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

Toolbox:
• Volume =$\pi r^2h$
Step 1:
Radius of the sphere$=R$
Let $h$ be the diameter of the base of the inscribed cylinder .
Then,
$h^2+x^2=(2R)^2$
$h^2+x^2=4R^2$------(1)
Volume of the cylinder $=\pi r^2h$
$V=\pi\big(\large\frac{x^2}{2}\big)^2.$$h \;\;=\pi\large\frac{x^4}{4}$$.h$
Volume=$\large\frac{1}{4}$$\pi x^2 h Substituting the value of x^2 we get V=\large\frac{1}{4}$$\pi h(4r^2-h^2)$
From (1),$x^2=4R^2-h^2$
$V=\pi R^2 h-\large\frac{1}{4}\pi h^3$
Step 2:
Differentiating with respect to $x$
$V=\pi R^2 h-\large\frac{1}{4}\pi h^3$
$\large\frac{dV}{dh}$$=\pi R^2-\large\frac{3}{4}\pi h^3 \quad\;\;=\pi[R^2-\large\frac{3}{4}h^2] \large\frac{dV}{dh}$$=0$
$\pi [R^2-\large\frac{3}{4}$$h^2]=0 R^2=\large\frac{3}{4}$$h^2$
$h=\large\frac{2R}{\sqrt 3}$
Step 3:
Also $\large\frac{d^2V}{dh^2}$$=-\large\frac{3}{4}$$.2\pi h$
$\Rightarrow \large\frac{-3}{2}$$\pi h At h=\large\frac{2R}{3} \large\frac{d^2V}{dh^2}=\frac{-3}{2}$$\pi\big(\large\frac{2R}{\sqrt 3}\big)=-ve$
$\Rightarrow V$ is maximum at $h=\large\frac{2R}{\sqrt 3}$
Maximum volume at $h=\large\frac{2R}{\sqrt 3}$
$\Rightarrow \large\frac{1}{4}$$\pi \big[\large\frac{2R}{\sqrt 3}$$\big]\big[4R^2-\large\frac{4R^2}{3}\big]$
$\Rightarrow \large\frac{\pi R}{2\sqrt 3}\big[\large\frac{8R^2}{3}\big]$
$\Rightarrow \large\frac{4\pi R^3}{3\sqrt 3}$ sq.units.
edited Sep 2, 2013

Toolbox:
• Volume =$\pi r^2h$
Step 1:
Radius of the sphere$=R$
Let $h$ be the diameter of the base of the inscribed cylinder .
Then,
$h^2+x^2=(2R)^2$
$h^2+x^2=4R^2$------(1)
Volume of the cylinder $=\pi r^2h$
$V=\pi\big(\large\frac{x^2}{2}\big)^2.$$h \;\;=\pi\large\frac{x^4}{4}$$.h$
Volume=$\large\frac{1}{4}$$\pi x^2 h Substituting the value of x^2 we get V=\large\frac{1}{4}$$\pi h(4r^2-h^2)$
From (1),$x^2=4R^2-h^2$
$V=\pi R^2 h-\large\frac{1}{4}\pi h^3$
Step 2:
Differentiating with respect to $x$
$V=\pi R^2 h-\large\frac{1}{4}\pi h^3$
$\large\frac{dV}{dh}$$=\pi R^2-\large\frac{3}{4}\pi h^3 \quad\;\;=\pi[R^2-\large\frac{3}{4}h^2] \large\frac{dV}{dh}$$=0$
$\pi [R^2-\large\frac{3}{4}$$h^2]=0 R^2=\large\frac{3}{4}$$h^2$
$h=\large\frac{2R}{\sqrt 3}$
Step 3:
Also $\large\frac{d^2V}{dh^2}$$=-\large\frac{3}{4}$$.2\pi h$
$\Rightarrow \large\frac{-3}{2}$$\pi h At h=\large\frac{2R}{3} \large\frac{d^2V}{dh^2}=\frac{-3}{2}$$\pi\big(\large\frac{2R}{\sqrt 3}\big)=-ve$
$\Rightarrow V$ is maximum at $h=\large\frac{2R}{\sqrt 3}$
Maximum volume at $h=\large\frac{2R}{\sqrt 3}$
$\Rightarrow \large\frac{1}{4}$$\pi \big[\large\frac{2R}{\sqrt 3}$$\big]\big[4R^2-\large\frac{4R^2}{3}\big]$
$\Rightarrow \large\frac{\pi R}{2\sqrt 3}\big[\large\frac{8R^2}{3}\big]$
$\Rightarrow \large\frac{4\pi R^3}{3\sqrt 3}$ sq.units.