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Let $M$ and $m$ denote resp. the max and min values of the func. $f(x,y,z) = xyz $ over the region defined by the interior and boundary of the sphere $x^2 + y^2 + z^2 = 3$. What is the value of $M + m$?

I tried using the method of lagrange multiplier to find $M$ and $m$, but I am having trouble solving the four equations in the four unknowns that are derived from it. I am doubting if Lagrange multipliers is the appropriate method.

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up vote 6 down vote accepted

Lagrange multipliers work just fine here. But the Arithmetic Mean/Geometric Mean Inequality is a useful alternative that is calculus-free. Let $a=x^2$, $b=y^2$, and $c=z^2$. Then by AM/GM $$\frac{a+b+c}{3}\ge \sqrt[3]{abc}.$$ Conveniently, in this case the inequality will give you both the maximum and the minimum.

Remark: To use Lagrange multipliers, first note that the max and min are positive and negative respectively. So each occurs on the boundary. (For if not, scale until we reach the boundary.) The usual partial derivative calculation leads to the equations $yz+2\lambda x=0$, $xz+2\lambda y=0$, $xy+2\lambda z=0$, and $x^2+y^2+z^2=3$.

Exploit the symmetry by multiplying the first equation through by $x$, the second by $y$, the third by $z$. We conclude that $2\lambda x^2=2\lambda y^2=2\lambda z^2$, and now things become simple.

If $\lambda=0$, then two of $x$, $y$, $z$ must be $0$, and the third must be $\pm\sqrt{3}$ (but these are irrelevant for the max or min). If $\lambda\ne 0$, then $x^2=y^2=z^2=1$.

Another way to look at the problem, fairly natural but somewhat less attractive than AM/GM or Lagrange multipliers, is to use spherical coordinates. Note that the general point at distance $r$ from the origin has coordinates $\,x=r\sin\theta\cos\varphi$, $\,y=r\sin\theta\sin\varphi$, $\,z=r\cos\theta$. Calculate the product $xyz$. The max/min problem now decomposes into two one-variable problems.

All of our arguments have been computational. Note that a non-computational approach that gives the answer in seconds has been posted.

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Could you please explain bit more how AM-GM inequality give us both the maximum and minimum value for $f$ –  Learner Aug 2 '12 at 8:50
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@Learner: From AM-GM we get $x^2y^2z^2\le 1$, with equality when $x^2=y^2=z^2$. So $-1 \le xyz \le 1$. This gives both upper and lower bound. And each is reached, upper at $x=y=z=1$ (and $3$ others, like $(-1,-1,1)$) and lower at $x=y=z=-1$ and $3$ others. –  André Nicolas Aug 2 '12 at 11:12
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Hint: If $(x,y,z)$ is on or inside the sphere, then so is $(-x,y,z)$, and $f(x,y,z) = -f(-x,y,z)$.

Maybe you don't even need Lagrange multipliers. But if I were your prof, I'd insist on a clean argument. :)

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Here is my proof that the answer is zero. Suppose $(x,y,z)$ maximizes $f$. Now, $M$ must be strictly positive, because it must be greater than or equal to the value at $f(1,1,1) = 1 $. I claim that $ m = - M$. Certainly $-M$ is achieved, for example at the point $(-x,y,z)$. And, if $ f(a,b,c) = R < -M$ for some valid point $(a,b,c)$, it follows that $f(-a,b,c) = -R > M $, contradicting $M$ maximum. Hence, $M + m = M - M = 0$. –  Peter Aug 2 '12 at 1:50
    
The only slightly subtle point is that the minimum and maximum exist (because $f$ is a continuous function on a compact set). –  Robert Israel Aug 2 '12 at 3:08
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