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

It is given that
$V=consant=\frac{\pi }{3}{r}^{2}h$
Then
$h=\frac{3V}{\pi .r^2}$ ...(i)
Now
$CSA$ $=\pi rl$
$=\pi r\left(\sqrt{h^2+r^2}\right)$
$=\pi r\sqrt{\frac{9V^2}{{\pi }^2{r}^4}+r^2}$
$=\frac{\pi .r}{\pi .r^2}\sqrt{9V^2+{\pi }^2{r}^6}$
$=\frac{\sqrt{9V^2+{\pi }^2{r}^6}}{r}$
Let $A=CS{A}^2$
$=\frac{9V^2+{\pi }^2{r}^6}{r^2}$
$=\frac{9V^2}{r^2}+{\pi }^2{r}^4$
$\frac{dA}{dr}=\frac{-18V^2}{r^3}+4{\pi }^2{r}^3$ $=0$
Hence
$\frac{18V^2}{r^3}=4{\pi }^2{r}^3$
$r^6=\frac{18V^2}{4{\pi }^2}$
$r^3=\frac{3\sqrt{2}V}{2\pi }$
$r^2.r=\frac{3\sqrt{2}V}{2\pi }$...(ii)
Now
$h=\frac{3V}{\pi .r^2}$ Or
$r^2=\frac{3V}{\pi h}$
Substituting in ii gives us
$r^2.r=\frac{3\sqrt{2}V}{2\pi }$
$\frac{3V}{\pi h}.r=\frac{3\sqrt{2}V}{2\pi }$
$\frac{r}{h}=\frac{\sqrt{2}}{2}$
$\frac{r}{h}=\frac{1}{\sqrt{2}}$
$h=\sqrt{2}r$.
Hence proved.