JEE Main 2018 — Application Of Derivatives Question with Solution

Sphere of radius r = 3 cm

Let b, h be base radius and height of cone respectively.

So, volume of cone = $$\frac{1}{2}$$ $$\pi$$b²h



In right $$\Delta$$ABC by Pythagoras theorem

(h $$-$$ r)² + b² = r²

$$\Rightarrow$$ b² = r² $$-$$ (h $$-$$ r)² = r² $$-$$ (h² $$-$$ 2hr + r²) = 2hr $$-$$ h²

$$\therefore$$ Volume (v) = $$\frac{1}{3}$$ $$\pi$$h[2hr $$-$$ h²] = $$\frac{1}{3}$$ [ 2h²r $$-$$ h³]

$$\frac{dv}{dh}$$ = $$\frac{1}{3}$$ [4hr $$-$$ 3h²] = 0

$$\Rightarrow$$ h (4r $$-$$ 3h) = 0

$$\frac{d^{2}v}{d{h}^2}$$ = $$\frac{1}{3}$$ [4r $$-$$ 6h]

At h = $$\frac{4r}{3}$$, $$\frac{d^{2}y}{d{h}^2}$$ = $$\frac{1}{3}$$ $$\left[4r-\frac{4r}{3}\times 6\right]=\frac{1}{3}\left[4r-8r\right]<0$$

$$\Rightarrow$$ maximum volume cours at h = $$\frac{4r}{3}$$ = $$\frac{4}{3}$$ $$\times$$ 3 = 4 cm

As from (1),

(h $$-$$ r)² + b² = r²

$$\Rightarrow$$ b² = 2hr $$-$$ h² = 2.$$\frac{4r}{3}$$ r $$-$$ $$\frac{16r^2}{9}$$ = $$\frac{8r^2}{3}$$ $$-$$ $$\frac{16r^2}{9}$$

= $$\frac{\left(24-16\right)r^2}{9}$$ = $$\frac{8r^2}{9}$$

$$\Rightarrow$$ b = $$\frac{2\sqrt{2}}{3}$$ r = 2 $$\sqrt{2}m$$

Therefore curved surface area = $$\pi bl$$

= $$\pi$$b$$\sqrt{h^2+r^2}$$ = $$\pi$$2$$\sqrt{2}$$ $$\sqrt{4^2+8}$$ = 8$$\sqrt{3}$$$$\pi$$ cm²