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प्रश्न
Of all the closed cylindrical cans (right circular), which enclose a given volume of 100 cm3, which has the minimum surface area?
उत्तर
\[\text { Let r and h be the radius and height of the cylinder, respectively . Then, } \]
\[\text { Volume }\left( V \right) \text { of the cylinder } = \pi r^2 h\]
\[ \Rightarrow 100 = \pi r^2 h\]
\[ \Rightarrow h = \frac{100}{\pi r^2}\]
\[\text { Surface area }\left( S \right) \text { of the cylinder} = 2\pi r^2 + 2\pi r h = 2\pi r^2 + 2\pi r \times \frac{100}{\pi r^2}\]
\[ \Rightarrow S = 2\pi r^2 + \frac{200}{r}\]
\[ \therefore \frac{dS}{dr} = 4\pi r - \frac{200}{r^2} \]
\[\text { For the maximum or minimum, we must have }\]
\[ \frac{dS}{dr} = 0\]
\[ \Rightarrow 4\pi r - \frac{200}{r^2} = 0\]
\[ \Rightarrow 4\pi r^3 = 200\]
\[ \Rightarrow r = \left( \frac{50}{\pi} \right)^\frac{1}{3} \]
\[\text { Now,} \]
\[ \frac{d^2 S}{d r^2} = 4\pi + \frac{400}{r^3}\]
\[ \Rightarrow \frac{d^2 S}{d r^2} > 0 \text { when r } = \left( \frac{50}{\pi} \right)^\frac{1}{3} \]
\[\text { Thus, the surface area is minimum when r =} \left( \frac{50}{\pi} \right)^\frac{1}{3} . \]
\[\text { At r }= \left( \frac{50}{\pi} \right)^\frac{1}{3} : \]
\[h = \frac{100}{\pi \left( \frac{50}{\pi} \right)^\frac{2}{3}} = 2 \left( \frac{50}{\pi} \right)^\frac{1}{3}\]
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