This is surely well known, but:

Q1. What is the $(d{-}1)$-dimensional polytope that realizes the maximum volume cross-section of a unit hypercube by a $(d{-}1)$-dimensional hyperplane?

Answered in the comments: Ball's theorem.

          (Image from MathWorld. This is not the maximum area slice [Noam Elkies].)
I am less interested in the numerical maximum volume as I am in a description of the polytope that achieves it.
          (4D slicing snapshots from eusebeia.dyndns.org.)

Q2. Is the polytope that answers Q1 is the same as that which achieves the maximum volume orthogonal projection (shadow) of the hypercube?

Answered by Paata Ivanisvili: No for $d \ge 3$.

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    $\begingroup$ Q1: As I recall the central section orthogonal to $(1,1,1,\ldots,1)$ gives a cross-section approaching $\sqrt{6/\pi}$ for large $n$, which is less than the $\sqrt 2$ of the central section orthogonal to $(1,1,0,0,0,\ldots,0)$. Certainly in dimension $3$ the hexagon is pretty but smaller (area $\sqrt{27/16} < 1.3 < 1.4 < \sqrt2$). That $\sqrt 2$ is supposed to be maximal but I don't know/remember whether this has been proved. $\endgroup$ – Noam D. Elkies Aug 20 '16 at 1:02
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    $\begingroup$ The somewhat strange constant $\sqrt{6/\pi}$ comes from the Gaussian approximation to $\int_{-\infty}^\infty \bigl(\frac{\sin x}{x}\bigr)^n \, dx$, with $\frac{\sin x}{x} = 1 - \frac16 x^2 + O(x^4)$. = 1 - \frac16 x^2 + O(x^4)$. $\endgroup$ – Noam D. Elkies Aug 20 '16 at 1:05
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    $\begingroup$ Q1: I think this has been proved, the maximal area is $\sqrt{2}$ and it is attained only if the hyperplane has a normal unit vector with two nonzero coordinates with absolute value $1/\sqrt{2}$ As far as I know this is Ball's theorem. Fedja wrote a simple proof once many years ago math.spbu.ru/analysis/f-doska/cube.pdf $\endgroup$ – Paata Ivanishvili Aug 20 '16 at 2:07
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    $\begingroup$ Thanks! Now I have to decide if I want to learn it enough to decipher the Russian . . . (in the title TEOREMA BOLLA = Ball's theorem, and EDINICHNOVO KUBA is presumably "unit cube", but it would take a long time to workout everything from context.) $\endgroup$ – Noam D. Elkies Aug 20 '16 at 2:26
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    $\begingroup$ Here is another exposition of Ball's theorem www2.warwick.ac.uk/fac/sci/maths/people/staff/pooley/… $\endgroup$ – Gjergji Zaimi Aug 20 '16 at 2:28

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