A tank contains 200 gallons of water in which 300 grams of salt is dissolved. A brine solution containing 0.4 kilograms of salt per gallon of water is pumped into the tank at the rate of 5 liters per minute, and the well-stirred mixture is pumped out at the same rate. Let $A ( t )$ represent the amount of salt in the tank at time t. The correct initial value problem for $A ( t )$ is
A) $\frac { d A } { d t } = 2 + A / 40 , A ( 0 ) = 0.3$
B) $\frac { d A } { d t } = 2 - A / 40 , A ( 0 ) = 0.3$
C) $\frac { d A } { d t } = 5 + A / 40 , A ( 0 ) = 300$
D) $\frac { d A } { d t } = 5 - A / 40 , A ( 0 ) = 300$
E) $\frac { d A } { d t } = 0.4 - A / 40 , A ( 0 ) = 300$

Question

Quizplus · Accepted Answer

The rate of salt entering the tank is 0.4 kg/gallon * 5 liters/minute * (1 gallon/3.78541 liters) = 2 grams/minute. The rate of salt leaving is proportional to the concentration of salt in the tank, which is $A(t)/200$ grams per gallon, times the outflow rate of 5 liters/minute * (1 gallon/3.78541 liters), simplifying to $A(t)/40$ grams/minute. The initial amount of salt is 300 grams. Thus, the correct model is $\frac{dA}{dt} = 2 - A/40, A(0) = 0.3$, after converting the initial salt amount to kilograms to match the units in the options.

A Tank Contains 200 Gallons of Water in Which 300 A(t)A ( t )A(t)

A Tank Contains 200 Gallons of Water in Which 300 $A ( t )$