Find the volume of the largest box with sides parallel to the $xy$, $xz$, and $yz$ planes that can fit inside the ellipsoid $(x/a)^2 + (y/b)^2 + (z/c)^2 = 1$.
My answer: We want to maximize $f(x,y,z) = xyz$ subject to $(x/a)^2 + (y/b)^2 + (z/c)^2 \leq 1$. So $\nabla f $ parallel to $\nabla g$, i.e.
$$(yz, xz, xy) = \lambda \left( \frac{x}{a}, \frac{y}{b}, \frac{z}{c} \right)\\ \Rightarrow \frac{ayz}{x}=\frac{bxz}{y}= \frac{cxy}{z}\\ \Rightarrow ay^2 = bx^2 ; bz^2 = cy^2 ; az^2 = cx^2 \\ \Rightarrow (x,y,z) \text{ parallel to } \left( 1, \sqrt{\frac{a}{b}}, \sqrt{\frac{a}{c}}\right)$$
so we need to find $t$ such that $\left( t, t\sqrt{\frac{a}{b}}, t\sqrt{\frac{a}{c}}\right)$ is on the ellipsoid, i.e.
$$\left( \frac{t}{a\sqrt{a}}\right)^2 + \left( \frac{t}{b\sqrt{b}}\right)^2 + \left( \frac{t}{c\sqrt{c}}\right)^2 = 1.$$
So the volume is $$8t^3 \frac{1}{\sqrt{abc}}\\ = 8\frac{1}{\left(\sqrt{ \frac{1}{a^3} + \frac{1}{b^3} + \frac{1}{c^3} }\right)^3}\cdot \frac{1}{\sqrt{abc}}$$
But the answer which the T.A. has indicated is $\frac{8abc}{3\sqrt{3}}$ (of course there could be a mistake). Where is my error?
Best Answer
It is not difficult to verify that the largest box in the unit sphere is the cube with sides $2/\sqrt3$, whence for such a sphere you get the volume $8/3\sqrt3$. Now notice that $f:\mathbb R^3\to\mathbb R^3$, $(x,y,z)\mapsto(ax,by,cz)$ maps unit sphere to your ellipsoid, and $V(f(X))=abc V(X)$ where $V$ is the volume and $X$ is any set. Moreover, $f$ brings your box to a box again. Therefore the maximum is necessarily attained for $f(X)$ where $X$ is the cube. And you get $V(f(X))=8abc/3\sqrt3$.
Your mistake is that $\partial_x (x/a)^2 = 2x/a^2$ and not $2x/a$.