Max volume of rectangular box in ellipsoid Can anyone give a solution to the problem below ? 

What is the maximum volume of rectangular box that can be placed in the $$36x^2 + 4y^2 + 9z^2 = 36$$ ellipsoid? The sides of the ellipsoid are parallel to the coordinate axes.


I write down $$ V = f(x,y,z) = x*y*z \\ f_x=yz=\lambda*72x \\ f_y=xz=\lambda*8y \\ f_z=xy=\lambda*18z \\ x=\frac{1}{\sqrt{3}} \\ y=\sqrt{3} \\ z=\frac{2}{\sqrt{3}} \\ V=\frac{2}{\sqrt{3}} \\$$
But I am not sure whether my solution is correct.
Thanks for your answers.
 A: Hint. When you write the equation as
$$
x^2 + \frac{1}{9}y^2 + \frac{1}{4}z^2 = 1
$$
you see that the ellipsoid is built by stretching the unit sphere by factors $1$, $3$ and $2$ along the axes. That stretching changes all volumes by a factor of $6$. So just find the volume of the cube inscribed in the unit sphere (which will have the largest volume among such parallelepipeds) and multiply its volume by $6$.
Viewing ellipsoids this way is often a good start to questions like this.
A: Hint. The ellipsoid is symmetric with respect to the coordinate planes. So we solve the problem only in the first octant. Let the dimensions of the part of this box in this octant be $x,y,z,$ so that for $0\le x\le 1,\,0\le y\le 3$ and $0\le z\le 2,$ we want to maximize the function $$xyz$$ where the variables are bound by the relationship $$x^2+y^2/9+z^2/4=1.$$ In view of the form of the relationship, you may maximize $x^2y^2z^2$ instead, which will then imply maximum volume. Substitute for $x^2$ in the expression for the squared volume. This gives you a polynomial function of only two variables $y,z,$ which you can differentiate to find the possible extreme points.
