A large mixing tank initially contains 100 gallons of water in which 30 pounds of salt have been dissolved. Another brine solution is pumped into the tank at the rate of 4 gallons per minute, and the resulting mixture is pumped out at the same rate. The concentration of the incoming brine solution is 2 pounds of salt per gallon. If $A ( t )$ represents the amount of salt in the tank at time t, the correct differential equation for A is
A) $\frac { d A } { d t } = 8 - .02 A$
B) $\frac { d A } { d t } = 8 - .04 A$
C) $\frac { d A } { d t } = 4 - .04 A$
D) $\frac { d A } { d t } = 2 - .04 A$
E) $\frac { d A } { d t } = 4 - .08 A$

Question

Quizplus · Accepted Answer

The rate of salt entering the tank is $4 \text{ gallons/minute} \times 2 \text{ pounds/gallon} = 8 \text{ pounds/minute}$. The rate of salt leaving the tank is proportional to its concentration in the tank, which is $\frac{A(t)}{100}$ pounds of salt per gallon (since the volume of the tank remains constant at 100 gallons). Since the solution is being pumped out at 4 gallons per minute, the rate of salt leaving is $4 \times \frac{A(t)}{100} = 0.04A(t)$ pounds per minute. Thus, the net rate of change of salt in the tank is the rate in minus the rate out, which gives $\frac{dA}{dt} = 8 - 0.04A$.

A Large Mixing Tank Initially Contains 100 Gallons of Water A(t)A ( t )A(t)

A Large Mixing Tank Initially Contains 100 Gallons of Water $A ( t )$