Question

A tank initially contains 40 ounces of salt mixed in 100 gallons of water. a solution containing 4 oz of salt per gallon is then pumped into the tank at a rate of 5 gal/min. the stirred mixture flows out of the tank at the same rate. how much salt is in the tank after 20 minutes ?

Answer 1

Let $A(t)$ be the amount of salt in the tank at time $t$. We're given that $A(0)=40$. The rate at which this amount changes is given by

$A'(t)=\dfrac{5\text{ gal}}{1\text{ min}}\cdot\dfrac{4\text{ oz}}{1\text{ gal}}-\dfrac{5\text{ gal}}{1\text{ min}}\cdot\dfrac{A(t)\text{ oz}}{100+(5-5)t\text{ gal}}$

$A'(t)+\dfrac{A(t)}{20}=20$

$e^{t/20}A'(t)+e^{t/20}\dfrac{A(t)}{20}=20e^{t/20}$

$\bigg(e^{t/20}A(t)\bigg)'=20e^{t/20}$

$e^{t/20}A(t)=400e^{t/20}+C$

$A(t)=400+Ce^{-t/20}$

Since $A(0)=40$, we get

$40=400+C\implies C=-360$

so that the amount of salt at time $t$ is

$A(t)=400-360e^{-t/20}$

After 20 minutes, the tank contains

$A(20)=400-360e^{-20/20}\approx267.56\text{ oz}$