Answer:
[tex]11304 \frac{in^{3}}{s}[/tex]
Explanation:
r = radius of right circular cone = 150 in
h = height of right circular cone = 144 in
[tex]\frac{dr}{dt}[/tex] = rate at which radius increase = 1.5 in/s
[tex]\frac{dh}{dt}[/tex] = rate at which height decrease = - 2.4 in/s
Volume of the right circular cone is given as
[tex]V = \frac{\pi r^{2}h}{3}[/tex]
Taking derivative both side relative to "t"
[tex]\frac{dV}{dt} = \frac{\pi }{3}\left ( r^{2}\frac{dh}{dt} \right + 2rh\frac{dr}{dt})[/tex]
[tex]\frac{dV}{dt} = \frac{3.14 }{3}\left ( (150)^{2}(- 2.4) + 2(150)(144)(1.5))[/tex]
[tex]\frac{dV}{dt} = 11304 \frac{in^{3}}{s}[/tex]
