Question

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

Answer 1

Let r be the radius of the base, h be the height, V be the volume and S be the curved surface area of the cone. Then,
$\begin{array}{ccc}& V=\frac{1}{3}\pi r^{2}h\Rightarrow 3V=\pi r^{2}h& \\ \Rightarrow & 9V^2={\pi }^2{r}^4{h}^2\Rightarrow h^2=\frac{9V^2}{{\pi }^2{r}^4}& \dots \left(i\right)\\ and& S=\pi rl\Rightarrow S=\pi r\sqrt{r^2+h^2}& (\because l=\sqrt{h^2+r^2})\\ \Rightarrow & S^2={\pi }^2{r}^2(r^2+h^2)={\pi }^2{r}^2\left(\frac{9V^2}{{\pi }^2{r}^4+r^2}\right)& (UsingEq.(i\left)\right)\\ \Rightarrow & S^2=\frac{9V^2}{r^2}+{\pi }^2{r}^4& \dots \left(ii\right)\end{array}$
When S is least, $S^2$ is also least.
Now, $\frac{d}{dr}\left(S^2\right)=-\frac{18V^2}{r^3}+4{\pi }^2{r}^3\dots \left(iii\right)$
For minima, put $\frac{d}{dr}\left(S^2\right)=0$
$\Rightarrow -\frac{18V^2}{r^3}+4{\pi }^2{r}^3=0\Rightarrow 18V^2=4{\pi }^2{r}^6\Rightarrow 9V^2=2{\pi }^2{r}^6\dots \left(iv\right)$
Again differentiating Eq. (iii) w.r.t. r, we get
$\frac{d^2}{d{r}^2}\left(S^2\right)=\frac{54V^2}{r^4}+12{\pi }^2{r}^2$
AT $9V^2=2{\pi }^2{r}^6,\frac{d^2}{d{r}^2}\left(S^2\right)=\frac{54}{r^4}\left(\frac{2{\pi }^2{r}^6}{9}\right)+12{\pi }^2{r}^2=\frac{12{\pi }^2{r}^{+}6}{r^4}+12{\pi }^2{r}^2=24{\pi }^2{r}^2>0$
Hence, $S^2$ and therefore S is minimum when $9V^2=2{\pi }^2{r}^6$
On putting $9V^2=2{\pi }^2{r}^6$ in Eq.(i), we get
$=2{\pi }^2{r}^6={\pi }^2{r}^4{h}^2$
$\Rightarrow 2r^2=h^2\Rightarrow h=\sqrt{2}r$
Hence, altitude of right circular cone is $\sqrt{2}$ times the radius of the base.
Hence, proved.