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प्रश्न
A closed cylinder has volume 2156 cm3. What will be the radius of its base so that its total surface area is minimum ?
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उत्तर
\[\text { Let the height, radius of the base and surface area of the cylinder be h, r and S, respectively . Then }, \]
\[\text { Volume =} \pi r^2 h\]
\[ \Rightarrow 2156 = \pi r^2 h\]
\[ \Rightarrow 2156 = \frac{22}{7} r^2 h\]
\[ \Rightarrow h = \frac{2156 \times 7}{22 r^2}\]
\[ \Rightarrow h = \frac{686}{r^2} . . . \left( 1 \right)\]
\[\text { Surface area } = 2\pi r h + 2\pi r^2 \]
\[ \Rightarrow S = \frac{4312}{r} + \frac{44 r^2}{7} \left[ \text { From eq } . \left( 1 \right) \right]\]
\[ \Rightarrow \frac{dS}{dr} = \frac{4312}{- r^2} + \frac{88r}{7}\]
\[\text { For maximum or minimum values of S, we must have }\]
\[\frac{dS}{dr} = 0\]
\[ \Rightarrow \frac{4312}{- r^2} + \frac{88r}{7} = 0\]
\[ \Rightarrow \frac{4312}{r^2} = \frac{88r}{7}\]
\[ \Rightarrow r^3 = \frac{4312 \times 7}{88}\]
\[ \Rightarrow r^3 = 343\]
\[ \Rightarrow r = 7 cm\]
\[\text { Now }, \]
\[\frac{d^2 s}{d r^2} = \frac{8624}{r^3} + \frac{88}{7}\]
\[ \Rightarrow \frac{d^2 s}{d r^2} = \frac{8624}{343} + \frac{88}{7}\]
\[ \Rightarrow \frac{d^2 s}{d r^2} = \frac{176}{7} > 0\]
\[\text{ So, the surface area is minimum when r = 7 cm }.\]
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