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Thread: Lagrange Multipliers

  1. #1
    Senior Member chella182's Avatar
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    Question Lagrange Multipliers

    (Don't worry, this isn't the same question as I posted the other day)

    Find the maximum value of

    $\displaystyle f(v,w,x,y)=(1,0,6,6)\left( \begin{array}{c} v \\ w \\ x \\ y \end{array} \right)$

    subject to the constraint

    $\displaystyle v^4+w^4+x^4+y^4=1$

    [NB: your answer should be given as a surd, not a decimal.]

    To begin with, I multiplied out the function to get

    $\displaystyle f(v,w,x,y)=v+6x+6y$

    Is that right? If not, can you tell me why? If it is, I then got the following partials...

    $\displaystyle \frac{\partial L}{\partial v}=1-4\lambda v^3$

    $\displaystyle \frac{\partial L}{\partial w}=-4\lambda w^3$

    $\displaystyle \frac{\partial L}{\partial x}=6-4\lambda x^3$

    $\displaystyle \frac{\partial L}{\partial y}=6-4\lambda y^3$

    $\displaystyle \frac{\partial L}{\partial\lambda}=-(v^4+w^4+x^4+y^4-1)$

    ...but then rearranging for $\displaystyle v$, $\displaystyle w$, $\displaystyle x$, $\displaystyle y$ and put them into $\displaystyle \frac{\partial L}{\partial\lambda}=0$ I get a really awful expression for $\displaystyle \lambda$ and consequently awful expressions for $\displaystyle v$, $\displaystyle x$ and $\displaystyle y$ (not $\displaystyle w$ 'cause I get that equal to $\displaystyle 0$)

    Can anyone maybe see where I've gone wrong? I can tell you what I got for $\displaystyle \lambda$ et cetera if needs be.
    Last edited by chella182; Aug 5th 2009 at 08:15 AM. Reason: missed something out
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  2. #2
    MHF Contributor
    Opalg's Avatar
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    If the going gets rough when doing a math exercise, it's always worth checking whether you copied the question correctly. I'm wondering whether those fourth powers should really be squares. That would make things a lot less unpleasant.

    Assuming that they really are fourth powers, you just have to roll up your sleeves and persevere with the calculations. Your partial derivative equations tell you that $\displaystyle v = 1/(4\lambda)^{1/3}$, $\displaystyle w=0$ and $\displaystyle x=y=6^{1/3}/(4\lambda)^{1/3}$. Then the equation $\displaystyle v^4+w^4+x^4+y^4=1$ becomes $\displaystyle 1/(4\lambda)^{4/3}(1 + 2\times6^{4/3}) = 1$, from which $\displaystyle 4\lambda = (1 + 2\times6^{4/3})^{3/4}$. Then $\displaystyle v = (1 + 2\times6^{4/3})^{-1/4}$, and x and y are the same expression multiplied by $\displaystyle 6^{1/3}$.

    So it's all a bit messy, but not impossibly hard.
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