Question

Show that the height of a closed right circular cylinder of given surface and maximum volume, is such that its height is equal to the diameter of its base.

Answer 1

Let $S$ be the given surface area of the closed cylinder where radius is $r$ and height is $h$.

Let $V$ be the volume

Surface area $S=2\pi r^2+2\pi rh$

$h=\frac{S-2\pi r^2}{2\pi r}$ ....(1)

Volume $=\pi r^{2}h$

Substitute the value of $h$ in above equation,

Volume $=\pi r^2\left[\frac{S-2\pi r^2}{2\pi r}\right]$

$=\frac{1}{2}r(S-2\pi r^2)$

$=\frac{1}{2}(Sr-2\pi r^3)$

Differentiating wrt $r$, we get

$\frac{dV}{dr}=\frac{1}{2}(S-6\pi r^2)$

For maxima and minima $\frac{dV}{dx}=0$

$S-6\pi r^2=0$

$S=6\pi r^2$

From (1), we havee

$h=\frac{S-2\pi r^2}{2\pi r}$

Substitute the value of $S$ in the above equation,

$h=\frac{6\pi r^2-2\pi r^2}{2\pi r}$

$=\frac{4\pi r^2}{2\pi r}$

Thus $h=2r$

On double differentiation of $V$, we get

$\frac{d^{2}V}{d{r}^2}=\frac{1}{2}(-12\pi r)$

$=-6\pi r$

$-ve$

Therefore, $V$ is maximum.

Thus volume is maximum when $h=2r$.

Or when height of cylinder is equal to diameter of base.