# Homework Help: Lagrange Multiplier Method

1. Apr 6, 2012

### hawaiifiver

1. The problem statement, all variables and given/known data

Find the stationary value of

$$f(u,v,w) = \left( \frac{c}{u} \right)^m + \left( \frac{d}{v} \right)^m + \left( \frac{e}{w} \right)^m$$

Constraint: $$u^2 + v^2 + w^2 = t^2$$

Note: $$u, v, w > 0$$. $$c,d, e, t > 0$$. $$m > 0$$ and is a positive integer.

2. Relevant equations

I have found the auxiliary function:

$$F = \left( \frac{c}{u} \right)^m + \left( \frac{d}{v} \right)^m + \left( \frac{e}{w} \right)^m - \lambda ( u^2 + v^2 + w^2 - t^2)$$

and

$$F_u = \frac{- m c^m}{ u^{m+1}} - 2\lambda u = 0$$

$$F_v = \frac{- m d^m}{ v^{m+1}} - 2\lambda v = 0$$

$$F_w = \frac{- m e^m}{ w^{m+1}} - 2\lambda w = 0$$

3. The attempt at a solution

Its after this I am having difficult. I don't understand what to do. I assume I have to rearrange the above equations to get $$u^2, v^2, w^2$$ and substitute into the constraint, but I can't see how to do that because I get expressions for $$u^{m + 2} , v^{m+2} , w^{m+2}$$ . Help please.

3. The attempt at a solution
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

2. Apr 6, 2012

### HallsofIvy

Okay, that's good. What I would do is switch all of those "$\lambda$" terms to the right:
$\frac{-m c^m}{u^{m+1}}= 2\lambda u$
$\frac{-m d^m}{v^{m+1}}= 2\lambda v$
$\frac{-m d^m}{w^{m+1}}= 2\lambda w$

(In fact, to find extrema of f with constraint g= constant, I tend to think of the Lagrange multiplier condition as $\nabla f= \lambda \nabla g$ rather than using the "auxiliary function" $f+ \lambda g$. It gives the same result, of course.)

Now, since a particular value of $\lambda$ is not part of the solution start by eliminating $\lambda$ by dividing one equation by another. Dividing the first equation by the second, for example, gives
$$\left(\frac{c}{d}\right)^m\left(\frac{v}{u}\right)^{m+1}= \frac{u}{v}$$
which gives
$$\left(\frac{v}{u}\right)^{m+2}= \left(\frac{d}{c}\right)^m$$
and similarly for the other equations.

3. Apr 6, 2012

### Ray Vickson

You have expressions for $u^{m + 2} , v^{m+2} , w^{m+2}$ in terms of λ. You can take (m+2) roots, to get $u, v, w$. You can then substitute these expressions into the constraint, to get a single, simple equation for λ. Actually, it is easier to re-write everything in terms of μ = -λ; that will eliminate a lot of minus signs in the equations and solutions.

RGV

4. Apr 6, 2012

### hawaiifiver

Thank you both. I'm understanding this problem now.