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Math Help - Easy Lagrange optimization problem help

  1. #1
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    Easy Lagrange optimization problem help

    Find the max/min for 3xy subject to x^2 + y^2 = 8

    L = 3xy - x^2\lambda - y^2\lambda + 8\lambda

    L'(x) = 3y - 2x\lambda = 0
    L'(y) = 3x -2y\lambda = 0
    L'(\lambda) = -x^2 - y^2 + 8 = 0

    3y=2x\lambda
    y=(2x\lambda)/3

    3x=3y\lambda
    3x=3((2x\lambda)/3)\lambda
    x=4\lambda

    Am I doing anything wrong so far? I can't seem to get the right answer. All help appreciated!
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  2. #2
    MHF Contributor Also sprach Zarathustra's Avatar
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    Re: Easy Lagrange optimization problem help

    I think that we can solve this problem without Lagrange optimization tools:

    We have a circle center (0,0) with radius of sqrt(8), our mission is to find point, (x,y), on that circle, with max{x*y}, try to show that such point have the following property -- x=y.

    Therefor, if x=y, we'll get,

    2x^2=8

    x=2
    y=2

    and 3xy=12.
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  3. #3
    MHF Contributor alexmahone's Avatar
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    Re: Easy Lagrange optimization problem help

    \lambda=\frac{3y}{2x}=\frac{3x}{2y}

    x^2=y^2

    Substituting this in L'(\lambda)=-x^2-y^2+8=0, we get x=\pm 2 and y=\pm 2. Clearly, the maximum is 12 and minimum is -12.
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  4. #4
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    Re: Easy Lagrange optimization problem help

    Quote Originally Posted by alexmahone View Post
    \lambda=\frac{3y}{2x}=\frac{3x}{2y}

    x^2=y^2

    Substituting this in L'(\lambda)=-x^2-y^2+8=0, we get x=\pm 2 and y=\pm 2. Clearly, the maximum is 12 and minimum is -12.
    Ah thanks a lot!
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