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Thread: Find the extreme values of the function subject to the given constraint

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    Find the extreme values of the function subject to the given constraint

    $\displaystyle f(x,y)=x^{2}y$, $\displaystyle x^{2}+2y^{2}=6$

    Ok, so I'm trying to do this using Lagrange multipliers

    So, first I need $\displaystyle \nabla f=\lambda \nabla g$, right?

    Which is:

    $\displaystyle 2xyi+x^{2}j=\lambda (2xi+4yj)$

    So:

    $\displaystyle 2xy=\lambda 2x$ and $\displaystyle x^{2}=\lambda 4y$

    From here, I have no idea what to do. Any help?

    Edit:

    Ok, I THINK I figured it out. Please someone double check my work

    So,

    $\displaystyle y=\lambda$ and $\displaystyle x^{2}=4\lambda^{2}$

    We plug these into the constraint and get:

    $\displaystyle 4\lambda^{2}+2\lambda^{2}=6$

    $\displaystyle =>\lambda=\pm 1$

    Then we plug $\displaystyle \lambda$ back into $\displaystyle y=\lambda$ and $\displaystyle x=2\lambda$

    So we get, $\displaystyle y=\pm 1$ and $\displaystyle x=\pm 2$

    Then, plugging these numbers into $\displaystyle f(x,y)=x^{2}y$

    $\displaystyle (-2)^{2}(-1)=-4$ and $\displaystyle (2)^{2}(1)=4$

    Is this all correct?
    Last edited by downthesun01; Oct 20th 2010 at 10:20 PM.
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