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Thread: lagrangian multiplier method help!

  1. #1
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    Dec 2011
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    Lagrange multiplier method help!

    Hi,

    I am trying to work on this problem, I have already sett up the Lagrange equation, i need to solve for x and y

    [tex]L(x,y)=\frac{1}{2}(\alpha(y^2)+x^2)+ \lambda[(x)+\theta (y)-\theta(z)-w][/tex]

    [tex]\frac{\partial(L)}{\partial(x)}=x+\lambda=0 [/tex]
    [tex]\frac{\partial(L)}{\partial(x)}=\alpha(y)+(\theta) (\lambda)=0 [/tex]
    [tex]\frac{\partial(L)}{\partial(\lambda)}=(x)+\theta (y)-\theta(z)-w=0 [/tex]

    [tex]\lambda=-x [/tex]
    [tex]\alpha(y)=-(\theta)(\lambda)[/tex]
    [tex]\alpha(y)=(\theta)(x)[/tex]
    [tex](y)=\frac{\theta}{\alpha}(x)[/tex]

    [tex](x)+\theta (y)=\theta(z)+w[/tex]
    [tex](x)+\theta (\frac{\theta}{\alpha}(x))=\theta(z)+w[/tex]
    >>>
    [tex](x)=\frac{\theta(z)+w}{[1+\theta (\frac{\theta}{\alpha})]}[/tex]

    [tex](y)=\frac{\theta}{\alpha}(\frac{\theta(z)+w}{[1+\theta (\frac{\theta}{\alpha})]})[/tex]

    ***
    then I have to sub it in back into the original equation

    [tex]f(x,y)=\frac{1}{2}(\alpha(y^2)+x^2)[/tex]
    [tex]f(x,y)=\frac{1}{2}(\alpha(\frac{\theta^2}{\alpha^2 }[/tex][tex](\frac{\theta(z)+w}{[1+\theta (\frac{\theta}{\alpha})]})^2)+(\frac{\theta(z)+w}{[1+\theta (\frac{\theta}{\alpha})]})^2)[/tex]

    [tex]f(x,y)=\frac{1}{2}[((\frac{\theta(z)+w}{[1+\theta (\frac{\theta}{\alpha})]})^2)((\frac{\theta}{\alpha})^2+1)][/tex]

    but how can I simply this further, have I made any mistakes?

    Any help will be greatly appreciated!
    Last edited by WlND; 04-12-2012 at 05:17 PM.

  2. #2
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    ..
    Last edited by WlND; 04-12-2012 at 03:47 PM.

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