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Maximize $f(x,y,z) = x^4 + y^4 + z^4$ subject to $g(x,y,z) = x^2 + y^2 + z^2 = 1$

it is required that $$\partial_xf = \lambda \partial_xg$$

$$4x^3 = (2x) \lambda \implies x^2 = y^2 = z^2 = \frac{\lambda}{2}$$

$$x^2 + y^2 + z^2 = \frac{ 3 \lambda}{2} = 1 \implies \lambda = \frac{2}{3}$$

$$\therefore x^2 = y^2 = z^2 = \frac{1}{3}$$

and my maximum value is: $\dfrac{3}{81}$

My question is why do I only have one extrema? Also, How do I show this is a max or min.

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Many computations of partials missing. After correct equations are obtained, careful about solving. For example, $4x^3=2\lambda x$ has $x=0$ as a solution. –  André Nicolas Jan 16 '13 at 23:14
    
What you want is to find $\lambda$ such that $\nabla f = \lambda \nabla g.$ More generally, you can show that if $g: \mathbb{R}^n \to \mathbb{R}^{m},$ then a local extremum $a$ of $f: \mathbb{R}^{n} \to \mathbb{R}$ subject to $g = 0$ must satisfy $Df(a) = \lambda_1 Dg_1 (a) + \ldots + \lambda_m Dg_m (a).$ Since $f$ is a functional, $Df(a) = (\nabla f(a))^{T},$ the transpose of the gradient (which is more commonly found in textbooks). –  lyj Jan 17 '13 at 0:53
    
In the 3 variable case you can avoid using $\lambda$ at all, since at the critical point(s), the cross-product of the two gradients willo be the zero vector $(0,0,0)$ This usually works quite well I've found. –  coffeemath Jan 17 '13 at 2:04
    
The statement of the constraint has now been changed from what it was originally. So I deleted my answer. –  coffeemath Jan 17 '13 at 17:39
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1 Answer 1

up vote 2 down vote accepted

Since $f$ is continuous and the constraint, a sphere, is closed and bounded, you have to have both a maximum and a minimum. So you're right to ask what happened, and @AndréNicolas pointed out in a comment some dropped solutions that lead to the maximum.

An alternate approach is to notice that the problem can rewritten in terms of $u=x^2$, $v=y^2$, $w=z^2$, and is equivalent to maximizing $u^2+v^2+w^2$, which is the square of the distance from the origin, over the equilateral triangle $u+v+w=1$, $0\le u,v,w \le 1$. The extrema occur at the points of the triangle farthest and closest to the origin, which are the vertices and the center respectively.

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