A certain cylindrical tank holds 20,000 gallons of water, which can be drained from the bottom of the tank in 20 minutes. The volume V of water remaining in the tank after t minutes is given by the function V(t) = 20,000(1- t/20)^2, where V is in gallons, t = 0 represents the instant the tank starts draining. The average rate of change in volume of water in the tank is -1,000 from t=0 to t = 20. At what time t is the instantaneous rate of the water draining from the tank at 1,000 gallons per minute?

Question

a. t = 12 minutes
b. t = 11 minutes
c. t = 10 minutes
d. t = 8 minutes
e. t = 9 minutes

Answer 1

Huh ?
dv /dt = -1000

v = 20,000 (1 - t/20)^2
dv/dt = 20,000 * 2 (1-t/20) (-1/20)
= -2000*(1 - t/20) = -1000 we want
0.5 = 1-t/20
0.5 = t/20
t = 10