The maximum volume (in cu.m) of the right circular cone having slant height 3 m is :

Question

Accepted Answer

$$ \therefore $$   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 $$\pi $$ $$ \times $$ $${2 \over 3} \times {1 \over {\sqrt 3 }}$$

= $$2\sqrt 3 \,\pi $$

Solutions · 1