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? 

You can cheat here, because the property of being a maximum-volume axis-parallel box is preserved by "stretching" transformations of the form $(x,y,z) \mapsto (ax,by,cz)$ (because such a transformation preserves the property of being axis-parallel, and multiplies all volumes by the constant $abc$).
So start with a sphere of radius 1: then the biggest box has corners
$$\left(\pm\sqrt \frac{1}{3},\pm\sqrt \frac{1}{3},\pm\sqrt \frac{1}{3}\right)$$
and a volume of $\left(2\sqrt \frac{1}{3}\right)^3 = \frac{8}{3}\sqrt \frac{1}{3}$ (see the comments for more about this). Now stretch it by a factor of $2$ along the $x$-axis, $3$ along the $y$-axis, and $4$ along the $z$-axis. The corners go to
$$\left(\pm2\sqrt \frac{1}{3},\pm3\sqrt \frac{1}{3},\pm4\sqrt \frac{1}{3}\right)$$
and the volume is $64\sqrt \frac{1}{3}$.
