The radius of the base of a cone is increasing at the rate 3 ~cm / minute and the altitude is decreasing at the rate 4 ~cm / minute. The rate at which the lateral surface area is changing, when the radius is 7 cm and altitude is 24 cm . is

Step-by-step solution

Given a right circular cone with base radius r , altitude h , and slant height l = r^2 + h^2 , the lateral surface area is S = r l . Differentiating with respect to time t : dS dt = [ dr dt l + r 1 2 r^2 + h^2 ( 2r dr dt + 2h dh dt ) ] At the instant r = 7 cm, h = 24 cm, the slant height is l = 7^2 + 24^2 = 25 cm. With dr dt = 3 cm/min and dh dt = -4 cm/min: dS dt = [ 3 25 + 7 25 ( 7 3 + 24 (-4) ) ] = [ 75 + 7 25 (21 - 96) ] dS dt = [ 75 + 7 25 (-75) ] = (75 - 21) = 54 cm²/min Final answer: D