Question

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

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

Answer 1

The correct option is B

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

Explanation for the correct option:

Step 1: Solve for the first derivative of Volume.

If $r$ is the radius and $h$ is the height then the from the figure,

$r^2+{\left(\frac{h}{2}\right)}^2=R^2\Rightarrow h^2=4(R^2-r^2)$

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

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

For maximum or minimum, $\frac{dV}{dr}=0$

Step 2: Solve for the maximum volume.

$$\begin{gathered} \Rightarrow 4\mathrm{\pi r}\sqrt{{\mathrm{R}}^2-{\mathrm{r}}^2}=\frac{2{\mathrm{\pi r}}^3}{\sqrt{{\mathrm{R}}^2-{\mathrm{r}}^2}} \\ \Rightarrow {\mathrm{R}}^2=3{\mathrm{r}}^2 \\ \Rightarrow \mathrm{r}=\sqrt{\frac{2}{3}}\mathrm{R} \\ and\frac{{\mathrm{d}}^2\mathrm{V}}{{\mathrm{dr}}^2}=-\mathrm{ve} \end{gathered}$$

So, volume is maximum when,$r=\sqrt{\frac{2}{3}}R$

Hence, option(B) is the correct answer