#### Question

A closed cylinder has volume 2156 cm^{3}. What will be the radius of its base so that its total surface area is minimum ?

#### Solution

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

\[\text { Volume =} \pi r^2 h\]

\[ \Rightarrow 2156 = \pi r^2 h\]

\[ \Rightarrow 2156 = \frac{22}{7} r^2 h\]

\[ \Rightarrow h = \frac{2156 \times 7}{22 r^2}\]

\[ \Rightarrow h = \frac{686}{r^2} . . . \left( 1 \right)\]

\[\text { Surface area } = 2\pi r h + 2\pi r^2 \]

\[ \Rightarrow S = \frac{4312}{r} + \frac{44 r^2}{7} \left[ \text { From eq } . \left( 1 \right) \right]\]

\[ \Rightarrow \frac{dS}{dr} = \frac{4312}{- r^2} + \frac{88r}{7}\]

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

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

\[ \Rightarrow \frac{4312}{- r^2} + \frac{88r}{7} = 0\]

\[ \Rightarrow \frac{4312}{r^2} = \frac{88r}{7}\]

\[ \Rightarrow r^3 = \frac{4312 \times 7}{88}\]

\[ \Rightarrow r^3 = 343\]

\[ \Rightarrow r = 7 cm\]

\[\text { Now }, \]

\[\frac{d^2 s}{d r^2} = \frac{8624}{r^3} + \frac{88}{7}\]

\[ \Rightarrow \frac{d^2 s}{d r^2} = \frac{8624}{343} + \frac{88}{7}\]

\[ \Rightarrow \frac{d^2 s}{d r^2} = \frac{176}{7} > 0\]

\[\text{ So, the surface area is minimum when r = 7 cm }.\]