# Show that the height of the cylinder of maximum volume

Question:

Show that the height of the cylinder of maximum volume that can be inscribed in a sphere of radius $R$ is $\frac{2 R}{\sqrt{3}}$. Also find the maximum volume.

Solution:

A sphere of fixed radius (R) is given.

Let r and h be the radius and the height of the cylinder respectively.

From the given figure, we have $h=2 \sqrt{R^{2}-r^{2}}$.

The volume $(V)$ of the cylinder is given by,

$V=\pi r^{2} h=2 \pi r^{2} \sqrt{R^{2}-r^{2}}$

$\therefore \frac{d V}{d r}=4 \pi r \sqrt{R^{2}-r^{2}}+\frac{2 \pi r^{2}(-2 r)}{2 \sqrt{R^{2}-r^{2}}}$

$=4 \pi r \sqrt{R^{2}-r^{2}}-\frac{2 \pi r^{3}}{\sqrt{R^{2}-r^{2}}}$

$=\frac{4 \pi r\left(R^{2}-r^{2}\right)-2 \pi r^{3}}{\sqrt{R^{2}-r^{2}}}$

$=\frac{4 \pi r^{2}-6 \pi r^{3}}{\sqrt{R^{2}-r^{2}}}$

Now, $\frac{d V}{d r}=0 \Rightarrow 4 \pi r R^{2}-6 \pi r^{3}=0$

$\Rightarrow r^{2}=\frac{2 R^{2}}{3}$

Now, $\frac{d^{2} V}{d r^{2}}=\frac{\sqrt{R^{2}-r^{2}}\left(4 \pi R^{2}-18 \pi r^{2}\right)-\left(4 \pi r R^{2}-6 \pi r^{3}\right) \frac{(-2 r)}{2 \sqrt{R^{2}-r^{2}}}}{\left(R^{2}-r^{2}\right)}$

$=\frac{\left(R^{2}-r^{2}\right)\left(4 \pi R^{2}-18 \pi r^{2}\right)+r\left(4 \pi r R^{2}-6 \pi r^{3}\right)}{\left(R^{2}-r^{2}\right)^{\frac{3}{2}}}$

$=\frac{4 \pi R^{4}-22 \pi r^{2} R^{2}+12 \pi r^{4}+4 \pi r^{2} R^{2}}{\left(R^{2}-r^{2}\right)^{\frac{3}{2}}}$

Now, it can be observed that at $r^{2}=\frac{2 R^{2}}{3}, \frac{d^{2} V}{d r^{2}}<0$.

$\therefore$ The volume is the maximum when $r^{2}=\frac{2 R^{2}}{3}$.

When $r^{2}=\frac{2 R^{2}}{3}$, the height of the cylinder is $2 \sqrt{R^{2}-\frac{2 R^{2}}{3}}=2 \sqrt{\frac{R^{2}}{3}}=\frac{2 R}{\sqrt{3}}$.

Hence, the volume of the cylinder is the maximum when the height of the cylinder is $\frac{2 R}{\sqrt{3}}$