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Thread: Lagrange multiplier

  1. #1
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    Lagrange multiplier

    Decide, using Lagrange multiplier, the extremes for
    $\displaystyle f(x,y)=z^3+yz$
    where
    $\displaystyle y+3z+24=0$

    I've tried, but I just couldn't solve the system of equations that arise. Maybe it's z^3 that confuses me.
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  2. #2
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    Re: Lagrange multiplier

    Do you mean $\displaystyle f(y,z) $ ?
    Also, can you show your working so far, maybe we can help you spot your errors.
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  3. #3
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    Re: Lagrange multiplier

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  4. #4
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    Re: Lagrange multiplier

    how do you do it?
    Last edited by jacob93; Sep 14th 2012 at 07:13 AM.
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    Re: Lagrange multiplier

    Attempt:
    $\displaystyle 3\cdot \frac{\partial L}{\partial y}-\frac{\partial L}{\partial z}=0\; \Rightarrow\; 3z\left( 1-z \right)=y\; \Rightarrow\; \left( 1,0 \right)\; and\; \left( 0,0 \right).$

    The book says $\displaystyle \left( 4,-36 \right)\; and \; \left( -2,-18 \right)$
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  6. #6
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    Re: Lagrange multiplier

    We are asked to used Lagrange multipliers to find the extrema for:

    $\displaystyle f(y,z)=z^3+yz$

    subject to the constraint:

    $\displaystyle g(y,z)=y+3z+24=0$

    By the theorem of Lagrange, we need to solve the system:

    $\displaystyle z=\lambda$

    $\displaystyle 3z^2+y=3\lambda$

    These two equations imply:

    $\displaystyle y=3z-3z^2$

    Substituting for y into the constraint yields:

    $\displaystyle z^2-2z-8=0$

    $\displaystyle (z-4)(z+2)=0$

    Hence, the extrema occur at:

    $\displaystyle (y,z)=(-18,-2),\,(-36,4)$
    Thanks from jacob93
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    Re: Lagrange multiplier

    thanks!
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