Question

A 200 gallon tank initially contains 100 gallons of water with 20 pounds of salt. A salt solution with 1/4 pound of salt per gallon is added to the tank at 4 gal/min, and the resulting mixture is drained out at 2 gal/min. Find the quantity of salt in the tank as it’s about to overflow.

Answer 1

The volume of the solution in the tank is

$V(t)=100+(4-2)t=100+2t$

and so the solution will overflow when $V(t)=200$. This happens at $t=50$.

Now the rate of change of the amount of salt in the tank at time $t$, $A(t)$, is described by

$\dfrac{\mathrm dA}{\mathrm dt}=\left(\dfrac14\text{ lb/gal}\right)\left(4\text{ gal/min}\right)-\left(\dfrac{A(t)}{100+2t}\text{ lb/gal}\right)\left(2\text{ gal/min}\right)$
$A'+\dfrac A{50+t}=1$

which is linear in $A$. Multiplying both sides by $50+t$, we have

$(50+t)A'+A=50+t$
$\bigg((50+t)A\bigg)'=50+t$
$(50+t)A=\displaystyle\int(50+t)\,\mathrm dt=50t+\dfrac12t^2+C$
$A=\dfrac{100t+t^2}{100+2t}+\dfrac C{50+t}$

There are 20 lbs of salt in the tank at the start, so $A(0)=20$ and we get that

$20=0+\dfrac C{50}\implies C=1000$

so that the amount of salt in the tank is given by the function

$A(t)=\dfrac{100t+t^2}{100+2t}+\dfrac{1000}{50+t}=\dfrac{2000+100t+t^2}{100+2t}$

At $t=50$, the tank contains

$A(50)=\dfrac{2000+100(50)+50^2}{100+2(50)}=\dfrac{95}2$

pounds of salt.