JEE Main 2018 — Mathematics Calculus

In \triangle AMC,AM=3\mathrm{sin}2\theta &MC=3\mathrm{cos}2\theta

$V=\frac{1}{3}\pi {r}^{2}h$V=31​πr2h where, $r$r is radius and $h$h is height of cone

$$\Rightarrow V=\frac{1}{3}\pi {(3\mathrm{sin}⁡2\theta )}^{2}(3+3\mathrm{cos}2\theta )$$⇒V=31​π(3sin⁡2θ)2(3+3cos2θ)

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

\Rightarrow V=\pi (36{\mathrm{sin}}^{2}⁡\theta {\mathrm{cos}}^{2}⁡\theta )(2{\mathrm{cos}}^{2}⁡\theta ) [\because \mathrm{sin}⁡2\theta =2\mathrm{sin}⁡\theta \mathrm{cos}⁡\theta &\mathrm{cos}⁡2\theta =2{\mathrm{cos}}^{2}⁡\theta -1]

$$=72\pi {\mathrm{sin}}^{2}⁡\theta {\mathrm{cos}}^{4}⁡\theta$$=72πsin2⁡θcos4⁡θ

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

$$\frac{dv}{d\theta }=72\pi [2\mathrm{sin}⁡\theta {\mathrm{cos}}^{5}⁡\theta -4{\mathrm{sin}}^{3}⁡\theta {\mathrm{cos}}^{3}⁡\theta ]$$dθdv​=72π[2sin⁡θcos5⁡θ−4sin3⁡θcos3⁡θ]

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

$$\Rightarrow 72\pi [2\mathrm{sin}⁡\theta {\mathrm{cos}}^{5}⁡\theta -4{\mathrm{sin}}^{3}⁡\theta {\mathrm{cos}}^{3}⁡\theta ]=0\Rightarrow {\mathrm{tan}}^{2}⁡\theta =\frac{1}{2}$$⇒72π[2sin⁡θcos5⁡θ−4sin3⁡θcos3⁡θ]=0⇒tan2⁡θ=21​

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

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

$$=\pi r\sqrt{{(3+3\mathrm{cos}⁡2\theta )}^{2}+{(3\mathrm{sin}⁡2\theta )}^{2}}[\because l=\sqrt{{r}^{2}+{h}^{2}}]$$=πr(3+3cos⁡2θ)2+(3sin⁡2θ)2​[∵l=r2+h2​]

$$=\pi (3\mathrm{sin}⁡2\theta )\sqrt{36{\mathrm{cos}}^{2}⁡\theta }=18\pi (2\mathrm{sin}⁡\theta {\mathrm{cos}}^{2}⁡\theta )$$=π(3sin⁡2θ)36cos2⁡θ​=18π(2sin⁡θcos2⁡θ)

$$=36\pi \frac{1}{\sqrt{3}}.\frac{2}{3}=\frac{24\pi }{\sqrt{3}}=8\sqrt{3}\pi$$=36π3​1​.32​=3​24π​=83​π