#### Question

Of all the closed cylindrical cans (right circular), which enclose a given volume of 100 cm^{3}, which has the minimum surface area?

#### Solution

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

\[\text { Volume }\left( V \right) \text { of the cylinder } = \pi r^2 h\]

\[ \Rightarrow 100 = \pi r^2 h\]

\[ \Rightarrow h = \frac{100}{\pi r^2}\]

\[\text { Surface area }\left( S \right) \text { of the cylinder} = 2\pi r^2 + 2\pi r h = 2\pi r^2 + 2\pi r \times \frac{100}{\pi r^2}\]

\[ \Rightarrow S = 2\pi r^2 + \frac{200}{r}\]

\[ \therefore \frac{dS}{dr} = 4\pi r - \frac{200}{r^2} \]

\[\text { For the maximum or minimum, we must have }\]

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

\[ \Rightarrow 4\pi r - \frac{200}{r^2} = 0\]

\[ \Rightarrow 4\pi r^3 = 200\]

\[ \Rightarrow r = \left( \frac{50}{\pi} \right)^\frac{1}{3} \]

\[\text { Now,} \]

\[ \frac{d^2 S}{d r^2} = 4\pi + \frac{400}{r^3}\]

\[ \Rightarrow \frac{d^2 S}{d r^2} > 0 \text { when r } = \left( \frac{50}{\pi} \right)^\frac{1}{3} \]

\[\text { Thus, the surface area is minimum when r =} \left( \frac{50}{\pi} \right)^\frac{1}{3} . \]

\[\text { At r }= \left( \frac{50}{\pi} \right)^\frac{1}{3} : \]

\[h = \frac{100}{\pi \left( \frac{50}{\pi} \right)^\frac{2}{3}} = 2 \left( \frac{50}{\pi} \right)^\frac{1}{3}\]