QJEEM19Ma4fe7: JEE Main 2019 Mathematics - Class 12 JEEM19Ma4fe7 Solution: The maximum volume (in cu.m) of the right circular cone having…

$\therefore$    h = 3 cos $\theta$

r = 3 sin $\theta$

We know volume of right circular cone,

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

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

= 9 $\pi$ sin² $\theta$ cos $\theta$

For maximum or minimum value of volume,

${{dv} \over {d\theta }}$ = 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 $\theta$ = ${2 \over 3}$

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

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

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

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

= {2 \over {\sqrt 3 }} - 3\sqrt 2 \, < 0

$\therefore$   Volume is maximum

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

$\therefore$   Maximum volume is

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

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

= $2\sqrt 3 \,\pi$