Maximizing hyper-cylinder volume I would like to calculate the volume of a $n$-dimensional hyper-cylinder which is inside a unit $n$-dimensional hyper-sphere. Any ideas on how to approach this problem?
 A: The general formula for the volume of a cylinder in $n$-dimensions with (non necessarily circular) base of $(n-1)$-area $B$ and height $h$ is
$$V_{\text{cone}} = Bh.$$
The base will be an $(n-1)$-sphere. The volume of such a sphere with radius $r$ is given by
$$V_{\text{sphere}} = \frac{\pi^{\frac{n-1}{2}}}{\Gamma\left(\frac{n-1}{2}+1\right)}r^{n-1}.$$
The height of our cylinder determines the radius by $r^2+\frac{h^2}{4} = 1$. Therefore, the optimization problem becomes:

Maximize
  $$\frac{\pi^{\frac{n-1}{2}}}{4\Gamma\left(\frac{n-1}{2}+1\right)}r^{n-1}h$$
  subject to
  $$r^2+\frac{h^2}{4} = 1.$$

Of course, the constant doesn't really matter for the optimization, so the function to optimize is in fact $r^{n-1}h$.

Using Lagrange multipliers, what I get is $r=\sqrt{1-\frac{1}{n}}$, $h=2\sqrt{\frac{1}{n}}$.
A: Volume of a n-hypersphere:
$V = \frac {\pi^{\frac n2}}{\Gamma(\frac n2)} R^n$
Volume of a n-hypercyilnder: volume of the n-1 hpersphere - height.
equation of a hyperspehre:  
$x_1^1 + x_2^2 + \cdots x_n^2 = R^2$
Hyper-cylinder:
$x_2^2 + x_3^2 + \cdots x_{n}^2 = r^2$
subtracting one from the other
$x_1^2=R^2-r^2$
with $x_1$ as our height.
$V = \frac {\pi^{\frac {n-1}2}}{\Gamma(\frac {n-1}{2})} r^{n-1} \sqrt {R^2 - r^2}$
To maximize V.
$\frac {\pi^{\frac {n-1}2}}{\Gamma(\frac {n-1}{2})}$ is a constant.  And we can treat $R$ as constant and find $r$ in terms of $R.$
$\frac {dv}{dt} = $$(n-1) r^{n-2} \sqrt {R^2 -r^2} - r^{n}(\sqrt {R^2 -r^2})^{-1} = 0\\
 (n-1) (R^2 -r^2) = r^2\\
r = \frac {R}{\sqrt n}$
$V = \frac {\pi^{\frac {n-1}2}}{\Gamma(\frac {n-1}{2})} \frac {R^n}{n^{\frac{n}2}} \sqrt {n-1}$
Since this is a unit hypersphere $R = 1$
$V = \frac {\pi^{\frac {n-1}2}}{\Gamma(\frac {n-1}{2})} \frac { (n-1)^\frac 12}{n^{\frac{n}2}} $
