What is the minimum amount of water in the tank from t=0 to t=12?

Question

A tank has 5 gallons 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 N(t)=−4(1.06)^t+16, where N(t) is measured in gallons per hour.Water is being pumped into a tank at a rate modeled by the function Z(t)=−2(1.19)^t+18, where Z(t) is measured in gallons per hour.What is the minimum amount of water in the tank from t=0 to t=12? You may use a calculator and round to the nearest thousandth. Use proper units and show how you got your answer.

Answer 1

Because there are 5 gallons initially, and because the net rate of change of volume is given by Z(t) - N(t) (rate_in - rate_out), the volume at any time t, for 0 ≤ t ≤ 12, is given by V(t) = 5 + ∫₀^t (Z(x) - N(x))dx.

It is easy to see by comparing Z(0) and N(0), Z₀ < N₀, that the volume is decreasing at t = 0. It will continue to do so until Z(t) = N(t), and thereafter the volume will increase (because Z(t) > N(t).

So, graph Z(t) and N(t) on the same axes, use a calculator to solve fo the t-coordinate of the point of intersection. Then, plug that t-value in for the upper bound of integration in the integral expression given above for V(t). That number will be the minimum volume of water in the tank (in gallons) on t ∈ [0 , 12].