Question

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

Answer 1

Let r and h be the radius and height of the cylinder respectively.

Then, the surface area of the cylinder is given by

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

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

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

Let V be the volume of the cylinder. Then,

$V=\pi r^{2}h=\pi r^2\left[\frac{S}{2\pi }\right(\frac{1}{r})-r]=\frac{Sr}{2}-\pi r^3$

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

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

Now, $\frac{dV}{dr}=0⟹\frac{S}{2}=3\pi r^2⟹r^2=\frac{S}{6\pi }$

When, $r^2=\frac{S}{6\pi },\frac{d^{2}V}{d{r}^2}<0$

Therefore, by second derivative test, the volume is maximum when $r^2=\frac{S}{6\pi }$.

Now, when $r^2=\frac{S}{6\pi },h=2r$.

Hence, the volume is the maximum when the height is twice the radius, i.e., when height is equal to the diameter.