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

Huh ?
dv /dt = 1000
v = 20,000 (1  t/20)^2
dv/dt = 20,000 * 2 (1t/20) (1/20)
= 2000*(1  t/20) = 1000 we want
0.5 = 1t/20
0.5 = t/20
t = 10 👍
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answered by Anonymous
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