About rate, my ideia is wrong?

Question

The following problem: An oil tank should be drained for cleaning. V oil gallons are left in the tank T minutes after the drain started, where $V = 40*(50 - t^2)$

So, I change to: $f(t) = 40*(50 - t^2)$

a) the median rate which the oil is drained out of the tank during the first 20 minutes.

a) My I thought about to use the following ideia: $\dfrac{f(20) - f(0)}{20}$ Result would be: $\dfrac{-14000-2000}{20}$ Resulting in -800gal/min

The book give-me the answer = $3200 gal/min$, I'm wrong or my book is wrong?

In terms of applications, $f(20)$ occurs well after the tank has been fully drained (which happens at about $t=7$ minutes). Are you sure the question asked for the median rate? — Roger Burt, Jun 16 '14 at 18:31
Cálculo com Geometria Analítica, Volume 1 - George F. Simmons. Translated: Calculus with analytic geometry, Vol 1 - George F. Simmons. — mastergoo, Jun 16 '14 at 18:44

score 1 · Accepted Answer · answered Jun 16 '14 at 22:04

Technically, the median is the rate that the flow rate is greater than half the time an less than half the time. As the tank is empty after $\sqrt {50} \approx 7.071$ min, the flow rate function cannot be correct after that time. If we assume the flow rate is zero after the tank is empty, the median over $20$ minutes is $0$-it spends almost $13$ minutes with zero flow rate.

The flow rate is the (negative) derivative of the volume, so is $80t$ If you are looking for the median over the period of flow, that happens halfway (in time) through the flow-it flows slower the first half of the time and faster the second half (which is very strange-you would expect a full tank to have more pressure at the outlet and empty faster). At $t=\frac 12\sqrt {50}$, the flow rate is $40 \sqrt{50} \approx 280$

Clearly there is something wrong with the book answer as the maximum flow rate is about $561$.

About rate, my ideia is wrong?

1 Answers1