Rate In and Rate Out (Absolute Extrema)

Question

A tank has 25 liters of water in it at time t = 0 hours. Water begins to be pumped into the tank at time t = 0. A different pipe is draining water from the tank starting at t = 0.

Water is being removed from the tank at a rate modeled by the function P (t) = -5 (1.18)^t +19, where P(t) is measured in liters per hour.

Water is being pumped into a tank at a rate modeled by the function N(t) = 15(0.87)^t, where N(t), is measured in liters per hour.

What is the minimum amount of water in the tank from t=0 to t=8? You may use a calculator and round to the nearest thousandth.

Answer 1

Absolute min and max can occur at critical values or endpoints.

First get the critical values. Because you’re trying to get a minimum of the amount of water with respect to time, to get the critical values, we need to set the rate at which water is filling the tank to 0.

N(t) - P(t) = 0

15(0.87)^t + 5(1.18)^t -19 = 0

We get two critical values at t = 0.94138 and t = 5.09709

Now the functions they gave us were rates. To get a function for the amount of water in the tank at a given time, you need to take the antiderivative. We’ll call the function F(t)

F(t) = ∫N(t) - P(t)dt

F(t) = 15/ln(0.87)* (0.87)^t + 5/ln(1.18)* (1.18)^t -19t + C

F(t) = -107.71059* (0.87)^t + 30.20884* (1.18)^t -19t + C

At t=0, the tank has 25 liters, so F(0) = 25

Plugging this in, we get C = 102.50175

F(t)= -107.71059*(0.87)^t + 30.20884*(1.18)^t - 19t + 102.50175

Plug the critical values and endpoints in and find the minimum

Answer 2

First just write down what they gave you:

1. W(0) = 25
2. W_in = N = 15(.87)^t
3. W_out = P = -5(1.18)^t + 19

That's all.

Now, you always solve a problem backwards.

Extrema occur at critical points and endpoints.

They gave us the endpoints, but critical points are where W' is 0 or doesn't exist.

The fundamental equation for this subject is W' = W_in - W_out.

So set that equal to 0 and solve for t. You get two times between 0 and 8. You can use a calculator, but I'll just call them t₁ and t₂.

Yay we found all the critical points and endpoints.

We don't care which ones are max or min because they just want the minimum. We just plug all the times into W and see which one is the lowest.

FTC says that W(t) = W(0) + ∫₀^t W'(s) ds (s is just a fake variable so we don't get confused with t).

Use a calculator to plug the four different t values in.