#### Question

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

#### Solution

\[\text{ Let the height and radius of the base of the cylinder be h and r, respectively . Then }, \]

\[\frac{h^2}{4} + r^2 = R^2 \]

\[ \Rightarrow h = 2\sqrt{R^2 - r^2} ............. \left( 1 \right)\]

\[\text { Volume of cylinder }, V = \pi r^2 h\]

\[\text { Squaring both sides, we get }\]

\[ \Rightarrow V^2 = \pi^2 r^4 h^2 \]

\[ \Rightarrow V^2 = 4 \pi^2 r^4 \left( R^2 - r^2 \right) ..............\left[ \text { From eq. } \left( 1 \right) \right]\]

\[\text { Now,} \]

\[Z = 4 \pi^2 \left( r^4 R^2 - r^6 \right)\]

\[ \Rightarrow \frac{dZ}{dr} = 4 \pi^2 \left( 4 r^3 R^2 - 6 r^5 \right)\]

\[\text { For maximum or minimum values of Z, we must have} \]

\[\frac{dZ}{dr} = 0\]

\[ \Rightarrow 4 \pi^2 \left( 4 r^3 R^2 - 6 r^5 \right) = 0\]

\[ \Rightarrow 4 r^3 R^2 = 6 r^5 \]

\[ \Rightarrow 6 r^2 = 4 R^2 \]

\[ \Rightarrow r^2 = \frac{4 R^2}{6}\]

\[ \Rightarrow r = \frac{2R}{\sqrt{6}}\]

\[\text { Substituting the value of r in eq. }\left( 1 \right), \text { we get }\]

\[ \Rightarrow h = 2\sqrt{R^2 - \left( \frac{2R}{\sqrt{6}} \right)^2}\]

\[ \Rightarrow h = 2\sqrt{\frac{6 R^2 - 4 R^2}{6}}\]

\[ \Rightarrow h = 2\sqrt{\frac{R^2}{3}}\]

\[ \Rightarrow h = \frac{2R}{\sqrt{3}}\]

\[\text { Now,} \]

\[ \frac{d^2 Z}{d r^2} = 4 \pi^2 \left( 12 r^2 R^2 - 30 r^4 \right)\]

\[ \Rightarrow \frac{d^2 Z}{d r^2} = 4 \pi^2 \left( 12 \left( \frac{2R}{\sqrt{6}} \right)^2 R^2 - 30 \left( \frac{2R}{\sqrt{6}} \right)^4 \right)\]

\[ \Rightarrow \frac{d^2 Z}{d r^2} = 4 \pi^2 \left( 8 R^4 - \frac{80 R^4}{6} \right)\]

\[ \Rightarrow \frac{d^2 Z}{d r^2} = 4 \pi^2 \left( \frac{48 R^4 - 80 R^4}{6} \right)\]

\[ \Rightarrow \frac{d^2 Z}{d r^2} = 4 \pi^2 \left( - \frac{16 R^4}{3} \right) < 0\]

\[\text { So, volume of the cylinder is maximum when } h = \frac{2R}{\sqrt{3}} . \]

\[\text { Hence proved }.\]