A tank contains 160 liters of fluid in which 30 grams of salt is dissolved. Brine containing 1 gram of salt per liter is then pumped into the tank at a rate of 4 L/min; the well-mixed solution is pumped out at the same rate. Find the number A (t) of grams of salt in the tank at time t.

A (t) = 160 - 130 e^ (-t/40)

Explanation:

At the start, the tank contains A (0) = 30 g of salt.

Salt flows in at a rate of

(1 g/L) * (4 L/min) = 5 g/min

and flows out at a rate of

(A (t) / 160 g/L) * (4 L/min) = A (t) / 40 g/min

so that the amount of salt in the tank at time t changes according to

A' (t) = 4 - A (t) / 40

Solve the ODE for A (t):

A' (t) + A (t) / 40 = 4

e^ (t/40) A' (t) + e^ (t/40) / 40 A (t) = 4e^ (t/40)

(e^ (t/40) A (t)) ' = 4e^ (t/40)

e^ (t/40) A (t) = 160e^ (t/40) + C

A (t) = 160 + Ce^ (-t/40)

Given that A (0) = 30, we find

30 = 160 + C

C = - 130

so that the amount of salt in the tank at time t is

A (t) = 160 - 130 e^ (-t/40)