What is the semi vertical angle of a right circular cone of maximum volume and given slant height?

Concept:

To find maxima and minima of a function y = f(x), follow these steps.

Step 1

\(Find\;\frac{{dy}}{{dx}},\;and\;put\frac{{dy}}{{dx}} = 0.\)

Find the value of x and this value is said to be the stationary point, this is a necessary condition to find the extremum value of a function.

Step 2

\(Find\;\frac{{{d^2}y}}{{d{x^2}}}\;\)

Check the value at the stationary point obtained in Step 1.

A function f(x) has a maxima at x = a if f’(a) = 0 and f”(a) < 0

A function f(x) has a minima at x = a if f’(a) = 0 and f”(a) > 0

A function f(x) has no maxima and minima at x = a if f’(a) = 0 and f”(a) = 0.

Calculation:

In the above diagram

θ = semi-vertical angle, h = height, l = slant height, r = radius of the circular base

The volume of the right circular cone is given by -

\(V = \frac{1}{3}\;\pi {r^2}h\)

In ΔABC

r = l × sin θ and h = l × cos θ 

\(∴ V = \frac{1}{3}\;\pi \;{\left( {lsin\;\theta } \right)^2}\left( {lcos\;\theta } \right) \Rightarrow \frac{1}{3}\;\pi {l^3}{\sin ^2}\theta \cos \theta \)

Step 1:

\(\frac{{dV}}{{d\theta }} = 0\)

\(\left( {\frac{1}{3}\;\pi {l^3}{{\sin }^2}\theta } \right)\left( { - \sin \theta } \right) + \left( {\frac{1}{3}\;\pi {l^3}\cos \theta } \right)\left( {2\sin \theta \cos \theta } \right) = 0\)

\(\left( {\frac{1}{3}\;\pi {l^3}} \right)\left( { - si{n^3}\theta + 2\sin \theta {{\cos }^2}\theta } \right) = 0\)

\( - \sin \theta \left( {{{\sin }^2}\theta - 2{{\cos }^2}\theta } \right) = 0\)

If sin θ = 0

⇒ θ = 0° which is not possible

∴ sin2 θ – 2cos2 θ = 0

\(\frac{{{{\sin }^2}\theta }}{{{{\cos }^2}\theta }} = 2 \ldots \ldots \left( i \right)\)

tan2 θ = 2

\(\tan \theta = \sqrt 2 \)

\(\theta = {\tan ^{ - 1}}\sqrt 2 \)

Step 2:

\(\frac{{{d^2}V}}{{d{\theta ^2}}} = 0\)

\(\frac{{dV}}{{d\theta }} = \left( {\frac{1}{3}\;\pi {l^3}} \right)\left( { - si{n^3}\theta + 2\sin \theta {{\cos }^2}\theta } \right)\)

\(\frac{1}{3}\;\pi {l^3}\left( { - 3{{\sin }^2}\theta \cos \theta + 2\sin \theta \cdot 2\cos \theta \left( { - \sin \theta } \right) + 2{{\cos }^2}\theta \cdot \cos \theta } \right) = 0\)

\(\left( { - 3{{\sin }^2}\theta \cos \theta + 2\sin \theta \cdot 2\cos \theta \left( { - \sin \theta } \right) + 2{{\cos }^2}\theta \cdot \cos \theta } \right) = 0\)

\( - 3{\sin ^2}\theta \cos \theta - 4{\sin ^2}\theta \cos \theta + 2{\cos ^3}\theta = 0\)

\(Using\;\frac{{{{\sin }^2}\theta }}{{{{\cos }^2}\theta }} = 2\)

\(( (- 3 )\times (2{\cos ^2}\theta \cos \theta)) - ((4) \times (2{\cos ^2}\theta \cos \theta)) + 2{\cos ^3}\theta = 0\)

\( - 6{\cos ^3}\theta - 8{\cos ^3}\theta + 2{\cos ^3}\theta = 0\)

\( - 12{\cos ^3}\theta \)

Semi-vertical angle θ is less than 90° i.e it lies in 1st quadrant.

cos θ is positive in 1st quadrant

\(∴ \frac{{{d^2}V}}{{d{\theta ^2}}} < 0\)

∴ The volume will be maximum when \({\bf{\theta }} = {\tan ^{ - 1}}\sqrt 2 \)

Open in App

With usual notation, given that total surface area $S=\pi rl+\pi r^2$
$\Rightarrow S=\pi r\sqrt{r^2+h^2}+\pi r^2(\because l=\sqrt{r^2+h^2})$
$\Rightarrow \frac{S}{\pi r}-r=\sqrt{r^2+h^2}\Rightarrow \frac{S^2}{{\pi }^2{r}^2}-\frac{2S}{\pi }=h^2\Rightarrow h=\sqrt{\frac{S^2}{{\pi }^2{r}^2}-\frac{2S}{\pi }}(\because \frac{S^2}{{\pi }^2{r}^2}>\frac{2S}{\pi })\dots \left(i\right)$
and volume $V=\frac{1}{3}\pi r^{2}h=\frac{1}{3}\pi r^2\sqrt{(\frac{S^2}{{\pi }^2{r}^2}-\frac{2S}{\pi })}$
$\Rightarrow V=\frac{r}{3}\sqrt{S^2-2S\pi r^2},r^2<\frac{S}{2\pi }$ i.e., $0<r<\sqrt{\frac{S}{2\pi }}$
Since, V is maximum, then $V^2$ is maximum
Now, $V^2=\frac{S^2{r}^2}{9}-\frac{2S\pi r^4}{9},0<r<\sqrt{\frac{S}{2\pi }}$
$\therefore \frac{d}{dr}\left(V^2\right)=\frac{2r{S}^2}{9}-\frac{8S\pi r^3}{9}$
and $\frac{d}{d{r}^2}\left(V^2\right)=\frac{2S^2}{9}-\frac{24S\pi r^2}{9}$
For maxima put $\frac{dV}{dr}=0$
$\Rightarrow \frac{2r{S}^2}{9}-\frac{8S\pi r^3}{9}=0\Rightarrow r^2=\frac{S}{4\pi }\Rightarrow \frac{d^2\left(V^2\right)}{d{r}^2}<0forr=\sqrt{\frac{S}{4\pi }}$
From Eq. (i) $h=\sqrt{\frac{S^2}{{\pi }^2{r}^2}-\frac{2S}{\pi }}=\sqrt{\frac{S^2\left(4\pi \right)}{{\pi }^{2}S}-\frac{2S}{\pi }}=\sqrt{\frac{2S}{\pi }}$
If $\theta$ is semi-vertical angle of the cone when the volume is maximum, then in right triangle AOC,

$sin\theta =\frac{r}{\sqrt{r^2+h^2}}=-\frac{\sqrt{\frac{S}{4\pi }}}{\sqrt{\frac{S}{4\pi }+\frac{2S}{\pi }}}=\frac{1}{\sqrt{1+8}}i.e.,\theta =si{n}^{-1}\left(\frac{1}{3}\right)$