Question

Show that the height of the cylinder which is open at the top having a given surface area and greatest volume is equal to the radius of its base.

Answer 1

Let $r$ be the radius and $h$ be the height of the surfaces, Then

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

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

Let $V$ be the volume of the cylinder. Then

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

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

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

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

For maximum and minimum we have

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

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

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

$\Rightarrow r=h$

Differentiating wrt $r$ we get

$\frac{d^{2}V}{{dr}^2}=-3\pi r<0$

$V$ is maximum at $r=h$