A solid box is formed by placing a cylinder, having equal height and radius on top of a cube such that, the circular base of cylinder is the inscribed circle for square top of the cube. If the radius of cylinder is changing at the rate (1/2 π+16) cm / s, then the rate of change of volume of the box when radius is 2 cm, is (Assuming that box always remain in the given shape)

Question

A solid box is formed by placing a cylinder, having equal height and radius on top of a cube such that, the circular base of cylinder is the inscribed circle for square top of the cube. If the radius of cylinder is changing at the rate (1/2 π+16) cm / s, then the rate of change of volume of the box when radius is 2 cm, is (Assuming that box always remain in the given shape) (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

⇒ V=π r3+(2 r)3 (d v/d t)=(π+8) 3 r2 ⋅ (d r/d t) =(π+8) ⋅ 3 ⋅ (4/2(π+8)) =6 cc / sec

⇒V=πr3+(2r)3 dtdv​=(π+8)3r2⋅dtdr​ =(π+8)⋅3⋅2(π+8)4​ =6cc/sec

$\Rightarrow V = π r^{3} + (2 r)^{3}$
$\frac{d v}{d t} = (π + 8) 3 r^{2} \cdot \frac{d r}{d t}$
$= (π + 8) \cdot 3 \cdot \frac{4}{2 ( π + 8 )}$
$= 6 cc / sec$