Question

Show that the height of a closed right circular cylinder of given volume and least surface area is equal to its diameter.

Answer 1

Suppose that $r$ be the radius of the base and $h$ the height of a cylinder.

Given that,

The surface area is given by

$S=2\pi r(h+r)$

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

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

Let $V$ be the volume of the cylinder.

$\therefore V=\pi r^{2}h$

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

$V=\frac{Sr-2\pi r^3}{2}$

Differentiation this with respect to $x$ and we get,

$\frac{dV}{dr}=\frac{S}{2}-3\pi r^2$ …… (2)

For Maximum or minimum, We have

$\frac{dV}{dr}=0$

$\frac{S}{2}-3\pi r^2=0$

$S=6\pi r^2$

We know that,

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

$6\pi r^2=2\pi rh+2\pi r^2$

$6\pi r^2-2\pi r^2=2\pi rh$

$\Rightarrow h=2r$

Again differentiation equation (2), we get

$\frac{d^{2}V}{d{t}^2}=-6\pi r<0$

Hence, $V$ is maximum when $h=2r$.