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 \(tan^{-1}\sqrt{2}\)