QJEEM18Ma3b3f: JEE Main 2018 Mathematics - Class 12 JEEM18Ma3b3f Solution: If a right circular cone, having maximum volume, is inscribed…

Sphere of radius r = 3 cm

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

So, volume of cone = ${1 \over 2}$ $\pi$ 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) = ${1 \over 3}$ $\pi$ h[2hr $-$ h²] = ${1 \over 3}$ [ 2h²r $-$ h³]

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

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

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

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

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

As from (1),

(h $-$ r)² + b² = r²

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

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

$\Rightarrow$ b = ${{2\sqrt 2 } \over 3}$ 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