Question

Show that the right circular cone of least curved surface and given volume has an altitude equal to$\sqrt{2}$ time the radius of the base.

Answer 1

Let r and h be the radius and the height(altitude)of the cone respectively.

Then,the volume(V)of the cone is given as:

V=\frac{1}{3\pi}$$\pi r^{2}h⇒h=\frac{3V}{r^{2}}

The surface area$(S)$of the cone is given by,

$S=\pi rl$ (where $l$ is the slant height)

$=\pi r\sqrt{r^{2}+h^{2}}$

=\pi r\sqrt{r^{2}+\frac{9V^{2}\pi}{/π^{2}r^{4}}}$$=\frac{r\sqrt{9^{2}r^{6}+V^{2}}}{\pi r^{2}}

$=\frac{1}{r}\sqrt{π^{2}r^{6}+9V^{2}}$

∴\frac{dS}{dr}$$=\frac{r.\frac{6\pi ^{2}r^{5}}{2\pi^{2}\sqrt{r^{6}9V^{2}}}-\sqrt{\pi ^{2}r^{6}+9V^{2}}}{r^{2}}

$=\frac{3\pi^{2}r^{6}-\pi ^{2}r^{6}-9V^{2}}{r^{2}\sqrt{\pi 2r^{6}+9V^{2}}}$

$=\frac{2\pi^{2}r^{6}-9V^{2}}{r^{2}\sqrt{\pi^{2}r^{6}+9V^{2}}}$

$=\frac{2\pi^{2}r^{6}-9V^{2}}{r^{2}\sqrt{\pi^{2}r^{6}+9V^{2}}}$

Now,\frac{dS}{dr}$$=0⇒2\pi^{2}r^{6}=9V^{2}⇒r^{6}=\frac{9V^{2}}{2\pi^{2}}

Thus,it can be easily verified that when $r^{6}=\frac{9V^{2}}{2\pi^{2}},\frac{d^{2}S}{dr^{2}}>0.$

By second derivative test, the surface area of the cone is the least when $r^{6}=\frac{9V^{2}}{2\pi^{2}}$

When r^{6}=\frac{9V^{2}}{2\pi^{2}}$$,h=\frac{3V}{\pi r^{2}}=\frac{3}{\pi r^{2}}(\frac{2\pi ^{2}r^{6}}{9})^{\frac{1}{2}}$$=\frac{3}{\pi r^{2}}.\frac{\sqrt{2}\pi r^{3}}{3}=\sqrt{2}r.

Hence,for a given volume,the right circular cone of the least curved surface has an

altitude equal to$\sqrt{2}$ times the radius of the base.