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katrin [286]
2 years ago
5

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 ?
Mathematics
1 answer:
stellarik [79]2 years ago
6 0
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}
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