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Math Help - Lagrange multiplier

  1. #1
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    Lagrange multiplier

    Decide, using Lagrange multiplier, the extremes for
    f(x,y)=z^3+yz
    where
    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  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; September 14th 2012 at 08:13 AM.
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  5. #5
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    Re: Lagrange multiplier

    Attempt:
    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 \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:

    f(y,z)=z^3+yz

    subject to the constraint:

    g(y,z)=y+3z+24=0

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

    z=\lambda

    3z^2+y=3\lambda

    These two equations imply:

    y=3z-3z^2

    Substituting for y into the constraint yields:

    z^2-2z-8=0

    (z-4)(z+2)=0

    Hence, the extrema occur at:

    (y,z)=(-18,-2),\,(-36,4)
    Thanks from jacob93
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  7. #7
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    Re: Lagrange multiplier

    thanks!
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