Question

The semi vertical angle of right circular cone of maximum volume of a given slant height is

• A. ${\cos }^{-1}\sqrt{2}$
• B. ${\sin }^{-1}\sqrt{2}$
• C. ${\tan }^{-1}\sqrt{3}$
• D. ${\tan }^{-1}\sqrt{2}$

Answer 1

Answer: (D)

${\tan }^{-1}\sqrt{2}$

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

And let $\alpha$ be the semi vertical angle of the cone

$l^2=h^2+r^2$

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

$V=\frac{1}{3}\pi h(l^2-h^2)$

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

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

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

For critical point $\frac{dV}{dh}=0$

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

$l^2=3h^2$

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

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

So, volume of cone is maximum, when

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

$l^2=3h^2$

$h^2+r^2=3h^2$

$r=\sqrt{2}h$

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

$tan\alpha =\sqrt{2}$

$\alpha =ta{n}^{-1}\sqrt{2}$