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 ?

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

[tex]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}}[/tex]

[tex]A'(t)+\dfrac{A(t)}{20}=20[/tex]

[tex]e^{t/20}A'(t)+e^{t/20}\dfrac{A(t)}{20}=20e^{t/20}[/tex]

[tex]\bigg(e^{t/20}A(t)\bigg)'=20e^{t/20}[/tex]

[tex]e^{t/20}A(t)=400e^{t/20}+C[/tex]

[tex]A(t)=400+Ce^{-t/20}[/tex]

Since [tex]A(0)=40[/tex], we get

[tex]40=400+C\implies C=-360[/tex]

so that the amount of salt at time [tex]t[/tex] is

[tex]A(t)=400-360e^{-t/20}[/tex]

After 20 minutes, the tank contains

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