Lagrange Multipliers

**chella182** · August 5th 2009, 08:14 AM

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

Find the maximum value of

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

subject to the constraint

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

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

To begin with, I multiplied out the function to get

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

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

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

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

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

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

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

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

Can anyone maybe see where I've gone wrong? I can tell you what I got for $\lambda$ et cetera if needs be.

**Opalg** · August 5th 2009, 11:58 AM

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 $v = 1/(4\lambda)^{1/3}$ , $w=0$ and $x=y=6^{1/3}/(4\lambda)^{1/3}$ . Then the equation $v^4+w^4+x^4+y^4=1$ becomes $1/(4\lambda)^{4/3}(1 + 2\times6^{4/3}) = 1$ , from which $4\lambda = (1 + 2\times6^{4/3})^{3/4}$ . Then $v = (1 + 2\times6^{4/3})^{-1/4}$ , and x and y are the same expression multiplied by $6^{1/3}$ .

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

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