Find ratio between volume of optimal cuboid bound in ellipse, and volume of the ellipse.

Question

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?

Answer 1

The following assumes the more restrictive definition of the cuboid as a rectangular parallelepiped.

These would need to be established, first, perhaps more formally:

• scaling transforms cuboids into cuboids and preserves volume ratios of arbitrary bodies, so the optimal ratio is the same for an ellipsoid as is for a sphere (which is what this comment hinted at);

• for a sphere, the maximal (volume-wise) cuboid must have its vertices on the sphere, otherwise it could be "scaled" along each diagonal direction until the vertices "hit" the surface of the sphere;

• by symmetry, it follows that the maximal cuboid in the ellipsoid case can be assumed WLOG to have its vertices on the ellipsoid, and its faces parallel to the ellipsoid axes.

Let $\,(x,y,z)\,$ be the vertex of the cuboid lying on the ellipsoid in the first octant, then:

$$ R = \frac{V_{cuboid}}{V_{ellipsoid}} = \frac{8\,xyz}{\;\;4 \pi \dfrac{abc}{3}\;\;} = \frac{6}{\pi}\cdot \frac{xyz}{abc} $$

By the RMS-GM inequality:

$$ \sqrt[3]{\frac{x}{a}\cdot\frac{y}{b}\cdot \frac{z}{c}} \le \sqrt{\frac{\left(\dfrac{x}{a}\right)^2+\left(\dfrac{y}{b}\right)^2+\left(\dfrac{z}{c}\right)^2}{3}} = \frac{1}{\sqrt{3}} \;\;\iff\;\; \frac{xyz}{abc} \le \frac{1}{3\sqrt{3}} $$

It follows that $\,R \le \dfrac{6}{\pi}\cdot\dfrac{1}{3 \sqrt{3}}=\dfrac{2}{\pi \sqrt{3}}\,$, with equality attained for $\,\dfrac{x}{a}=\dfrac{y}{b}=\dfrac{z}{c}=\dfrac{1}{\sqrt{3}}\,$.

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P.S.   About this part of OP's post...

I have: $\quad V_{cuboid}=xyz$

There are two problems with this:

• it assumes a priori that the optimal cuboid has its faces parallel to the axes of the ellipsoid (otherwise the volume is not $\,8xyz\,$), but that was not given as a premise, nor is necessarily obvious upfront;

• it's missing a factor of $\,8\,$ even in the case of a rectangular cuboid, since $\,xyz\,$ would be the volume inside the first octant, only.

Answer 2

You wish to find the maximum of:

$$8 \cdot \frac{3}{4\pi} \cdot \frac{xyz}{abc}$$

Subject to the constraint:

$$(\frac{x}{a})^2+(\frac{y}{b})^2+(\frac{z}{c})^2 \leq 1$$

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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.

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Let $X=r \sin \phi \cos \theta$, $Y=r \sin \phi \sin \theta$, and $Z=r \cos \phi$ with $r \leq 1$. Then,

$$XYZ=r^3 \sin^2 \phi \cos \phi \sin \theta \cos \theta$$

$$=\frac{1}{2} r^3 (\cos^3 \phi-\cos \phi) \sin (2\theta)$$

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

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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.

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