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Math Help - Max/Min using Lagrange

  1. #1
    Junior Member
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    Max/Min using Lagrange

    Can anyone help with this problem please? I understand the concept of the langrange multiplier but this ones stumping me.

    F(x,y)=2x+y subject to the ellipse g(x,y)=X^2+25y^2=1

    I have f(x,y)=2i-1j. g(x,y)=2xi+50yj

    When I apply the lamb though I get 2=lambda 2x. Which means lambda=1/x which doesn't make sense.

    Can anyone tell me where I am wrong with is one?
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  2. #2
    Senior Member
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    Mar 2010
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    We have

    <br />
(2,1)=\lambda (2x,50y)<br />


    <br />
2=\lambda \; 2x \; \; (1)<br />

    <br />
1=\lambda \; 50y \; \; (2)<br />

    From (1) get \lambda and insert to (2)

    <br />
x=50y<br />

    and we can insert it to ellipse equation.
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  3. #3
    MHF Contributor

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    Because, in Lagrange multilie problems, the specific value of \lambda is not relevant, I find it often helpful to eliminate \lambda by dividing equations. From the two equations 2= 2\lambda x and 1= 50\lambda y we get \frac{2}{1}= \frac{2\lambda x}{50\lambda y}
    so that
    50y= x
    as zzzoak says.
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