Question

a tank contains 100 gal of fresh water. a solution containing 4 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 3 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 = \boxed{27.8} min
(type an integer or decimal rounded to the nearest tenth as needed.)
the maximum amount of fertilizer in the tank is y = \boxed{} lb
(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 lb) in the tank at time \(t\) (in min).
The volume of the tank at time \(t\) is \(V(t)=100+(1 - 3)t=100-2t\).
The rate of change of \(y\) is \(\frac{dy}{dt}=\text{input rate}-\text{output rate}\).
The input rate is \(4\times1 = 4\) lb/min.
The output rate is \(\frac{y}{V(t)}\times3=\frac{3y}{100 - 2t}\) lb/min.
So, \(\frac{dy}{dt}=4-\frac{3y}{100 - 2t}\).
This is a first - order linear differential equation of the form \(\frac{dy}{dt}+P(t)y = Q(t)\), where \(P(t)=\frac{3}{100 - 2t}\) and \(Q(t)=4\).
The integrating factor is \(\mu(t)=e^{\int P(t)dt}\).
\(\int\frac{3}{100 - 2t}dt=-\frac{3}{2}\ln|100 - 2t|\), so \(\mu(t)=(100 - 2t)^{-\frac{3}{2}}\).

Step2: Solve the differential equation

Multiply the differential equation by the integrating factor:
\((100 - 2t)^{-\frac{3}{2}}\frac{dy}{dt}-3(100 - 2t)^{-\frac{5}{2}}y = 4(100 - 2t)^{-\frac{3}{2}}\).
The left - hand side is the derivative of \((100 - 2t)^{-\frac{3}{2}}y\) with respect to \(t\).
Integrating both sides with respect to \(t\):
\((100 - 2t)^{-\frac{3}{2}}y=\int4(100 - 2t)^{-\frac{3}{2}}dt\).
Let \(u = 100 - 2t\), \(du=-2dt\), then \(\int4(100 - 2t)^{-\frac{3}{2}}dt=- 4(100 - 2t)^{-\frac{1}{2}}+C\).
So, \(y(t)=-4(100 - 2t)+C(100 - 2t)^{\frac{3}{2}}\).
Using the initial condition \(y(0) = 0\) (since the tank starts with fresh water), we get:
\(0=-400+C\times100^{\frac{3}{2}}\), \(C=\frac{400}{1000}=\frac{2}{5}\).
So, \(y(t)=-4(100 - 2t)+\frac{2}{5}(100 - 2t)^{\frac{3}{2}}\).

Step3: Find the maximum

Take the derivative of \(y(t)\) with respect to \(t\):
\(y^\prime(t)=8-\frac{6}{5}(100 - 2t)^{\frac{1}{2}}\).
Set \(y^\prime(t)=0\) (since at maximum, the derivative is zero):
\(8-\frac{6}{5}(100 - 2t)^{\frac{1}{2}} = 0\).
\(\frac{6}{5}(100 - 2t)^{\frac{1}{2}}=8\).
\((100 - 2t)^{\frac{1}{2}}=\frac{20}{3}\).
\(100 - 2t=\frac{400}{9}\).
\(2t = 100-\frac{400}{9}=\frac{900 - 400}{9}=\frac{500}{9}\).
\(t=\frac{250}{9}\approx27.8\) min.

Step4: Calculate the maximum amount

Substitute \(t = \frac{250}{9}\) into \(y(t)\):
\(y(\frac{250}{9})=-4(100-2\times\frac{250}{9})+\frac{2}{5}(100 - 2\times\frac{250}{9})^{\frac{3}{2}}\).
\(100-2\times\frac{250}{9}=\frac{900 - 500}{9}=\frac{400}{9}\).
\(y(\frac{250}{9})=-4\times\frac{400}{9}+\frac{2}{5}(\frac{400}{9})^{\frac{3}{2}}\).
\(y(\frac{250}{9})=\frac{-1600}{9}+\frac{2}{5}\times\frac{8000}{27}\).
\(y(\frac{250}{9})=\frac{-4800 + 16000}{27}=\frac{11200}{27}\approx414.8\) lb.

Answer:

The time required to reach the maximum amount of fertilizer in the tank \(t\approx27.8\) min.
The maximum amount of fertilizer in the tank \(y\approx414.8\) lb.