Question

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

Answer 1

Let, $r,l$ and $h$ be the radius, slant height and altitude of the cone respectively, then

$V=\frac{1}{3}\pi r^{2}h⟶\left(1\right)$

& $h=\frac{3V}{\pi r^2}⟶\left(2\right)$

$S=\pi rl$, squaring both sides

$S^2={\pi }^2{r}^2{l}^2=\pi r^2(h^2+r^2)[\because l^2=r^2+h^2]$

$={\pi }^2{r}^2[\frac{9V^2}{{\pi }^2{r}^4}+r^2]\left[from\right(2\left)\right]$

$S^2=\frac{9V^2}{r^2}+{\pi }^2{r}^4$, differentiating with respect to ${}^{\prime }{r}^{\prime }$

$2S\frac{ds}{dr}=\frac{-18V^2}{r^3}+4{\pi }^2{r}^3$

For maximum or minimum,

$\frac{ds}{dr}=0\Rightarrow \frac{-18V^2}{r^3}+4{\pi }^2{r}^3=0$

$\Rightarrow 4{\pi }^2{r}^3=\frac{18V^2}{r^3}\Rightarrow 2{\pi }^2{r}^6=9V^2=9\times \frac{1}{9}{\pi }^2{r}^4{h}^2\left[from\right(1\left)\right]$

$\Rightarrow 2{\pi }^2{r}^6={\pi }^2{r}^4{h}^2\Rightarrow {2r}^2=h^2\Rightarrow h=\sqrt{2}r$

$\frac{d^{2}S}{{dr}^2}=12{\pi }^2{r}^2+\frac{54V^2}{r^4}>0$ for all values of $V$ and $r.$

Hence, for the least surface area of a cone and given volume, altitude is equal to $\sqrt{2}$ times the radius of the base.