Answer:
Impossible. t = 30 minutes.
Step-by-step explanation:
We are given the function.
[tex]\displaystyle v(t)=20000\left(1-\frac{t}{20}\right)^2, \text{ } 0 \leq t \leq 20[/tex]
Where v(t) represents the amount of gallons remaining after t minutes.
We want to find at what time t is the instanteous rate of change from the tank 1000 gallons per minute.
Hence, find the derivative of the function with respect to time t:
[tex]\displaystyle \begin{aligned} \frac{d}{dt}[v(t)] &=\frac{d}{dt}\left[20000\left(1-\frac{t}{20}\right)^2\right] \\ \\ v'(t) & = 20000\frac{d}{dt}\left[\left(1-\frac{t}{20}\right)^2\right] \\ \\ & = 20000\left(2\left(1-\frac{t}{20}\right)^1\right)\left(-\frac{1}{20}\right) \\ \\ & = -2000\left(1-\frac{t}{20}\right) \\ \\ & = 100t - 2000\end{aligned}[/tex]
To find the time at which the instanteous rate is 1000 gallons/minute, set v'(t) = 1000 and solve for t:
Hence:
[tex]\displaystyle \begin{aligned} 1000 &= 100t - 2000 \\ \\ 100t & = 3000 \\ \\ t &= 30\text{ minutes}\end{aligned}[/tex]
Therefore, after 30 minutes, the instantaneous rate of change will be 1000 gallons per minute.
However, v(t) is only defined for 0 ≤ t ≤ 30.
In conclusion, it is impossible for our instantaneous rate of change to ever reach 1000 gallons per minute.
