Question

A tank contains 100 gal of fresh water. A solution containing 2 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 Find the maximum amount of fertilizer in the tank and the time required to reach the maximum.

Answer 1

Let $A(t)$ be the amount of fertilizer in the tank at time $t$. Then

$\dfrac{\mathrm dA}{\mathrm dt}=\underbrace{\dfrac{2\text{ lb}}{1\text{ gal}}\times\dfrac{1\text{ gal}}{1\text{ min}}}_{\text{rate of salt flowing in}}-\underbrace{\dfrac{A(t)\text{ lb}}{100+(1-1)t\text{ gal}}\times\dfrac{1\text{ gal}}{1\text{ min}}}_{\text{rate of salt flowing out}}$
$\dfrac{\mathrm dA}{\mathrm dt}+\dfrac1{100}A=2$

This ODE is separable, and you can write

$\dfrac{\mathrm dA}{2-\frac1{100}A}=\mathrm dt$

Integrating both sides gives

$\displaystyle\int\frac{\mathrm dA}{2-\frac1{100}A}=\int\mathrm dt$
$\displaystyle100\int\frac{\mathrm dA}{200-A}=\int\mathrm dt$
$-100\ln|200-A|=t+C$
$\ln|200-A|=-\dfrac t{100}+C$
$200-A=Ce^{-t/100}$
$A=200-Ce^{-t/100}$

At $t=0$, there is no fertilizer in the tank, so $A(0)=0$. This means

$0=200-Ce^{-0/100}\implies C=200$

so that the amount of fertilizer in the tank is given by

$A(t)=200-200e^{-t/100}$

This function is constantly increasing, so it never achieves its maximum, but there is a strict upper bound as $t\to\infty$. You have $\lim\limits_{t\to\infty}A(t)=200$ lbs of fertilizer in the tank, which makes sense. Given enough time, the tank will be completely saturated with the fertilizer solution, so 2 lb/gal*100 gal = 200 lbs.