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# Show that the Height of the Cylinder of Maximum Volume that Can Be Inscribed a Sphere of Radius R is 2 R √ 3 . - Mathematics

#### Question

Show that the height of the cylinder of maximum volume that can be inscribed a sphere of radius R is $\frac{2R}{\sqrt{3}} .$

#### Solution

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

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

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

$\text { Volume of cylinder }, V = \pi r^2 h$

$\text { Squaring both sides, we get }$

$\Rightarrow V^2 = \pi^2 r^4 h^2$

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

$\text { Now,}$

$Z = 4 \pi^2 \left( r^4 R^2 - r^6 \right)$

$\Rightarrow \frac{dZ}{dr} = 4 \pi^2 \left( 4 r^3 R^2 - 6 r^5 \right)$

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

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

$\Rightarrow 4 \pi^2 \left( 4 r^3 R^2 - 6 r^5 \right) = 0$

$\Rightarrow 4 r^3 R^2 = 6 r^5$

$\Rightarrow 6 r^2 = 4 R^2$

$\Rightarrow r^2 = \frac{4 R^2}{6}$

$\Rightarrow r = \frac{2R}{\sqrt{6}}$

$\text { Substituting the value of r in eq. }\left( 1 \right), \text { we get }$

$\Rightarrow h = 2\sqrt{R^2 - \left( \frac{2R}{\sqrt{6}} \right)^2}$

$\Rightarrow h = 2\sqrt{\frac{6 R^2 - 4 R^2}{6}}$

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

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

$\text { Now,}$

$\frac{d^2 Z}{d r^2} = 4 \pi^2 \left( 12 r^2 R^2 - 30 r^4 \right)$

$\Rightarrow \frac{d^2 Z}{d r^2} = 4 \pi^2 \left( 12 \left( \frac{2R}{\sqrt{6}} \right)^2 R^2 - 30 \left( \frac{2R}{\sqrt{6}} \right)^4 \right)$

$\Rightarrow \frac{d^2 Z}{d r^2} = 4 \pi^2 \left( 8 R^4 - \frac{80 R^4}{6} \right)$

$\Rightarrow \frac{d^2 Z}{d r^2} = 4 \pi^2 \left( \frac{48 R^4 - 80 R^4}{6} \right)$

$\Rightarrow \frac{d^2 Z}{d r^2} = 4 \pi^2 \left( - \frac{16 R^4}{3} \right) < 0$

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

$\text { Hence proved }.$

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