Maximize volume of box in ellipsoid I need to find the dimensions of the box with maximum volume (with faces parallel to the coordinate planes) that can be inscribed in ellipsoid 
$$\frac{x^2}{4} + \frac{y^2}{9} + \frac{z^2}{16} = 1$$
A hint given was: 

If vertex of box in first octants is (x,y,z) then volume is 8xyz. 

So I first find the partial derivatives
$V_x = 8yz$
$V_y = 8xz$
$V_z = 8xy$
Now the partial derivatives for constraint: 
$g_x = \lambda\frac{x}{2}$
$g_y = \lambda\frac{2y}{9}$
$g_z = \lambda\frac{z}{8}$
Then $f_\alpha = \lambda g_\alpha$ so: 
$8yz = \lambda\frac{x}{2} \rightarrow \lambda = \frac{16yz}{x}$
$8xz = \lambda\frac{2y}{9} \rightarrow \lambda = \frac{36xz}{y}$
$8xy = \lambda\frac{z}{8} \rightarrow \lambda = \frac{64xy}{z}$
$\frac{16yz}{x} = \frac{36xz}{y} \rightarrow y = \frac{3x}{2}$
$\frac{36xz}{y} = \frac{64xy}{z} \rightarrow y = \frac{3z}{4}$
$\frac{3x}{2} = \frac{3z}{4} \rightarrow x = \frac{z}{2}$
Now how do I continue? I seems to be missing something? 

Very recently, you asked for advice on a proof of the AM/GM inequality in three variables. This inequality says that for $u$, $v$, and $w$ all $\ge 0$, we have
$$\frac{u+v+w}{3}\ge \sqrt[3]{uvw},$$
with equality iff $u=v=w$. 
The result you want is a direct consequence of three variable AM/GM.
Let $u=\frac{x^2}{4}$, $v=\frac{y^2}{9}$, and $w=\frac{z^2}{16}$. We are told that $u+v+w=1$. It follows from AM/GM that
$$\frac{1}{3}\ge \sqrt[3]{uvw}=\sqrt[3]{\frac{x^2y^2z^2}{4\cdot 9\cdot 16}}\tag{$\ast$}$$
with equality if $\frac{x^2}{4}=\frac{y^2}{9}=\frac{z^2}{16}$.
From the inequality $(\ast)$, you can reach a conclusion about the maximum value of $xyz$ for $x$, $y$, $z$ non-negative. Eight times this maximum value gives the maximum volume, as per the hint. 
