JEE Main 2019 — Application Of Derivatives Question with Solution

$$\therefore$$   h = 3 cos$$\theta$$

r = 3 sin$$\theta$$

We know volume of right circular cone,

V = $$\frac{1}{3}\pi r^{2}h$$

= $$\frac{1}{3}\pi$$(3 sin$$\theta$$)² 3 cos$$\theta$$

= 9 $$\pi$$ sin²$$\theta$$ cos$$\theta$$

For maximum or minimum value of volume,

$$\frac{dv}{d\theta }$$ = 0

$$\therefore$$   (2sin$$\theta$$ cos$$\theta$$) cos$$\theta$$ + 3sin²$$\theta$$ .($$-$$ sin$$\theta$$) = 0

$$\Rightarrow$$  2 sin$$\theta$$ cos²$$\theta$$ $$-$$ sin³$$\theta$$ = 0

$$\Rightarrow$$  2 sin$$\theta$$(1 $$-$$ sin²$$\theta$$) $$-$$ sin³ $$\theta$$ = 0

$$\Rightarrow$$  2 sin$$\theta$$ $$-$$ 2 sin³$$\theta$$ $$-$$ sin³$$\theta$$ = 0

$$\Rightarrow$$  3 sin³$$\theta$$ = 2 sin$$\theta$$

$$\Rightarrow$$   sin²$$\theta$$ = $$\frac{2}{3}$$

$$\Rightarrow$$  sin$$\theta$$ = $$\sqrt{\frac{2}{3}}$$

$$\frac{d^{2}v}{d{\theta }^2}$$ = 2cos$$\theta$$ $$-$$ 3(3sin$$\theta$$ cos$$\theta$$)

= 2 cos$$\theta$$ $$-$$ 9 sin$$\theta$$ cos$$\theta$$

= 2 $$\times$$ $$\frac{1}{\sqrt{3}}$$ $$-$$ 9 $$\times$$ $$\frac{\sqrt{2}}{\sqrt{3}}$$ $$\times$$ $$\frac{1}{\sqrt{3}}$$

= $$\frac{2}{\sqrt{3}}-3\sqrt{2}<0$$

$$\therefore$$  Volume is maximum

when sin$$\theta$$ = $$\sqrt{\frac{2}{3}}$$

$$\therefore$$  Maximum volume is

= 9 $$\pi$$ $${\left(\sqrt{\frac{2}{3}}\right)}^2\times \frac{1}{\sqrt{3}}$$

= 9 $$\pi$$ $$\times$$ $$\frac{2}{3}\times \frac{1}{\sqrt{3}}$$

= $$2\sqrt{3}\pi$$