medaka rb 2016/02/21

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  • 7/24/2019 Medaka rb 2016/02/21

    1/3

    Solve:

    maximize f =xy

    subject to x+y= 2

    Soln : KKT condition f =xyh=x+y 2

    l

    f +h= 0 y

    x

    +

    11

    = 0

    y+ = 0x+= 0

    x+y= 2l

    x= 1y= 1= 1

    l

  • 7/24/2019 Medaka rb 2016/02/21

    2/3

    2(xy) =

    2fx2

    2fxy

    fyx

    fy2

    =

    0 11 0

    (x, y

    0 11 0

    xy

    = 2xy 0

    0 11 0

    0However,

    p=log(x)q=log(y)

    maximize f =exp(p+ q)

    subject to x+y= 2

    f =exp(p+ q)

  • 7/24/2019 Medaka rb 2016/02/21

    3/3

    2(xy) =

    2fx2

    2fxy

    fyx

    fy2

    =

    exp(p+ q) exp(p+ q)exp(p+ q) exp(p+ q)

    exp(p+ q) =exp(p) +exp(q) =x+y= 2

    2(xy) =

    2 22 2

    (x y

    2 22 2

    x

    y

    = 2(x+y)2 0