Find ratio between volume of optimal cuboid bound in ellipse, and volume of the ellipse. I am trying to find the 'general' ratio between the cuboid of maximum volume bound in an ellipse, and the volume of that same ellipse.
After doing some partial derivatives and optimisation, it seems possible to find the ratio but looks like it would take a LONG time.
I have:
$$V_{ellipse}=4\pi*abc/3 : \frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2}=1$$
$$V_{cuboid}=8xyz=8xyc\sqrt{1-x^2/a^2-y^2/b^2}$$ (Attained by rearranging the ellipse formula in terms of z, and then substituting.)
After several partial derivatives:
Optimally; 
$$x=\sqrt{\frac{a^2-\frac{a^2y^2}{b^2}}{2}}$$
$$y=\sqrt\frac{b^2-\frac{b^2x^2}{a^2}}{2}$$
Then you'd proceed to sub this into the cuboid equation, and then finally divide by the volume of the ellipse. This clearly would take a long time, the equation would be a huge mess, and it would be a pain to actually calculate.
Surely there is an easier way? Maybe using triple integrals or something like that?

You wish to find the maximum of:
$$8 \cdot \frac{3}{4\pi} \cdot \frac{xyz}{abc}$$
Subject to the constraint:
$$(\frac{x}{a})^2+(\frac{y}{b})^2+(\frac{z}{c})^2 \leq 1$$

Let $X=\frac{x}{a}$, $Y=\frac{y}{b}$ and $Z=\frac{z}{c}$. Then we wish to find the maximum of:
$$8 \cdot \frac{3}{4 \pi} \cdot XYZ$$
Subject to:
$$X^2+Y^2+Z^2 \leq 1$$
Which shows the problem reduces down to what was suggested in the comments. 

Let $X=r \sin \phi \cos \theta$, $Y=r \sin \phi \sin  \theta$, and $Z=r \cos \phi$ with $r \leq 1$. Then,
$$XYZ=r^3 \sin^2 \phi \cos \phi \sin \theta \cos \theta$$
$$=\frac{1}{2} r^3 (\cos^3 \phi-\cos \phi) \sin (2\theta)$$

To maximize $XYZ$ it’s enough to maximize $\cos \phi -\cos^3 \phi$ as well as $\cos (2\theta)$ and $r^2$. 

The maximum of $\cos (2\theta)$ is $1$, and the maximum of $r^2$ is $1$ since $r \leq 1$. To maximize $\cos \phi-\cos^3 \phi$, it’s enough to maximize $f(x)=x-x^3$ where $-1 \leq x \leq 1$. The maximum is $\frac{2}{3\sqrt{3}}$ by methods of single variable calculus. 

This shows a maximum ratio of $8 \cdot \frac{3}{4 \pi} \cdot \frac{1}{3 \sqrt{3}}$.
