Given that the volume of water remaining in the tank after t minutes is given by the function
[tex]V(t)=20,000\left(1- \frac{t}{20} \right)^2[/tex]
where V is in gallons, 0 ≤ t ≤ 20 is in minutes, and t = 0 represents the instant the tank starts draining.
The rate at which water is draining four and a half minutes after it begins is given by
[tex]\left.
\frac{dV}{dt} \right|_{t=4 \frac{1}{2} = \frac{9}{2}
}=\left[40,000\left(1- \frac{t}{20} \right)\left(- \frac{1}{20}
\right)\right]_{t= \frac{9}{2} } \\ \\ =\left[-2,000\left(1-
\frac{t}{20} \right)\right]_{t= \frac{9}{2} }=-2,000\left(1-
\frac{4.5}{20} \right) \\ \\ =-2,000(1-0.225)=-2,000(0.775)=-1,550\,
gallons\ per\ minute[/tex]
Therefore, the water is draining at a rate of 1,550 gallons per minute four ans a half minutes after it begins.
Answer option E is the correct answer.
