Let $C_n\subset\mathbb{R}^n$ be the $n$-dimensional cube with side $1$, and let $P_k$ be any $k$-dimensional plane, $k\leq n$. What is the maximal $k$-volume $V_{n,k}$ of the projection of $C_n$ on $P_k$?
Quite obviously, the minimal area should be $1$, obtained by taking $C_n = [0,1]^n$ and projecting it on $\{\mathbf{x}\in\mathbb{R}^n|x_{k+1}=\ldots=x_n=0\}$. I think the maximum should be obtained by projecting onto something orthogonal to one of the maximal diagonals of the cube, but I haven't found any proof of this, nor a formula for the volume so obtained.
I am particularly interested in the case $k = n-1$.
I got an upper bound for $V_{n,k}$.
We can inscribe $C_n$ in the $n$-ball of radius $\sqrt{n}$. The projection of such a sphere on a $k$-plane is a $k$-ball of radius $\sqrt{n}$ containing the projection of $C_n$. Its volume is
$$V(n,k) = \frac{(n\pi)^\frac{k}{2}}{\Gamma\left(1+\frac{k}{2}\right)}\geq V_{n,k}$$
where $\Gamma$ is the Gamma function.
Conjecture: As $n,k$ become big we have the asymptotical behavior $V(n,k)\sim V_{n,k}$.
Would anyone care to try to prove this, if not to solve the initial problem?
Assuming the conjecture to be true, we have the asymptotical behavior for $V(n,k)$ given by the estimate of the volume of the $k$-ball for $k\gg 1$:
$$V_{n,k}\sim V(n,k)\sim\frac{1}{\sqrt{\pi}}\left(\frac{2\pi e}{k}\right)^\frac{k}
{2}n^\frac{k}{2}$$
as $n,k\rightarrow\infty$.
Best Answer
It happens that the area of the orthogonal projection of a unit $n$-dimensional cube onto a $k$ dimensional space is equal to the area of the projection of the cube on the $(n-k)$-dimensional orthogonal plane.
Hence for $k=n-1$ the answer is very simple: the maximal area of the projection is equal to the diagonal of the cube: $\sqrt{n}$.
Reference: https://londmathsoc.onlinelibrary.wiley.com/doi/abs/10.1112/blms/16.3.278