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# Solution for Show that the Maximum Volume of the Cylinder Which Can Be Inscribed in a Sphere of Radius 5 √ 3 C M is 500 π C M 3 . - CBSE (Commerce) Class 12 - Mathematics

#### Question

Show that the maximum volume of the cylinder which can be inscribed in a sphere of radius $5\sqrt{3 cm} \text { is }500 \pi {cm}^3 .$

#### Solution

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

$\frac{h^2}{4} + r^2 = R^2$

$\Rightarrow h^2 = 4\left( R^2 - r^2 \right)$

$\Rightarrow r^2 = R^2 - \frac{h^2}{4} . . . \left( 1 \right)$

$\text { Now },$

$V = \pi r^2 h$

$\Rightarrow V = \pi\left( h R^2 - \frac{h^3}{4} \right) \left[\text { From eq } . \left( 1 \right) \right]$

$\Rightarrow \frac{dV}{dh} = \pi\left( R^2 - \frac{3 h^2}{4} \right)$

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

$\frac{dV}{dh} = 0$

$\Rightarrow \pi\left( R^2 - \frac{3 h^2}{4} \right) = 0$

$\Rightarrow R^2 - \frac{3 h^2}{4} = 0$

$\Rightarrow R^2 = \frac{3 h^2}{4}$

$\Rightarrow h = \frac{2R}{\sqrt{3}}$

$\frac{d^2 V}{d h^2} = \frac{- 3\pi h}{2}$

$\frac{d^2 V}{d h^2} = \frac{- 3\pi}{2} \times \frac{2R}{\sqrt{3}}$

$\Rightarrow \frac{d^2 V}{d h^2} = \frac{- 3\pi R}{\sqrt{3}} < 0$

$\text { So, the volume is maximum when h } = \frac{2R}{\sqrt{3}} .$

$\text { Maximum volume } = \pi h\left( R^2 - \frac{h^2}{4} \right)$

$= \pi \times \frac{2R}{\sqrt{3}}\left( R^2 - \frac{4 R^2}{12} \right)$

$= \frac{2\pi R}{\sqrt{3}}\frac{8 R^2}{12}$

$= \frac{4\pi R^3}{3\sqrt{3}}$

$= \frac{4\pi \left( 5\sqrt{3} \right)^3}{3\sqrt{3}}$

$= 500\pi {cm}^3$

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Solution Show that the Maximum Volume of the Cylinder Which Can Be Inscribed in a Sphere of Radius 5 √ 3 C M is 500 π C M 3 . Concept: Graph of Maxima and Minima.
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