Question

Show that the height of a cylinder open at the top of given volume and minimum total surface area, is equal to the radius of the base.

Answer 1

Let $r$ be the radius and $h$ be the height of the cylinder of given volume, $v$.

Now,

$v={\pi }^{2}h$

$h=\frac{v}{\pi r^2}$ …… (1)

As the cylinder is open at the top, so the total surface area is,

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

$s=2\pi \left(\frac{v}{\pi r^2}\right)+\pi r^2$

$s=\frac{2v}{r}+\pi r^2$

$\frac{ds}{dr}=-\frac{2v}{r}+2\pi r$ …… (2)

For total surface area to be minimum,

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

$\Rightarrow -\frac{2v}{r^2}+2\pi r=0$

$v=\pi r^3$

$\pi r^{2}h=\pi r^3$

$\Rightarrow h=r$

Differentiate equation (2) with respect to $r$.

$\frac{d^{2}s}{d{r}^2}=\frac{4v}{r^3}+2\pi$

$\Rightarrow {\left(\frac{d^{2}s}{d{r}^2}\right)}_{r=h}=\frac{4v}{h^3}+2\pi >0$

Hence, when the height of the cylinder is equal to the radius of the base of the cylinder, the surface area is minimum.