Let $S_n=\{x\in \mathbb R^n: \sum_{j=1}^n (x_j/a_j)^2\leq1\}$ be a filled ellipsoid, and $C_n=\{x\in \mathbb R^n: 0\leq x_j \leq 1,\forall j\}$. Let $$\mathcal T= \{U\in M_n(\mathbb R): \text{eigenvalues of $U$ are orthogonal, and }\\UC_n+ y\subseteq S_n \text{ for some } y\in\mathbb R^n \},$$ and $$ T=\{|\det U|: U\in \mathcal T\}, $$ i.e. $T$ is the set of volumes of a (hyper-)rectangular parallelpiped inside the ellipsoid.
How can I show rigorously that the maximal volume is given by $$ \sup T= \frac{2^n}{n^{n/2}} \prod_{1\leq j \leq n} a_j, $$ and that this can only be achieved when $U$ has eigenvalues $\lambda_j= a_j n^{-1/2}$?
All the results above are obvious if we only consider parallelpipeds with sides parallel to some coordinate axis. But when I also take other parallelpiped into consideration, the problem seem to get pretty messy. Is there any neat way to formulate this and prove it?
You can view a hyper-ellipsoid as a hyper-sphere stretched by $a_k$ along axis $x_k$. For a hyper-sphere the inscribed hyper-rectangle of maximum volume is of course a hyper-cube. Hence the maximum volume hyper-rectangle inscribed in a hyper-ellipsoid is nothing but a stretched hypercube, because stretching preserves the ratio of volumes.
A non axis-aligned hyper-rectangle cannot have maximum volume, because if you apply the inverse stretching (the one carrying the hyper-ellipsoid to a hyper-sphere) it doesn't become a hyper-cube.