In three dimensions, maximising volume of cylinder inside a sphere (denote $B_3(R)$ , wo.l.o.g centered around the origin) is straightforward. We get constraints to the radius of the cylinder via good ol' Pythagorean:
$$r^2 + \left (\frac{h}{2}\right )^2 = R^2.\tag{1} $$
How does one make sense of general constraints in $\mathbb R^n$? For instance, if we consider a four dimensional cylinder in a $B_4(R)$, the base of the cylinder is now a sphere $B_3(r)$ (we get the $4$-dimensional volume if we multiply by height).
If it is safe to assume that maximal cylinder is also centered around the origin, what would the constraints on $r$ (the radius of the base) look like?
Relevant formulae for volume of $n$-balls can be found here, so I'm not worried about the optimisation process itself.
The following intuition might be as wrong as wrong gets, but here goes.
For instance, if we look at the projection for $n=3$, we get $(1)$. For $n=4$, the projection would be a $B_3(R)$ with the maximal cylinder inside it, so we project it further and the constraints remain the same. Assume the constraint $(1)$ holds, in general.
The volume $V_{n+1}^C$ of the cylinder would be
$$V_{n+1}^C = V_{B_n(r)}\cdot h = \frac{\pi ^{n/2}\cdot r^nh}{\Gamma (1+n/2)} := C_n \cdot r^nh. $$
With the constraint applied we have
$$r^n = \left (R^2 – \frac{h^2}{4}\right )^{n/2}. $$
Thus
$$V_{n+1}^C(h) = C_n h\left (R^2 – \frac{h^2}{4}\right )^{n/2} $$
This yields the maximal height of
$$h_m = \frac{2R}{\sqrt{n+1}}, $$
which agrees with the maximal height for three dimensional case! (curiously, the volume shrinks to zero as $n$ gets larger )
Is this inductive process of taking projections until we reach the case $n=2$ justified?
If the 'size' of the object gets smaller and smaller as $n$ increases, by taking projections the 'shadow' of the object becomes larger?!
I think this works as follows. In $n+1$-dimensional case the points in the sphere can be characterised as follows:
$$x = (x_1,\ldots, x_n, H)\qquad \|x\|\leq R $$
When we take the projection, we observe the set of points of dimension $n$:
$$\bar{x} = (x_1,\ldots , x_{n-1},H) $$
The projection of $B_{n+1}(R)$ still has radius $R$ thus
$$x_1^2+\ldots + x_{n-1}^2 + H^2 \leq R^2 $$
So, we would necessarely arrive at $(1)$.
Feel free to change tags to more adequate ones.
Best Answer
We should be able to argue by induction on the dimension. Taking projection of dimension 3 case (with respect to anything other than than the height component - we don' want two circles as projection in 2 dimensions).
So, if the center of cylinder is the center of sphere for dimension $k$, then taking any non-height-component-projection of the $k+1$ dimensional case, the necessary condition (1) is preserved.
I don't know how to formalise this idea or if it's correct, though.