Answer :
The rate of change of volume is 418.6 inch³/s .
Rate of increase of radius with time is ( dr/dt ) = 5
rate of decrease of height with time is (dh/dt ) = -3
volume of a cone is given by (V) = [tex]\frac{1}{3} \pi r^{2}h[/tex]
Now the rate of change of volume with respect to time is given by dV/dt
∴ dV/dt = d( [tex]\frac{1}{3} \pi r^{2}h[/tex]) / dt
dV/dt = [tex]\frac{1}{3} \pi (2rh\frac{dr}{dt} + r^{2}\frac{dh}{dt} )[/tex]
value of radius and height when volume is changing are :
r = 20 inches
h = 40 inches
dV/dt = [tex]\frac{1}{3} \pi ([/tex] 2×20×40 ×5 - [tex]20^{2}[/tex]×3 )
dV/dt = [tex]\frac{1}{3} \pi[/tex](1600 - 1200)
dV/dt = (400/3) 3.14
dV/dt = 418.66 inch³/s
So the rate of change of the volume is 418.66 inch³/s.
learn more about rate of change problems here :
https://brainly.com/question/22716418
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