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Math Help - Lagrange multipliers

  1. #1
    Junior Member
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    Lagrange multipliers

    What points on the the curve x^2+2xy+2y^2= 100 are closest to the origin? I think that  F(x,y,z,\lambda)=x^2+y^2+z^2+\lambda(x^2+2xy+2y^2-100)
    Then F_{x}=2x+2\lambda x+2\lambda y
    F_{y}=2y+2\lambda x+4\lambda y
    F_{z}=0
    F_{\lambda}=x^2+2xy+2y^2-100
    If I set those equal to zero I get
    2x=-2\lambda x-2\lambda y
    2y=-2\lambda x-4\lambda y
    x^2+2xy+2y^2=100
    I'm having trouble solving that system. I don't know if I've got the partials wrong, or am just doing the algebra wrong. Any helpful thoughts?
    Thanks
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  2. #2
    MHF Contributor
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    Re: Lagrange multipliers

    partials are correct. Solve for x in the first equation.
    Note the 3rd equation is

    $$(x+y)^2=100 \Rightarrow (x+y)=\pm 10$$

    So plug what you have for x into these two eqs and solve for y.

    Then solve for lambda in the final eq.

    There are two minima and 2 maxima.
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