Question

Show that the semi-vertical angle of the cone of the maximum volume and of given slant height is $tan^{-1}\sqrt{2}.$

Answer 1

Let $θ$ be the semi-vertical angle of the cone.

It is clear that θ$$∈[0,\frac{\pi}{2}].

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

The slant height of the cone is given as constant.

Now$,r=lsin θ$ and $h=lcos θ$

The volume$(V)$of the cone is given by,

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

$=\frac{1}{3}\pi (l^{2}sin^{2}θ)(lcosθ)$

$=\frac{1}{3}\pi l^{2}sin^{2}θcosθ$

∴\frac{dV}{dθ}$$=l^{3}\frac{\pi}{3}$$[sin^{2}θ(-sinθ)+cosθ(2sinθcosθ)]

$=\frac{l^{3}\pi}{3}[-sin^{3}+2sinθcos^{2}θ]$

$\frac{d^{2}V}{dθ}$==\frac{l^{3}\pi}{3}$$[-3sin^{2}θcosθ+2cos^{3}θ-4sin^{2}θcosθ]

=\frac{l^{3}\pi}{3}$$[2cos^{3}θ-7sin^{2}θcosθ]

Now,$\frac{dV}{dθ}=0$

$⇒sin^{3}θ=2sinθcos^{2}θ$

$⇒tan^{2}θ=2$

$⇒tanθ=\sqrt{2}$

$⇒θ=tan^{-1}\sqrt{2}$

Now,when $⇒θ=tan^{-1}\sqrt{2}$,then $tan^{2}θ=2$ or $sin^{2}θ=2cos^{2}θ.$

Then,we have:

$\frac{d^{2}V}{dθ}$==\frac{l^{3}\pi}{3}$$[2cos^{3}θ-14cos^{3}θ]$$=-4\pi l^{3}cos^{3}θ<0 for θ∈[0,\frac{\pi}{2}]

By second derivative test,the volume(V)is the maximum when $θ=tan^{-1}\sqrt{2}$

Hence,for a given slant height,the semi-vertical angle of the cone of the maximum

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