# Homework Help: Lagrangian multiplier question

1. Jul 17, 2011

### autre

So I have to find the min and max values of f(x,y,z) = x^4 + y^4 + z^ 4 given the constraint x^2 + y^2 + z^2 = 1. I've found the points (+-1/sqrt(3),1/sqrt(3) ,1/sqrt(3)), (+-1/sqrt(3),-1/sqrt(3) ,1/sqrt(3)) ... etc all of which have the f-value of 1/3 when x =/= 0 & y =/= 0 & z =/= 0 (this will probably turn out to be the min?)

My issue is the case when one value is zero and the other two are nonzero. I have 4x^3 = 2Lx (I'm using L for lambda here), then 2x^3 - Lx = 0, x(2x^2 - L) = 0....L = +-1/sqrt(2). From there, I have 2y^2 = L = 1/sqrt(2), y^2 = 1/2sqrt(2), and I'm not sure where to go from there for this case.

2. Jul 17, 2011

### Ray Vickson

I don't know how you found those points; they do not follow from the Lagrange conditions. For Lagrangian L = x^4 + y^4 + z^4 + u*(x^2 + y^2 + z^2 -1) [using u as the Lagrange multiplier and L as the Lagrangian] the conditions dL/dx = dL/dy = dL/dz = 0 give 2x - 2ux = 0, 2y - 2uy = 0, 2z - 2uz = 0. Thus, when x,y,z =/= 0 we have u = -1, and we have NO other conditions except g = 0. Although the Lagrange multiplier rule does hold at a max or min in this case (because the constraint qualification always holds), it gives no information. Instead, you can look at x1 = x^2, x2 = y^2, x3 = z^2 to get the problem max/min x1^2 + x2^2 + x3^2, subject to x1 + x2 + x3 = 1 and x1, x2, x3 >= 0. Using the Karush-Kuhn-Tucker conditions we find that x1 = x2 = x3 = 1/2 is the min, and that there are three maxima: (x1 = 0, x2 = x3 = 1/2) and permutations of this. Of course, now you can take square roots to get x, y and z.

RGV

3. Jul 17, 2011

### autre

Isn't <4x^3, 4y^3, 4z^3> = u<2x, 2y, 2z>?

4. Jul 18, 2011

### Ray Vickson

Of course. Sorry: I really screwed that one up (I mistakenly wrote the Lagrangian for the problem of max/min x^2 + y^2 + z^2, subject to x^2 + y^2 + z^2 = 1, and of course I got no solution!). For Lagrangian L = x^4 + y^4 + z^4 + u*(x^2 + y^2 + z^2 - 1) the actual equations are 4x^3 + 2ux = 0, etc, so the solution is either: Case(1) x = 1, y = z = 0, u = -2 (and permutations thereof, or a sign change on x); Case(2) x = y = 1/sqrt(2), z = 0, u = -1 (and permutations, or sign changes on each x,y,z variable separately); Case(3) x = y = z = 1/sqrt(3) , u = -2/3 (or sign changes on each x,y,z variable separately). We have f(Case1) > f(Case2) > f(case(3). Second order tests show that Case(1) is a strict local max, Case(3) is a strict local min and Case(2) is neither. (Here, we look at the projection of the Hessian of the *Lagrangian* on the tangent subspace of the constraint and test for positive or negative definiteness therein.)

RGV