Question

a tank contains 150 gal of fresh water. a solution containing 5 lb/gal of soluble lawn fertilizer runs into the tank at the rate of 1 gal/min, and the mixture is pumped out of the tank at the rate of 4 gal/min. find the maximum amount of fertilizer in the tank and the time required to reach the maximum. find the time required to reach the maximum amount of fertilizer in the tank. t=□ min (type an integer or decimal rounded to the nearest tenth as needed.)

Explanation:

Step1: Set up the differential equation

Let \(y(t)\) be the amount of fertilizer (in pounds) in the tank at time \(t\) (in minutes).
The volume of the tank at time \(t\) is \(V(t)=150+(1 - 4)t=150 - 3t\).
The rate of change of \(y(t)\) is given by \(\frac{dy}{dt}=\text{rate in}-\text{rate out}\).
The rate in is \(5\times1 = 5\) lb/min.
The rate out is \(\frac{y(t)}{V(t)}\times4=\frac{4y}{150 - 3t}\) lb/min.
So, \(\frac{dy}{dt}=5-\frac{4y}{150 - 3t}\), which can be rewritten as \(\frac{dy}{dt}+\frac{4}{150 - 3t}y = 5\).
The integrating factor is \(\mu(t)=e^{\int\frac{4}{150 - 3t}dt}\). Let \(u = 150-3t\), then \(du=-3dt\) and \(\int\frac{4}{150 - 3t}dt=-\frac{4}{3}\ln|150 - 3t|\), so \(\mu(t)=(150 - 3t)^{-\frac{4}{3}}\).
Multiply the differential equation by the integrating factor: \((150 - 3t)^{-\frac{4}{3}}\frac{dy}{dt}+(150 - 3t)^{-\frac{4}{3}}\frac{4}{150 - 3t}y = 5(150 - 3t)^{-\frac{4}{3}}\).
The left - hand side is the derivative of \(y(150 - 3t)^{-\frac{4}{3}}\) with respect to \(t\).
Integrating both sides: \(y(150 - 3t)^{-\frac{4}{3}}=\int5(150 - 3t)^{-\frac{4}{3}}dt\).
Let \(u = 150-3t\), \(du=-3dt\), then \(\int5(150 - 3t)^{-\frac{4}{3}}dt=-\frac{5}{3}\int u^{-\frac{4}{3}}du=5u^{-\frac{1}{3}}+C = 5(150 - 3t)^{-\frac{1}{3}}+C\).
Since \(y(0) = 0\) (initial amount of fertilizer is \(0\)), when \(t = 0\), \(0\times(150)^{-\frac{4}{3}}=5\times(150)^{-\frac{1}{3}}+C\), so \(C=-5\times(150)^{-\frac{1}{3}}\).
Then \(y(t)=5(150 - 3t)-5\times(150 - 3t)^{\frac{4}{3}}\times(150)^{-\frac{1}{3}}\).

Step2: Find the maximum

Take the derivative of \(y(t)\) with respect to \(t\): \(y^\prime(t)=- 15+20(150 - 3t)^{\frac{1}{3}}\times(150)^{-\frac{1}{3}}\).
Set \(y^\prime(t)=0\):
\(-15 + 20(150 - 3t)^{\frac{1}{3}}\times(150)^{-\frac{1}{3}}=0\).
\(20(150 - 3t)^{\frac{1}{3}}\times(150)^{-\frac{1}{3}}=15\).
\((150 - 3t)^{\frac{1}{3}}=\frac{15\times(150)^{\frac{1}{3}}}{20}=\frac{3\times(150)^{\frac{1}{3}}}{4}\).
Cube both sides: \(150-3t=\frac{27\times150}{64}\).
\(3t = 150-\frac{27\times150}{64}=\frac{150\times(64 - 27)}{64}=\frac{150\times37}{64}\).
\(t=\frac{150\times37}{64\times3}=\frac{185}{32}\approx5.8\)

Answer:

\(t = 5.8\) min