Question

A right circular cylinder which is open at the top and has a given surface area, will have the greatest volume, if its height $h$ and radius $r$ are related by

• A. $2h=r$
• B. $h=4r$
• C. $h=2r$
• D. $h=r$

Answer 1

The correct option is D

$h=r$

Explanation for the correct option:

Step 1. Find the maximum value of volume of cylinder:

Let, Radius of the cylinder $=r$

Height of the cylinder $=h$

Volume of the cylinder $=V$

Surface of the cylinder $=S$

As we know,

Volume of the cylinder $\mathit{V}\mathbf{=}\mathit{\pi }{\mathit{r}}^{\mathbf{2}}\mathit{h}$

Surface area of the cylinder $\mathit{S}\mathbf{=}\mathit{\pi }{\mathit{r}}^{\mathbf{2}}\mathbf{+}\mathbf{2}\mathit{\pi }\mathit{r}\mathit{h}$

$\Rightarrow$ $h=\frac{\left(S–\pi r^2\right)}{2\pi r}$

Step 2. Put the value of $h$ in $V$:

$V=\pi r^2\frac{\left(S–\pi r_2\right)}{2\pi r}$

$=\left(\frac{r}{2}\right)\left(S–\pi r^2\right)$

$=\left(\frac{1}{2}\right)(Sr–\pi r^3)$

Step 3. Differentiate it with respect to $r$:

$\frac{dV}{dr}=\left(\frac{1}{2}\right)(S–3\pi r^2)$

Step 4. For maxima or minima of $v$, $\frac{dV}{dr}=0$

$\Rightarrow$ $\frac{1}{2}(S-3\pi r^2)=0$ ​

$\Rightarrow$ ​$r=\sqrt{\frac{S}{3\pi }}$

Now, $\frac{d^{2}V}{d^{2}r}=\frac{1}{2}(-6\pi r)$

$⟹\frac{d^{2}V}{d^{2}r}<0$

$V$ is maximum when $r=\sqrt{\frac{S}{3\pi }}$

$\Rightarrow$ $3\pi r^2=S$

Step 5. Putting it in surface area $S$:

$3\pi r^2=\pi r^2+2\pi rh$

$⟹2\pi r^2=2\pi rh$

$\Rightarrow$ $\mathit{r}\mathbf{=}\mathit{h}$

Hence, Option ‘D’ is Correct.