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# A Closed Cylinder Has Volume 2156 Cm3. What Will Be the Radius of Its Base So that Its Total Surface Area is Minimum ? - Mathematics

#### Question

A closed cylinder has volume 2156 cm3. What will be the radius of its base so that its total surface area is minimum ?

#### Solution

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

$\text { Volume =} \pi r^2 h$

$\Rightarrow 2156 = \pi r^2 h$

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

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

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

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

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

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

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

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

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

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

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

$\Rightarrow r^3 = 343$

$\Rightarrow r = 7 cm$

$\text { Now },$

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

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

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

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

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