Question

Statement I The semi-vertical angle of the cone of the maximum volume and of given slant height is ${\tan }^{-1}\left(\frac{1}{3}\right)$.
Statement II The semi-vertical angle of right circular cone of given surface area and maximum volume is ${\sin }^{-1}\sqrt{2}$.

• A. Only statement I is true
• B. Only statement II is true
• C. Both the statements are true
• D. Both the statements are false

Answer 1

Answer: (D)

I. Let $\theta$ be the semi-vertical angle of the cone.
It is clear that $\theta \in \left(0,\frac{\pi }{2}\right)$.
Let $r,h$ and $I$ be the radius, height and the slant height of the cone respectively.
The slant height of the cone is given i.e., consider as constant.
Now, $r=I\sin \theta$ and $h=I\cos \theta$
Let $V$ be the volume of the cone; $V=\frac{\pi }{3}{r}^{2}h$

$\Rightarrow V=\frac{1}{3}\pi \left(I^2{\sin }^2\theta \right)(l\cos \theta )=\frac{1}{3}\pi l^3{\sin }^2\theta \cos \theta$
On differentiating w.r.t. $\theta$, we get
$\frac{dV}{d\theta }=\frac{I^3\pi }{3}\left[{\sin }^2\theta (-\sin \theta )+\cos \theta (2\sin \theta \cos \theta )\right]$
$=\frac{I^3\pi }{3}\left(-{\sin }^3\theta +2\sin \theta {\cos }^2\theta \right)$
and $\frac{d^{2}V}{d{\theta }^2}=\frac{l^3\pi }{3}\left(-3{\sin }^2\theta \cos \theta +2{\cos }^3\theta -4{\sin }^2\theta \cos \theta \right)$
$=\frac{l^3\pi }{3}\left(2{\cos }^3\theta -7{\sin }^2\theta \cos \theta \right)$
For maxima put $\frac{dV}{d\theta }=0$
$\Rightarrow {\sin }^3\theta =2\sin \theta {\cos }^2\theta$
$\Rightarrow {\tan }^2\theta =2$
$\Rightarrow \tan \theta =\sqrt{2}$
$\Rightarrow \theta ={\tan }^{-1}\sqrt{2}$
Now, when $\theta ={\tan }^{-1}\sqrt{2}$, then ${\tan }^2\theta =2$ or ${\sin }^2\theta =2{\cos }^2\theta$
Then, we have
$\frac{d^{2}V}{d{\theta }^2}=\frac{l^3\pi }{3}\left(2{\cos }^3\theta -14{\cos }^3\theta \right)=-4\pi l^3{\cos }^3\theta <0$ for
$\theta \in \left(0,\frac{\pi }{2}\right)$
$\therefore$ By second derivative test, the volume $V$ is maximum when $\theta ={\tan }^{-1}\sqrt{2}$
Hence, for a given slant height, the semi-vertical angle of the cone of the maximum volume is ${\tan }^{-1}\sqrt{2}$.
II. With usual notation, given that total surface area
$S=\pi rl+\pi r^2$
$\Rightarrow S=\pi r\sqrt{r^2+h^2}+\pi r^2\left[\because I=\sqrt{r^2+h^2}\right]$
$\Rightarrow \frac{S}{\pi r}-r=\sqrt{r^2+h^2}\Rightarrow \frac{S^2}{{\pi }^2{r}^2}-\frac{2S}{\pi }=h^2$
$\Rightarrow h=\sqrt{\frac{S^2}{{\pi }^2{r}^2}-\frac{2S}{\pi }},\left(\because \frac{S^2}{{\pi }^2{r}^2}>\frac{2S}{\pi }\right).....$(i)
and volume $V=\frac{1}{3}\pi r^{2}h$
$=\frac{1}{3}\pi r^2\sqrt{\frac{S^2}{{\pi }^2{r}^2}-\frac{2S}{\pi }}$
$\Rightarrow V=\frac{r}{3}\sqrt{S^2-2S\pi r^2},r^2<\frac{S}{2\pi }i\cdot e.,0<r<\sqrt{\frac{S}{2\pi }}$
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 $\frac{d^2}{d{r}^2}\left(V^2\right)=\frac{2S^2}{9}-\frac{24S\pi r^2}{9}$
For maxima, put $\frac{d}{dr}\left(V^2\right)=0$
$\Rightarrow \frac{2r{S}^2}{9}-\frac{8S\pi r^3}{9}=0$
$\Rightarrow r^2=\frac{S}{4\pi }$
$\Rightarrow r=\sqrt{\frac{S}{4\pi }}$
Here, $\frac{d^2\left(V^2\right)}{d{r}^2}<0$ for $r=\sqrt{\frac{S}{4\pi }}$
So, $V^2$ and hence, $V$ is maximum, when $r=\sqrt{\frac{S}{4\pi }}$
From Eq. (i), $h=\sqrt{\frac{S^2}{{\pi }^2{r}^2}-\frac{2S}{\pi }}=\sqrt{\frac{S^2(4\pi )}{{\pi }^{2}S}-\frac{2S}{\pi }}=\sqrt{\frac{2S}{\pi }}$
If $\theta$ is the 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.,$\theta ={\sin }^{-1}(\frac{1}{3})$
So, the both the given statements are false.