JEE Main 2018 — Application of Derivatives Question with Solution

In $△AMC, AM=3\sin 2\theta \& MC=3\cos 2\theta$

$V=\frac{1}{3}\pi r^{2}h$ where, $r$ is radius and $h$ is height of cone

$\Rightarrow V=\frac{1}{3}\pi {\left(3\sin 2\theta \right)}^2\left(3+3\cos 2\theta \right)$

(since, radius of cone $=AM$ and height of cone $=MC$)

$$\begin{gathered} \Rightarrow V=\pi \left(36{\sin }^2\theta {\cos }^2\theta \right)\left(2{\cos }^2\theta \right) \\ \left[\because \sin 2\theta =2\sin \theta \cos \theta \& \cos 2\theta =2{\cos }^2\theta -1\right] \end{gathered}$$

$=72\pi {\sin }^2\theta {\cos }^4\theta$

Differentiating both sides with respect to $\theta$, we get

$\frac{dv}{d\theta }=72\pi \left[2\sin \theta {\cos }^5\theta -4{\sin }^3\theta {\cos }^3\theta \right]$

For maximum value, $\frac{dV}{d\theta }=0$

$\Rightarrow 72\pi \left[2\sin \theta {\cos }^5\theta -4{\sin }^3\theta {\cos }^3\theta \right]=0\Rightarrow {\tan }^2\theta =\frac{1}{2}$

Thus, volume is maximum when $\tan \theta =\frac{1}{\sqrt{2}}$

Hence, curved surface area $S=\pi rl$

$=\pi r\sqrt{{\left(3+3\cos 2\theta \right)}^2+{(3\sin 2\theta )}^2} \left[\because l=\sqrt{r^2+h^2}\right]$

$=\pi \left(3\sin 2\theta \right)\sqrt{36{\cos }^2\theta }=18\pi \left(2\sin \theta {\cos }^2\theta \right)$

$=36\pi \frac{1}{\sqrt{3}}.\frac{2}{3}=\frac{24\pi }{\sqrt{3}}=8\sqrt{3}\pi$