What is the maximum volume of a cylinder that can fit in a sphere of a constant radius? The first question that comes into my mind here is whether any cylinder that touches(at 4 pts) the circumference of the sphere and does not go out of it, has equal volume? 
Second, how do i mathematically limit the volume of the cylinder to be less than that of a sphere? Squeeze theorem? 
Please help, thanks!
 A: Let $R$ be the radius of the sphere and let $h$ be the height of the cylinder centered on the center of the sphere. By the Pythagorean theorem, the radius of the cylinder is given by
$$
r^2 = R^2 - \left(\frac{h}{2}\right)^2.
$$
The volume of the cylinder is hence
$$
\begin{align}
V &= \pi r^2 h\\ 
&= \pi \left(h R^2 - \frac{h^3}{4}\right).
\end{align}
$$
Differentiating with respect to $h$ and equating to $0$ to find extrema gives
$$
\frac{dV}{dh}=\pi \left(R^2 - \frac{3h^2}{4}\right) = 0\\
\therefore h_0 = \frac{2R}{\sqrt{3}}
$$
The second derivative of the volume with respect to $h$ is negative if $h>0$ such that the volume is maximal at $h = h_0$. Substituting gives
$$
V_{max}=\frac{4 \pi R^3}{3\sqrt{3}}.
$$
A: Hint: In the context of a calculus course, I think you are first expected to argue informally that such a maximal cylinder must have axis that goes through the center of the circle, and that without loss of generality that axis is the $z$-axis.
So now suppose that the cylinder meets the $x$-$y$ plane in a circle of radius $t$. Find the height of the cylinder in terms of $t$, and hence the volume. Now use the ordinary tools to maximize. 
