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?