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 \(\frac{2R}{\sqrt3}.\)