Sphylinder

Let $R$ be the radius of the sphere, $h$ be the height of the cylinder and $r$ be the radius of the cylinder. By the Pythagorean theorem, the radius of the cylinder is given by

$r^2=R^2-\left(\dfrac{h}{2}\right)^2=R^2-\dfrac{h^2}{4}$

The volume of the cylinder is hence

$V=\pi r^2 h = \pi \left(R^2-\dfrac{h^2}{4}\right)h=\pi \left(R^2h-\dfrac{h^3}{4}\right)$

Differentiating with respect to $h$ and equating to $0$ to find extrema gives

$\dfrac{dV}{dh}=\pi \left(R^2-\dfrac{1}{4} \times 3h^2\right)=\pi \left(R^2-\dfrac{3}{4}h^2\right)=0$

$R^2=\dfrac{3}{4}h^2$

$h^2=\dfrac{4}{3}R^2$

$h=\sqrt{\dfrac{4}{3}R^2}=\dfrac{2R}{\sqrt{3}}$

Substituting, we get

$V=\pi \left(R^2-\dfrac{4}{3} \times \dfrac{1}{4} R^2 \right)\left(\dfrac{2R}{\sqrt{3}}\right)=\boxed{\dfrac{4}{3} \times \dfrac{\pi R^3}{\sqrt{3}}}$

Note: Do not forget the second derivative check.

Let the base radius and height of the cylinder fit in the sphere be $r$ and $h$ respectively. If the polar angle between the zenith and $R$ is $\theta$ , then we have $r=R\sin \theta$ and $\dfrac h2 = R\cos \theta$ . The volume $V$ of the cylinder is given by:

$\begin{aligned} V & = \pi r2h = 2 \pi R^3 \sin^2 \theta \cos \theta \\ \implies \frac {dV}{d \theta} & = 2 \pi R^3 (2\sin \theta \cos^2 \theta - \sin^3 \theta) & \small \color{#3D99F6} \text{Note that } \cos^2 \theta = 1 - \sin^2 \theta \\ & = 2 \pi R^3 (2\sin \theta - 3\sin^3 \theta) & \small \color{#3D99F6} \text{Equating }\frac {dV}{d\theta} = 0 \\ 2\sin \theta & = 3\sin^3 \theta & \small \color{#3D99F6} \text{Note that }\sin \theta = 0 \text{ gives minimum }V \\ \implies \sin \theta & = \sqrt {\frac 23} & \small \color{#3D99F6} \implies \cos \theta = \frac 1{\sqrt 3}, \ \tan \theta = \sqrt 2 \\ \frac {d^2V}{d \theta^2} & = 2 \pi R^3 (2\cos \theta - 9\sin^2 \theta \cos \theta) \end{aligned}$

$\begin{aligned} \implies \frac {d^2V}{d \theta^2}\bigg|_{\theta = \tan^{-1}\sqrt 2} & = 2 \pi R^3 \left(\frac 2{\sqrt 3} - 3\sqrt 2 \right) \color{#D61F06} < 0 \end{aligned}$

This means that $V$ is maximum when $\theta = \tan^{-1}\sqrt 2$ and $V_{\text{max}} = 2\pi R^3 \times \dfrac 23 \times \dfrac 1{\sqrt 3} = \boxed{\dfrac {4\pi R^3}{3\sqrt 3}}$ .