Question

The height of the cylinder of maximum volume inscribed in a sphere of radius '$a$' is

• A. $\frac{3a}{2}$
• B. $\frac{\sqrt{2}a}{3}$
• C. $\frac{a}{\sqrt{3}}$
• D. $\frac{2a}{\sqrt{3}}$

Answer 1

Answer: (D)

$\frac{2a}{\sqrt{3}}$

Solution:

Let $a$ be the radius of a sphere.

Let $r$ and $h$ be the radius and height of the cylinder respectively.

From figure, we get

$h=2\sqrt{a^2-r^2}$

The volume of the cylinder $v=\pi r^{2}h$

$=\pi r^{2}2\sqrt{a^2-r^2}=2\pi r^2\sqrt{a^2-r^2}$

$\therefore \frac{dV}{dr}=4\pi r\sqrt{a^2-r^2}+\frac{2\pi r^2(-2r)}{2\sqrt{a^2-r^2}}$

$=4\pi r\sqrt{a^2-r^2}-\frac{2\pi r^3}{\sqrt{a^2-r^2}}$

$=\frac{4\pi r(a^2-r^2)-2\pi r^3}{\sqrt{a^2-r^2}}$

$=\frac{4\pi r{a}^2-6\pi r^3}{\sqrt{a^2-r^2}}$

Now, $\frac{dV}{dr}=0⟹4\pi r{a}^2-6\pi r^3=0$

$⟹r^2=\frac{2a^2}{3}$

Now, $\frac{d^{2}V}{d{r}^2}=\frac{\sqrt{a^2-r^2}(4\pi a^2-18\pi r^2)-(4\pi r{a}^2-6\pi r^3)\frac{-2r}{2\sqrt{a^2-r^2}}}{(a^2-r^2)}$

$=\frac{(a^2-r^2)(4\pi a^2-18\pi r^2)+r(4\pi r{a}^2-6\pi r^3)}{(a^2-r^2{)}^{3/2}}$

$=\frac{4\pi a^4-22\pi r^2{a}^2+12\pi r^4+4\pi r^2{a}^2}{(a^2-r^2{)}^{3/2}}$

Now, it can be observed that at

$r^2=\frac{2a^2}{3},\frac{d^{2}V}{d{r}^2}<0$

$\therefore$ the volume is maximum, when

$r^2=\frac{2a^2}{3}$

Now, height of the cylinder $=2\sqrt{a^2-\frac{2a^2}{3}}=2\sqrt{\frac{a^2}{3}}=\frac{2a}{\sqrt{3}}$