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# Solution for Of All the Closed Cylindrical Cans (Right Circular), Which Enclose a Given Volume of 100 Cm3, Which Has the Minimum Surface Area? - CBSE (Commerce) Class 12 - Mathematics

#### Question

Of all the closed cylindrical cans (right circular), which enclose a given volume of 100 cm3, which has the minimum surface area?

#### Solution

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

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

$\Rightarrow 100 = \pi r^2 h$

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

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

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

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

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

$\frac{dS}{dr} = 0$

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

$\Rightarrow 4\pi r^3 = 200$

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

$\text { Now,}$

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

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

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

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

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

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Solution Of All the Closed Cylindrical Cans (Right Circular), Which Enclose a Given Volume of 100 Cm3, Which Has the Minimum Surface Area? Concept: Graph of Maxima and Minima.
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