - -ê úë û

(a) The 1 in column 4 and the 1 in column 5 indicate that the constraints involve .£ The problem being solved with this tableau is:

Maximize 1 2 36 7 5z x x x= + +

subject to: 1 2 3

1 2 3

4 2 3 95 4 10

x x xx x x

+ + £

+ + £

with 1 2 30, 0, 0.x x x³ ³ ³

(b) If the 1 in row 1, column 4 was 1- rather than 1, then the first constraint would have a surplus variable rather than a slack variable, which means the first constraint would be

1 24 2x x+ + 33 9x ³ instead of

1 2 34 2 3 9.x x x+ + £

(c) 1 2 3 1 2

3 0 5 2 1 0 811 10 0 1 3 0 2147 0 0 13 11 10 227

x x x s s z

é ù-ê úê ú-ê úê úê úë û

From this tableau, the solution is 1 0,x =

212 10 2.1,x = = 8

3 51.6,x = = and

22710 22.7.z = =

(d) The dual of the original problem is as follows:




subject to: 1 2

1 2

1 2

4 5 62 4 73 5

y yy yy y

+ ³

+ ³

+ ³

with 1 0,y ³ 2 0.y ³

(e) From the tableau in part (c), the solution of the dual in part (d) is 13

1 10 1.3,y = = 11

2 10 1.1,y = = and 22710 22.7.w = =

36. (a) Find matrices A, B, C, and X such that the problem

Maximize 1 2 33 2z x x x= + +

subject to: 1 2 3

1 2 3

1 2 3

2 1502 2 8 2002 3 320

x x xx x xx x x

+ + £

+ + £

+ + £

with 1 0,x ³ 2 0,x ³ 3 0x ³

can be written as

Maximize CX

subject to: AX B£

with .X O³

2 1 1 1502 2 8 , 200 ,2 3 1 320

A Bé ù é ùê ú ê úê ú ê ú= =ê ú ê úê ú ê úë û ë û

[ ]1

2

3

3 2 1 ,x

C X xx

é ùê úê ú= = ê úê úë û

(b) To write the dual, write the augmented matrix for the given problem.

2 1 1 1502 2 8 2002 3 1 3203 2 1 0


Now form the transpose.

2 2 2 31 2 3 21 8 1 1

150 200 320 0


The dual is now stated as:

Minimize 1 2 3150 200 320w y y y= + +

subject to: 1 2 3

1 2 3

1 2 3

2 2 2 32 3 28 1

y y yy y yy y y

+ + ³

+ + ³

+ + ³

with 1 0,y ³ 2 0,y ³ 3 0.y ³

This can be stated as: For [ ]1 2 2 ,Y y y y=

Minimize YB

subject to: YA C³

with .Y O³

37. (a) Let 1x = the number of cake plates,

2x = the number of bread plates,

and 3x = the number of dinner plates.

(b) The objective function to maximize is 1 2 315 12 5 .z x x x= + +

(c) The constraints are

1 2 3

1 2 3

1 2 3

15 10 8 15005 4 4 27006 5 5 1200.

x x xx x xx x x

+ + £

+ + £

+ + £

38. (a) Let 1x = the amount invested in oil leases;

2x = the amount invested in stocks;

and 3x = the amount invested in bonds.

(b) We want to maximize

1 2 30.15 0.09 0.05 .z x x x= + +


1 2 3

1 2

1 3

50,00015,00025,000.

x x xx xx x

+ + £

+ £

+ £

39. (a) Let 1x = number of gallons of Fruity wine

and 2x = number of gallons of Crystal wine.

(b) The profit function is

1 212 15 .z x x= +

(c) The ingredients available are the limitations; the constraints are



1 2

1 2

1 2

2 1102 3 1252 90.

x xx xx x

+ £

+ £

+ £

40. (a) Let 1y = the number of kilograms of canned whole tomatoes produced

and 2y = the number of kilograms of tomato sauce produced.

(b) The minimum cost function is

1 24 3.25 .w y y= +


1 2

1

2

1 2

3,000,000800,000

80,0006 3 6,600,000.

y yy

yy y

+ £

³

³

+ ³

41. Maximize 1 2 315 12 5z x x x= + +

subject to: 1 2 3

1 2 3

1 2 3

15 10 8 15005 4 4 27006 5 5 1200

x x xx x xx x x

+ + £

+ + £

+ + £

with 1 0,x ³ 2 0,x ³ 3 0.x ³


1 2 3 1 2 3

15 10 8 1 0 0 0 15005 4 4 0 1 0 0 2700

.6 5 5 0 0 1 0 1200

15 12 5 0 0 0 1 0



1 2 3 1 2 3

1 2 2

1 3 3

1 4 4

15 10 8 1 0 0 0 15000 2 4 1 3 0 0 6600R 3R R0 5 9 2 0 5 0 30002R 5R R0 2 3 1 0 0 1 1500 R R R

x x x s s s zé ùê úê ú-- + ê úê ú-- + ê úê ú

-+ ê úë û


1 2 3 1 2 3

1 2 2

1 3 3

1 4 4

15 10 8 1 0 0 0 1500R 5R R 15 0 12 6 15 0 0 31,500R 2R R 15 0 10 5 0 10 0 4500R 5R R 15 0 23 6 0 0 5 9000

x x x s s s zé ùê ú

- + ê ú- -ê ú- + ê ú- -ê ú

ê ú+ ê úë û

Create a 1 in the columns corresponding to 2,x

2,s 3,s and .z

1 2 3 1 2 33 4 11

1 1 2 5 10104 21

2 2 5 5153 11

3 3 2 21023 61

4 4 5 55

1 0 0 0 150R R

1 0 1 0 0 2100R R

0 1 0 1 0 450R R

3 0 0 0 1 1800R R

x x x s s s zé ù ê úê úê ú- - ê úê úê ú- - ê úê úê ú ê úë û

The maximum profit of $1800 when no cake plates, 150 bread plates, and no dinner places are produced.

42. Based on the information given in Exercise 38, the initial tableau is as follows.

1 2 3 1 2 3

1 1 1 1 0 0 0 50,000

1 1 0 0 1 0 0 15,0001 0 1 0 0 1 0 25,000

0.15 0.09 0.05 0 0 0 1 0


Continue by pivoting on each indicated entry.

1 2 3 1 2 3

2 1 1

2 3 3

2 4 4

0 0 1 1 1 0 0 35,000R R R1 1 0 0 1 0 0 15,000

0 1 1 0 1 1 0 10,000R R R0 0.06 0.05 0 0.15 0 1 22500.15R R R

x x x s s s zé ù-- + ê úê úê úê ú- -- + ê úê ú

-+ ê úë û

1 2 3 1 2 3

3 1 1

3 4 4

0 1 0 1 0 1 0 25,000R R R1 1 0 0 1 0 0 15,0000 1 1 0 1 1 0 10,0000 0.01 0 0 0.1 0.05 1 27500.05R R R

x x x s s s zé ù-- + ê úê úê úê ú- -ê úê ú+ ê úë û

The maximum value is 2750z = when 1 15,000,x = 2 0,x = and 3 10,000.x =

He should invest $15,000 in oil leases and $10,000 in stock for a maximum return of $2750.

43. Based on Exercise 39, the initial tableau is

1 2 1 2 3

2 1 1 0 0 0 110

2 3 0 1 0 0 125.

2 1 0 0 1 0 9012 15 0 0 0 1 0




Locating the first pivot in the usual way, it is found to be the 3 in row 2, column 2. After row transformations, we get the next tableau.

1 2 1 2 3

2 1 1

2 3 3

2 4 4

4 0 3 1 0 0 205R 3R R2 3 0 1 0 0 125

4 0 0 1 3 0 145R 3R R2 0 0 5 0 1 6255R R R

x x s s s zé ù-- + ê úê úê úê ú-- + ê úê ú-+ ê úë û


1 2 1 2 3

3 1 1

3 2 2

3 4 4

0 0 3 0 3 0 60R R R0 6 0 3 3 0 105R 2R R4 0 0 1 3 0 1450 0 0 9 3 2 1395R 2R R

x x s s s zé ù-- + ê úê ú-- + ê úê ú-ê úê ú+ ê úë û

1 2 1 2 31

1 13351 11 2 2 22 26

3 14511 4 4 43 349 3 13951 2 2 24 42

0 0 1 0 1 0 20R R0 1 0 0R R1 0 0 0R R0 0 0 1R R

x x s s s zé ù- ê úê ú-ê úê úê ú- ê úê úê ú ê úë û

The final tableau gives the solution 1451 4 ,x =

352 2 ,x = and 1395

2 697.5.z = = 36.25 gal of Fruity wine and 17.5 gal of Crystal wine should be produced for a maximum profit of $697.50.

44. Based on Exercise 40, the initial tableau is as follows.

1 2 1 2 3 4

1 1 1 0 0 0 0 3,000,000

1 0 0 1 0 0 0 800,0000 1 0 0 1 0 0 80,0006 3 0 0 0 1 0 6,600,0004 3.25 0 0 0 0 1 0

y y s s s s zé ùê úê ú-ê úê ú-ê úê ú

-ê úê úê úë û

Pivot on the 1 in row 2, column 1 since the basic solution is not feasible.

1 2 1 2 3 4

2 1 1

2 4 4

2 5 5

0 1 1 1 0 0 0 2,200,000R R R1 0 0 1 0 0 0 800,000

0 1 0 0 1 0 0 80,0000 3 0 6 0 1 0 1,800,0006R R R0 3.25 0 4 0 0 1 3,200,0004R R R

y y s s s s zé ù- + ê úê ú-ê úê ú-ê úê ú

-- + ê úê úê ú-- + ë û

Pivot on the 1 in row 3, column 2 since the basic solution is not feasible.

1 2 1 2 3 4

3 1 1

3 4 4

3 5 5

0 0 1 1 1 0 0 1,400,000R R R1 0 0 1 0 0 0 800,0000 1 0 0 1 0 0 80,000

0 0 0 6 3 1 0 1,560,0003R R R0 0 0 4 3.25 0 1 3,460,0003.25R R R

y y s s s s zé ù- + ê úê ú-ê úê ú-ê úê ú

-- + ê úê úê ú-- + ë û

Pivot on the 6 in row 4, column 4 since basic solution is not feasible.

1 2 1 2 3 4

4 1 1

4 2 2

4 5 5

0 0 6 0 3 1 0 6,840,000R 6R R6 0 0 0 3 1 0 6,360,000R 6R R0 1 0 0 1 0 0 80,000

0 0 0 6 3 1 0 1,560,000

0 0 0 0 3.75 2 3 13,500,0002R 3R R

y y s s s s z

- +

-+

-

-

- - -+


Create a 1 in the columns corresponding to 1,y 2,y 1,s 2,s and .z

1 2 1 2 3 41 11

1 1 2 661 112 62 26

1 112 64 46

2135 53

0 0 1 0 0 1,140,000R R1 0 0 0 0 1,060,000R R0 1 0 0 1 0 0 80,0000 0 0 1 0 260,000R R0 0 0 0 1.25 1 4,500,000R R

y y s s s s zé ù ê úê ú-ê úê úê ú-ê úê ú- ê úê ú-ê ú- ë û

The final tableau gives the solution 2 1,060,000,y = 2 80,000,y = and

4,500,000.z = Use 1,060,000 kg of whole tomatoes and 80,000 kg for sauce for a minimum cost of $4,500,000.

45. (a) Let 1y = the number of cases of corn,

2y = the number of cases of beans

and 3y = the number of cases of carrots.

Minimize 1 2 310 15 25w y y y= + +

subject to: 1 2 3

1 2

3

10002340

y y yy yy

+ + ³

³

³

with 1 20, 0.y y³ ³

The second constraint can be rewritten as 1 22 0.y y- ³ Change this to a

maximization problem by letting 1 2 310 15 25 .z w y y y= - = - - - Now

maximize 110z y= - 2 315 25y y- - subject to the constraints above. Begin by inserting surplus variables to set up the first tableau.



1 2 3 1 2 3

1 1 1 1 0 0 0 10001 2 0 0 1 0 0 00 0 1 0 0 1 0 340

10 15 25 0 0 0 1 0

y y y s s s zé ù-ê úê ú- -ê úê ú-ê úê úê úë û

Multiply row 2 by 1- so that 2s is positive.

1 2 3 1 2 3

2 2

1 1 1 1 0 0 0 10001 2 0 0 1 0 0 0R R0 0 1 0 0 1 0 340

10 15 25 0 0 0 1 0

y y y s s s z

é ù-ê úê ú-- ê úê ú-ê úê úê úë û


1 2 3 1 2 3

1 2 2

1 4 4

1 1 1 1 0 0 0 10000 3 1 1 1 0 0 1000R R R0 0 1 0 0 1 0 3400 5 15 10 0 0 1 10,00010R R R

y y y s s s zé ù-ê úê ú-+ ê úê ú-ê úê ú

-- + ê úë û


1 2 3 1 2 3

3 1 1

3 2 2

3 4 4

1 1 0 1 0 1 0 660R R R0 3 0 1 1 1 0 660R R R0 0 1 0 0 1 0 3400 5 0 10 0 15 1 15,10015R R R

y y y s s s zé ù-- + ê úê ú-- + ê úê ú-ê úê ú-- + ê úë û

The maximum value of z is 15,100- when 1 660,y = 2 0,y = and 3 340.y = Hence

the minimum value of w is 15,100 when 1 660,y = 2 0,y = and 3 340.y =

Produce 660 cases of corn and 340 cases of carrots for a minimum cost of $15,100.

(b) The dual problem is as follows.

Maximize 1 31000 340z x x= +

subject to: 1 2

1 2

1 3

102 15

25

x xx xx x

+ £

- £

+ £

with 1 2 30, 0, 0.x x x³ ³ ³


1 2 3 1 2 3

1 1 0 1 0 0 0 101 2 0 0 1 0 0 151 0 1 0 0 1 0 25

1000 0 340 0 0 0 1 0

x x x s s s zé ùê úê ú-ê úê úê úê ú- -ê úë û


1 2 3 1 2 3

1 2 2

1 3 3

1 4 4

1 1 0 1 0 0 0 100 3 0 1 1 0 0 5R R R0 1 1 1 0 1 0 15R R R0 1000 340 1000 0 0 1 10,0001000R R R

x x x s s s zé ùê úê ú- -- + ê úê ú- -- + ê úê ú

-+ ê úë û


1 2 3 1 2 3

3 4 4

1 1 0 1 0 0 0 100 3 0 1 1 0 0 50 1 1 1 0 1 0 150 660 0 660 0 340 1 15,100340R R R

x x x s s s zé ùê úê ú- -ê úê ú- -ê úê ú+ ê úë û

The minimum value of w is 15,100 when 1 660,y = 2 0,y = and 3 340,y = that is,

660 cases of corn, 0 cases of beans, and 340 cases of carrots should be produced to minimize costs, and the minimum cost is $15,100.

(c) The final tableau for the dual solution shows that the shadow cost of acreage 1( )x is $10 acre, so increasing the number of acres planted by 100 will increase the minimum cost by ($10)(100) or $1000, so the new minimum will be $15,100 $1000 $16,100.+ =

46. Let 1y = the number of packages of Sun Hill

and 2y = the number of packages of Bear Valley.

The problem is:


subject to: 1 2

1 2

1 2

10 2 204 4 242 8 24

y yy yy y

+ ³

+ ³

+ ³

with 1 20, 0.y y³ ³

(a) Change this to a maximization problem by letting 1 23 2 .z w y y= - = - - Now maximize 1 23 2z y y= - - subject to the constraints above. Begin by inserting surplus variables to set up the first tableau.




1 2 1 2 3

10 2 1 0 0 0 204 4 0 1 0 0 242 8 0 0 1 0 243 2 0 0 0 1 0

y y s s s zé ù-ê úê ú-ê úê ú-ê úê úê úë û


1 2 1 2 3

1 2 2

1 3 3

1 4 4

10 2 1 0 0 0 200 32 4 10 0 0 1604R 10R R0 38 1 0 5 0 100R 5R R0 14 3 0 0 10 603R 10R R

y y s s s zé ù-ê úê ú-- + ê úê ú-- + ê úê ú

-- + ê úë û


1 2 1 2 3

3 1 1

3 2 2

3 4 4

190 0 20 0 5 0 280R 19R R16R 19R R 0 0 60 190 80 0 1440

0 38 1 0 5 0 1007R 19R R 0 0 50 0 35 190 1840

y y s s s z

-- +

- + -

-

- + -



1 2 1 2 3

2 1 1

2 3 3

2 4 4

570 0 0 190 95 0 2280R 3R R0 0 60 190 80 0 14400 2280 0 190 380 0 4560R 60R R0 0 0 950 190 1440 18, 2405R 6R R

y y s s s zé ù-+ ê úê ú-ê úê ú

-- + ê úê úê ú- -- + ë û


1 2 1 2 3

2 1 1

2 3 3

2 4 4

9120 0 1140 570 0 0 912019R 16R R0 0 60 190 80 0 1440

0 9120 1440 2850 0 0 45, 60019R 4R R0 0 1440 3990 0 9120 118,56019R 8R R

y y s s s z

-- +

-

-+

-+


Create a 1 in the columns corresponding to 1,y

2,y and .z

1 2 1 2 31 11

1 1 8 169120

5113 3 8 169120

7114 4 8 169120

1 0 0 0 1R R0 0 60 190 80 0 1440

0 1 0 0 5R R

0 0 0 1 13R R

y y s s s zé ù- ê úê ú

-ê úê úê ú- ê úê úê ú- ê úë û

The maximum value of z is 13- when 1 1y = and 2 5.y = Hence the minimum value of w is 13 when 1 1y = and 2 5.y = Thus the minimum

cost is $13 for 1 package of Sun Hill and 5 packages of Bear Valley.

(b) The dual problem is as follows.

Maximize 1 2 3 20 24 24z x x x= + +

subject to: 1 2 3

1 2 3

10 4 2 32 4 8 2x x xx x x

+ + £

+ + £

with 1 2 30, 0, 0.x x x³ ³ ³


1 2 3 1 2

10 4 2 1 0 0 3

2 4 8 0 1 0 220 24 24 0 0 1 0


Pivot about the 4 in row 2, column 2.

1 2 3 1 2

2 1 1

2 3 3

R R R 8 0 6 1 1 0 12 4 8 0 1 0 2

6R R R 8 0 24 0 6 1 12

x x x s s zé ù- + - -ê úê úê úê ú+ -ê úë û

Pivot about the 8 in row 1, column 1.

1 2 3 1 2

1 2 2

1 3 3

8 0 6 1 1 0 1R 4R R 0 16 38 1 5 0 7

R R R 0 0 18 1 5 1 13

x x x s s zé ù- -ê úê ú- + -ê úê ú+ ê úë û

1 2 3 1 23 1 1 11

1 1 4 8 8 8819 5 71 1

2 216 8 16 16 16

1 0 0R R

R R 0 1 0

0 0 18 1 5 1 13

x x x s s zé ù- - ê úê úê ú -ê úê úê úë û

The minimum value of w is 13 when 1 1y = and 2 5,y = that is, you should buy 1 package of Sun Hill and 5 packages of Bear Valley for a minimum cost of $13.

(c) The shadow cost for peanuts is based on the variable 1x and can be read from the final

simplex tableau: 18 . To get 8 more ounces of

peanuts will cost an additional ( )188 $1,= so

the total cost will then be $14.

47. (a) Let 1x = the number of hours doing tai chi,

2x = the number of hours riding a unicycle,



and 3x = the number of hours fencing.

If Ginger wants the total time doing tai chi to be at least twice as long as she rides a unicycle, then

1 2

1 2

2or 2 0.

x xx x

³

- + £

The problem can be stated as follows.

Maximize 1 2 3236 295 354z x x x= + +

subject to: 1 2 3

3

1 2

102

2 0

x x xx

x x

+ + £

£

- + £

with 1 2 30, 0, 0.x x x³ ³ ³


1 2 3 1 2 3

1 1 1 1 0 0 0 10

0 0 1 0 1 0 0 21 2 0 0 0 1 0 0

236 295 354 0 0 0 1 0

x x x s s s zé ùê úê úê úê ú-ê úê ú- - -ê úë û


1 2 3 1 2 3

2 1 1

2 4 4

1 1 0 1 1 0 0 8R R R0 0 1 0 1 0 0 2

1 2 0 0 0 1 0 0236 295 0 0 354 0 1 708354R R R

x x x s s s zé ù-- + ê úê úê úê ú-ê úê ú- -+ ê úë û


1 2 3 1 2 3

3 1 1

3 4 4

R 2R R 3 0 0 2 2 1 0 160 0 1 0 1 0 0 21 2 0 0 0 1 0 0

295R 2R R 767 0 0 0 708 295 2 1416

x x x s s s zé ù- + - -ê úê úê úê ú-ê úê ú

+ -ê úë û


1 2 3 1 2 3

1 3 3

1 4 4

3 0 0 2 2 1 0 160 0 1 0 1 0 0 20 6 0 2 2 2 0 16R 3R R0 0 0 1534 590 118 6 16,520767R 3R R

x x x s s s zé ù- -ê úê úê úê ú-+ ê úê ú+ ê úë û


1 2 3 1 2 3162 2 11

1 1 3 3 3 33

81 1 113 3 3 3 3 36

767 295 59 826014 4 3 3 3 36

1 0 0 0R R0 0 1 0 1 0 0 2

0 1 0 0R R

0 0 0 1R R

x x x s s s zé ù- - ê úê úê úê úê ú- ê úê úê ú ê úë û

Ginger will burn a maximum of 132753

calories if she does 163 hours of tai chi, 8

3 hours riding a unicycle, and 2 hours fencing.

(b) Since fencing burns the most calories, she should do as much fencing as possible, which is 2 hours. This leaves 8 hours to divide between tai chi and the unicycle. The unicycle burns more calories, so she wants as much of unicycle as possible subject to the tai chi getting at least twice as much time as the unicycle. This requires devoting 1

3 of the

remaining 8 hours to the unicycle and 23 of

the 8 hours to tai chi. So the times are: 163 hours of tai chi, 8

3 hours of unicycle, and 2 hours of fencing.

Extended Application: Using Integer Programming in the Stock-Cutting Problem

1. (a) With Plan A you will need to buy 8 timbers: one cut will give you the two 4-ft lengths, two more cuts will give you two each of the 3-ft and 5-ft lengths, the two 3-ft pieces will come out of another full length, leaving 2 ft over, and all four 6-ft lengths will come out of 8-ft pieces. Your total waste amounts to 10 ft.

(b) There’s no advantage to Plan B; you still need 8 pieces of lumber: two cuts will give you four 4-ft lengths, two more cuts will give you two each of the 3-ft and 5-ft lengths, the two 5-ft lengths will come out of 8-ft pieces as will each of the two 6-ft lengths, leaving a total waste of 10 ft.

(c) If the original timbers were 9 ft in length, you could cut 6 timbers and cut the lengths for either Plan A or Plan Β with no waste.

2. Four patterns not in the minimizer’s list are, for example, 14|14|33|33|, 31|33|33|, 33|33|33|, and 14|17|31|33|.

3. The patterns are 31|33|36, 17|17|33|33, 14|17|33|36, and 14|14|36|36.

Extended Application 299


4. 808 2.3%35,600

»

5. A leftover piece less than 14 inches wide can’t be used for any standard width, but a leftover piece of 22 inches, for example, could be cut to make either the 14-inch or the 17-inch standard width. So it might be better to reserve the choice about how to cut this leftover until more orders come in.

6. The highest value is 34: choose weights 2, 2.5 and 4.5.

7. Minimize 1 2 3 4 5 6 7w y y y y y y y= + + + + + +

subject to:

1 2

3 4

5 6

6

1 2 3 4 5 6

2 4 7

2 100123

2 239121

2 2 4442 87

y yy y

y yy

y y y y y yy y y

+ ³

+ ³

+ ³

³

+ + + + + ³

+ + ³

with

1 2 3 4 5 6 70, 0, 0, 0, 0, 0, 0.y y y y y y y³ ³ ³ ³ ³ ³ ³

8. Minimum is 355.5 when 1 50,y = 2 0,y = 3 41,y = 4 82,y =

5 59,y = 6 121,y = and 7 2.5,y = or

1 50,y = 2 0,y = 3 38.5,y = 4 87,y =

5 59,y = 6 121,y = and 7 0.y = Minimum is 356 when

1 50,y = 2 0,y = 3 41,y = 4 82,y =

5 59,y = 6 121,y = and 7 3,y = or

1 50,y = 2 3,y = 3 38,y = 4 85,y = 5 59,y = 6 121,y = and 7 0.y =

linear programming: the simplex method ... 4 linear programming: the simplex method 4.1 slack...

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