Question

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

Answer 1

The correct answer is $\frac{2R}{\sqrt3}.$
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,
$V=πr^2h=2πr^2\sqrt{R^2-r^2}$
$∴\frac{dV}{dr}=4πr\sqrt{R^2-r^2}+\frac{2πr^2(-2r)}{2\sqrt{R^2-r^2}}$
$=\frac{4πr(R^2-r^2)-2πr^3}{\sqrt{R^2-r^2}}$
$=\frac{4πrR^2-6πr^3}{\sqrt{R^2-r^2}}$
Now,$\frac{dV}{dr}=0$
$⇒4πrR^2-6πr^3=0$
$⇒r^2=\frac{2R^2}{3}$
Now $\frac{\frac{d^2V}{dr^2}=\sqrt{R^2-r^2}(4πR^2-18πr^2)-(4πR^2-6πr^3)-\frac{(-2r)}{2\sqrt{R^2-r^2}}}{(r^2-R^2)}$
$=\frac{4πR^4-22πr^2R^2+12πr^4+4πr^2R^2}{(R^2-r^2)^\frac{3}{2}}$
Now, it can be observed that at $r^2=\frac{2R^2}{3},\frac{d^2V}{dr^2}<0$
∴The volume is the maximum when $r^2=\frac{2R^2}{3}$
When $r^2=\frac{2R^2}{3}$ , the height of the cylinder is $2\sqrt{R^2-\frac{2R^2}{3}}$
$=2\sqrt{\frac{R^2}{3}}=\frac{2R}{\sqrt3}.$
Hence, the volume of the cylinder is the maximum when the height of the cylinder is $\frac{2R}{\sqrt3}.$