Question

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

Answer 1

Radius of the sphere = R
Let h be the height and x be the diameter of the base of the inscribed cylinder. Then,
$h^2+x^2=(2R{)}^2\Rightarrow h^2+x^2=4R^2$ ...(i)
Volume of the cylinder $=\pi (radius{)}^2\times height$
$\Rightarrow V=\pi (\frac{x}{2}{)}^2.h=\frac{1}{4}\pi x^{2}h\Rightarrow V=\frac{1}{4}\pi h(4R^2-h^2)$ ...(ii)
[from Eq. (i) $x^2=4R^2-h^2$]
$\Rightarrow V=\pi R^{2}h-\frac{1}{4}\pi h^3$
On differentiating w.r.t.h, we get
$\frac{dV}{dh}=\pi R^2-\frac{3}{4}\pi h^2=\pi (R^2-\frac{3}{4}{h}^2)Put\frac{dV}{dh}=0\Rightarrow R^2=\frac{3}{4}{h}^2\Rightarrow h=\frac{2R}{\sqrt{3}}Also,\frac{d^{2}V}{d{h}^2}=-\frac{3}{4}\times 2\pi hAth=\frac{2R}{\sqrt{3}},\frac{d^{2}V}{d{h}^2}=-\frac{3}{4}\times 2\pi \left(\frac{2R}{\sqrt{3}}\right)=-\sqrt{3}\pi R=-ve$
$\Rightarrow$ V is maximum at $h=\frac{2R}{\sqrt{3}}$
Maximum volume at $h=\frac{2R}{\sqrt{3}}$ is $V=\frac{1}{4}\pi \left(\frac{2R}{\sqrt{3}}\right)(4R^2-\frac{4R^2}{3})$ [Using Eq (ii)]
$=\frac{\pi R}{2\sqrt{3}}\left(\frac{8R^2}{3}\right)=\frac{4\pi R^3}{3\sqrt{3}}sq.unit$
Thus, volume of the cylinder is maximum when $h=\frac{2R}{\sqrt{3}}.$