Question

The semi-vertical angle of a right circular cone of the maximum volume of a given slant height is ${\tan }^{-1}\sqrt{(}a)$. Then a equals

Answer 1

we know that

$l^2=h^2+r^2$ ---(1)

Volume of cone $=\frac{1}{3}\pi r^{2}h$

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

$\Rightarrow V=\frac{\pi }{3}\left(l^2-h^2\right)h$ ( from (i))

$\Rightarrow V=\frac{\pi }{3}\left(l^{2}h-h^3\right)$

$\Rightarrow \frac{dV}{dh}=\frac{\pi }{3}\left(l^2-3h^2\right)$

$\Rightarrow \frac{d^{2}V}{d{h}^2}=\frac{\pi }{3}\left(-6h\right)$

for critical points

$\frac{dV}{dh}=0$

$\Rightarrow \frac{\pi }{3}\left(l^2-3h^2\right)=0$

$\Rightarrow l^2-3h^2=0$

$\Rightarrow l=\sqrt{3}h$

$\Rightarrow h=\frac{l}{\sqrt{3}}$

${\left(\frac{d^{2}V}{d{h}^2}\right)}_{h=\frac{l}{\sqrt{3}}}=\frac{\pi }{3}\left(\frac{-6l}{\sqrt{3}}\right)<0$

Therefore volume of cone is maximum when

$h=\frac{l}{\sqrt{3}}$

$\Rightarrow l=\sqrt{3}h$

$\Rightarrow h^2+r^2=3h^2$ ( from (1))

$\Rightarrow r^2=2h^2$

$\Rightarrow r=\sqrt{2}h$

$\Rightarrow \frac{r}{h}=\sqrt{2}$

$\Rightarrow \tan \alpha =\sqrt{2}$

$\Rightarrow \alpha ={\tan }^{-1}\sqrt{2}$