Question

The radius of the cylinder of maximum volume, which can be inscribed a sphere of radius R is

• A. $\frac{2}{3}R$
• B. $\sqrt{\frac{2}{3}}R$
• C. $\frac{3}{4}R$
• D. $\sqrt{\frac{3}{4}}R$

Answer 1

Answer: (B)

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

Answer 2

If r be the radius and h the height, the from the figure, $r^{2} + \left( \frac{h}{2} \right)^{2} = R^{2}$⇒ $h^{2} = 4(R^{2} - r^{2})$

Now, $V = \pi r^{2}h = 2\pi r^{2}\sqrt{R^{2} - r^{2}}$

$\therefore$ $\frac{dV}{dr} = 4\pi r\sqrt{R^{2} - r^{2}} + 2\pi r^{2}.\frac{1}{2}\frac{( - 2r)}{\sqrt{R^{2} - r^{2}}}$

For max. or min., $\frac{dV}{dr} = 0$

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

⇒ $2R^{2} = 3r^{2}$ ⇒ $r = \sqrt{\frac{2}{3}}R$ ⇒ $\frac{d^{2}V}{dr^{2}} = - ve$.

Hence V is max. when $r = \sqrt{\frac{2}{3}}R$.