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Thread: LaGrange Multiplier Problem

  1. #1
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    LaGrange Multiplier Problem

    I'm trying to find the minimum and maximum values of
    $\displaystyle f(x,y,z,) = (4x - 1/2y + 27/2 z)$

    on the surface of $\displaystyle g=x^4+y^4+z^4=1$

    I have already put $\displaystyle \bigtriangledown f = \lambda \bigtriangledown g $
    Getting the following 4 equations... Where do I go from here?
    $\displaystyle 4\lambda x^3=4$
    $\displaystyle 4\lambda y^3=-1/2$
    $\displaystyle 4\lambda z^3=27/2$
    $\displaystyle x^4+y^4+z^4=1$
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  2. #2
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    Quote Originally Posted by keysar7 View Post
    Getting the following 4 equations... Where do I go from here?
    $\displaystyle 4\lambda x^3=4$
    $\displaystyle 4\lambda y^3=-1/2$
    $\displaystyle 4\lambda z^3=27/2$
    $\displaystyle x^4+y^4+z^4=1$
    Solve the system, you have equations and 4 unknowns.
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  3. #3
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    I know I need to solve...

    There are 4 unknowns (counting λ) but variables x, y, z are raised to the power of 4.

    I don't know how to solve this system... Any help?
    Last edited by keysar7; May 26th 2010 at 05:20 PM. Reason: typo
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  4. #4
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    $\displaystyle 4\lambda x^3=4\implies x^3= \frac{4}{4\lambda }\implies x^3= \frac{1}{\lambda }\implies x= \frac{1}{\sqrt[3]{\lambda}}$

    Do this for the next 2 equations

    Then $\displaystyle x^4+y^4+z^4=1$ becomes $\displaystyle \left( \frac{1}{\sqrt[3]{\lambda}}\right)^4+y^4+z^4=1$

    Your work will replace $\displaystyle y$ and $\displaystyle z$ with a function of $\displaystyle \lambda$

    You will then have a solution for $\displaystyle \lambda$ and use this to find $\displaystyle x,y,z$
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  5. #5
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    Quote Originally Posted by keysar7 View Post
    I'm trying to find the minimum and maximum values of
    $\displaystyle f(x,y,z,) = (4x - 1/2y + 27/2 z)$

    on the surface of $\displaystyle g=x^4+y^4+z^4=1$

    I have already put $\displaystyle \bigtriangledown f = \lambda \bigtriangledown g $
    Getting the following 4 equations... Where do I go from here?
    $\displaystyle 4\lambda x^3=4$
    $\displaystyle 4\lambda y^3=-1/2$
    $\displaystyle 4\lambda z^3=27/2$
    $\displaystyle x^4+y^4+z^4=1$
    Since the value of $\displaystyle \lambda$ is not part of the solution, I often find it best to eliminate $\displaystyle \lambda$ first by dividing one equation by another.

    Dividing the first equation by the second gives $\displaystyle \frac{4\lambda x^3}{4\lambda y^3}= \frac{4}{-1/2}$ or $\displaystyle \frac{x^3}{y^3}= -8$ so $\displaystyle x= -2y$.

    Dividing the third equation by the second gives $\displaystyle \frac{4\lambda z^3}{4\lambda y^3}= \frac{27/2}{-1/2}= \frac{z^3}{y^3}= -27$ so $\displaystyle z= -3y$.

    Putting x= -2y and z= -3y into the last equation gives a single equation for y.
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