Question

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

Answer 1

Let the radius of the cone = r

Height of the cone = h

${\displaystyle \therefore V=\frac{1}{3}\pi r^2\text{h}}$ = constant quantity

${\displaystyle \therefore r^{2}h=\frac{3\times \text{constant quantity}}{\pi }=k}$ (assumed)

${\displaystyle r^{2}h=k,h=\frac{k}{r^2}}$                ...(1)

Curved surface S = ${\displaystyle \pi rl=\pi r\sqrt{h^2+r^2}}$

S = ${\displaystyle \pi r\sqrt{\frac{k^2}{r^4}+r^2}}$

${\displaystyle =\pi r\sqrt{\frac{k^2+r^6}{r^4}}}$

${\displaystyle =\frac{\pi }{r}\sqrt{k^2+r^6}}$

On differentiating,

${\displaystyle \therefore \frac{dS}{dr}=\pi \left[\frac{\frac{6r^5}{2\sqrt{r^6+k^2}}\times r-\sqrt{r^6+k^2}\cdot 1}{r^2}\right]}$

${\displaystyle =\pi \cdot \frac{3r^6-(r^6+k^2)}{r^2\sqrt{r^6+k^2}}}$

${\displaystyle =\frac{2r^6-k^2}{r^2\sqrt{r^6+k^2}}}$

For maximum and minimum, ${\displaystyle \frac{dS}{dr}=0}$

${\displaystyle \Rightarrow 2r^6-k^2=0}$

${\displaystyle \Rightarrow r^6=\frac{k^2}{2}}$

${\displaystyle \Rightarrow r^6=\frac{h^2{r}^4}{2}}$          ...(2)

⇒ h² = 2r²

∴ h = ${\displaystyle \sqrt{2}r}$

At h = ${\displaystyle \sqrt{2}r}$, as r passes through ${\displaystyle \sqrt{2}r}$

${\displaystyle \frac{dS}{dr}}$ changes from -ve to + ve.

∴ S is minimum when h = ${\displaystyle \sqrt{2}r}$

Therefore, the height of the right circular cone with the minimum curved surface is ${\displaystyle \sqrt{2}}$ times the radius.