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Math Help - Vector calc minimum problem

  1. #1
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    Vector calc minimum problem

    I am given that the summation from i = 1 to 3 = 1
    I am to find x sub 1, xsub2, xsub3 so that the summation from i = 1 to 3 of x sub i ^2 is at a minimum. Then I am suppose to come up for a general formula based on my findings. If anyone is willing to give me a push in the right direction it would be much appreciated! Frostking
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  2. #2
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    Quote Originally Posted by Frostking View Post
    I am given that the summation from i = 1 to 3 = 1
    I am to find x sub 1, xsub2, xsub3 so that the summation from i = 1 to 3 of x sub i ^2 is at a minimum. Then I am suppose to come up for a general formula based on my findings. If anyone is willing to give me a push in the right direction it would be much appreciated! Frostking

    So what we know is that we are trying to minimize

    m=x_1^2+x_2^2+x_3^2 subject to

    x_1+x_2+x_3=1 \iff F(x_1,x_2,x_3)=x_1+x_2+x_3-1

    with these we can use Lagrange multipliers

    \nabla{m}=\nabla \lambda F

    (2x_1)\vec i +(2x_2) \vec j +(2x_3)\vec k =(\lambda)\vec i+(\lambda)\vec j +(\lambda) \vec k

    so x_1=x_2=x_3=\frac{\lambda}{2}

    subbing this into the constraint equation
    \frac{\lambda}{2}+\frac{\lambda}{2}+\frac{\lambda}  {2}=1 \iff \lambda =\frac{2}{3}

    This then implies that x_1=x_2=x_3=\frac{1}{3}
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  3. #3
    MHF Contributor matheagle's Avatar
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    Correct, but may I add two comments.
    First of all, F(x_1,x_2,x_3) doesn't have to equal zero, so you can drop the -1.
    And you don't need to solve for \lambda.
    Once you know that the x's are equal, you can substitute into
    your constraint to obtain what you really want, the x_i's.
    Last edited by matheagle; February 23rd 2009 at 08:16 PM.
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