Question

Show that the height of the cylinder of maximum volume that can be inscribed in a sphere of radius $20cm$ is $\frac{40}{\sqrt{3}}cm$.Also find the maximum volume.

Answer 1

Let $R,h$ and $x$ be the radius of the sphere, height of cylinder and radius of cylinder respectively.

Now, in $△ABC$

${AC}^2={AB}^2+{BC}^2$

${\left(2R\right)}^2=h^2+{\left(2x\right)}^2$

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

Let $V$ be the volume of cylinder.

$V=\pi x^{2}h$

$\Rightarrow V=\pi (R^2-\frac{h^2}{4})h$

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

Differentiating above equation w.r.t. $h$, we get

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

Putting $\frac{dV}{dh}=0$, we have

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

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

Differentiating equation $\left(2\right)$ w.r.t. $h$, we have

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

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

$\frac{d^{2}V}{d{x}^2}=-\frac{3\pi }{2}\left(\frac{2R}{\sqrt{3}}\right)=-\sqrt{3}\pi R<0$

Thus at $h=\frac{2R}{\sqrt{3}}$, the volume will be maximum.

$\because R=20\left(\text{Given}\right)$

$\therefore h=\frac{2\times 20}{\sqrt{3}}=\frac{40}{\sqrt{3}}cm$

Thus the height of the cylinder of maximum volume that can be inscribed in a sphere of radius $20cm$ is $\frac{40}{\sqrt{3}}cm$.

Hence proved.

Now, substituting the value of $R$ and $h$ in equation $\left(1\right)$, we have

$V=\pi {\left(20\right)}^2\times \frac{40}{\sqrt{3}}-\frac{\pi {\left(\frac{40}{\sqrt{3}}\right)}^3}{4}$

$\Rightarrow V=\frac{2\pi }{\sqrt{3}}{\left(20\right)}^3-\frac{2\pi }{3\sqrt{3}}{\left(20\right)}^3=\frac{4\pi {\left(20\right)}^3}{3\sqrt{3}}{cm}^3$

Hence the maximum volume will be $\frac{4\pi {\left(20\right)}^3}{3\sqrt{3}}{cm}^3$.