Basic questions about finding a volume formula for a box inside of an ellipsoid,

Question

I want to maximize the volume of a box, with sides parallel to the $xy$, $xz$ and $yz$-planes, with the box inside of an ellipsoid

$$\frac{x^2}{a^2} + \frac{y^2}{b^2} + \frac{z^2}{c^2}=1$$

So, my answer, using Lagrange multipliers, turns out to be wrong, at least comparing my work to the solution given.

I had thought to maximize "length times height times width", so I figured that the objective function should be $f(x,y,z)=xyxzyz = x^2y^2z^2$.

The solution instead maximizes a different function: $f(x,y,z) = 2x2y2z$

I am guessing that the solution is indeed correct? And the point I probably missed was this: the ellipsoid is centered at $0$. So, sketching out the box on paper, "length times height times width" does look like $2x 2y 2z$.

What do you think?

Also, how does an equation of an ellipsoid not centered at $0$ look like? Would it be something like this:

$$\frac{(x-1)^2}{a^2} + \frac{(y-2)^2}{b^2} + \frac{(z-3)^2}{c^2}=1?$$

...would this be an ellipsoid centered at $(1,2,3)$?

Sorry for the simple questions - my geometric intuition is a bit weak.

Thanks.

Answer 1

The correct equation for the ellispoid centered at the origin is

$$\frac{x^2}{a^2} +\frac{y^2}{b^2}+\frac{z^2}{c^2}=1.$$

It is obvious that the resulting (optimal) box would be centered at the origin.

If we denote one of the vertices of the box by $(x,y,z)$, then the sides have lengths

$$2x,\ 2y,\ 2z$$ assuming $x,y,z>0$. With this, the volume is $8xyz$.

Answer 2

A dimension check would’ve told you that your expression for the volume of the inscribed box was incorrect. A volume should have a dimension of $\text{length}^3$, but $x^2y^2z^2$ has a dimension of $\text{length}^6$.

As user170231 commented, your equation for an ellipsoid centered somewhere other than the origin is correct. Note that the axes of this translated ellipsoid are still parallel to the coordinate axes.