Volume of the projection of the unit cube on a hyperplane 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 $n,k\rightarrow\infty$.
 A: 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
A: As suggested by @AlexanderShamov, there is the following formula for $k=n-1$: let $u$ be a unit normal vector to the hyperplane, then the area of the projection is given by the formula:
$$\sum_{i=1}^n|\left<u,e_i\right>|=\sum_{i=1}^n|u_i|=\|u\|_1$$
where $e_i$ denotes the $i$th basis vector (this is an adaptation of the more general formula (1.2) found in the paper Projection Bodies, by J. Bourgain and J. Lindenstrauss to our simple case). We want to find the extrema for this formula for $u\in S^n$. We'll use Lagrange multipliers:
$$L(u,\lambda)=\|u\|_1-\lambda(\|u\|_2^2-1)$$
$$\frac{\partial}{\partial u_i}L(u,\lambda)=\operatorname{sgn}(u_i)-2\lambda u_i$$
where we assumed $u_i\neq0$. Thus $\lambda = \frac{1}{2|u_i|}$ must be true for all $i$, and by $\|u\|_2^2=1$ we get that $|u_i|=\frac{1}{\sqrt{n}}$. Thus the maximal value of the area of such a projection is:
$$V_{n,n-1}=\frac{n}{\sqrt{n}}=\sqrt{n}$$
The problem for $k\neq n-1$ remains open.
A: For $k \ll n$ it looks like a reasonable idea to split the coordinates into $k$ blocks of length roughly equal $n/k$ and project onto the $k$-space generated by $e_{in/k}+e_{in/k+1}+\dots+e_{in/k+(n/k-1)},i=1,\dots,k$. This projection will be isometric to the cube $[0,\sqrt{n/k}]^k$ (since the whole thing decomposes into the direct sum of $k$ instances of taking the longest diagonal), so $V_{n,k} \ge (n/k)^{k/2}$. Obviously, for larger $k$ - namely, at scale $k \sim \lambda n$, $\lambda$ fixed - we need to allow non-equal lengths of blocks - with this modification this idea gives an exponential lower bound there.
A similarly straightforward upper bound is given by considering the diameter, which is $\sqrt n$, and using an isodiametric inequality. The gap between the two bounds grows exponentially in $k$, which doesn't look like a big deal up to scale $k \sim \lambda n$, at which point it confirms exponential growth of $V_{n,k}$, but leaves the precise exponent unknown.
However, at $k$ really close to $n$ something funny happens. The upper bound becomes hopelessly inefficient. In the regime $k = (1 - \frac{1}{r})n$, $r$ constant or large, I would split the coordinates into $n/r$ blocks of length $r$, take inside each $r$-dimensional subspace the largest $(r-1)$-dimensional projection, which has volume $\sqrt{r}$, and take the direct sum of $(n/r)$ of these, which gives $V_{n,k} \ge r^{n/2r}$. This glues well with the $k = \lambda n$ regime, and provides the correct value for $k = n-\mathrm{const}$. I don't have a nontrivial upper bound here.
Edit: Actually, by expressing the cube as an orthogonal sum of two cubes, and taking the corresponding projections onto direct sums, we get the following inequality:
$V_{n_1+n_2,k_1+k_2} \ge V_{n_1,k_1} V_{n_2,k_2}$
All of the lower bounds in my answer follow from this inequality, together with the known values $V_{n,1} = V_{n,n-1} = \sqrt n$.
