what is the correct differential equation

Question

Tank A contains 80 gallons of water in which 20 pounds of salt has been dissolved. Tank B contains 30 gallons of water in which 5 pounds of salt has been dissolved. A brine mixture with a concentration of 0.5 pounds of salt per gallon of water is pumped into tank A at the rate of 4 gallons per minute. The well-mixed solution is then pumped from tank A to tank B at the rate of 6 gallons per minute. The solution from tank B is also pumped through another pipe into tank A at the rate of 2 gallons per minute, and the solution from tank B is also pumped out of the system at the rate of 4 gallons per minute. The correct differential equations with initial conditions for the amounts, x(t) and y(t), of salt in tanks A and B respectively, at time t are: A) dx/dt=2-x/40+y/5,dy/dt=x/40-y/3,x(0)=20,y(0)=5 B) dx/dt=2-3x/40+y/15,dy/dt=3x/40-y/5,x(0)=20,y(0)=5 C) dx/dt=4-3x/40+y/15,dy/dt=3x/40-y/5,x(0)=20,y(0)=5 D) dx/dt=4-x/40+y/5,dy/dt=x/40-y/3,x(0)=20,y(0)=5

Answer 1

Answer B.

The rates of flow keep the tank volumes constant.

Tank A is gaining 2lb of salt per gallon per minute, losing 6/80 lb = 3x/40 lb of salt to tank B, and gaining 2y/30=y/15 lb of salt from tank B per minute. So net change for tank A is dx/dt=2-3x/40+y/15.

Tank B is gaining 3x/40 lb of salt from tank A per minute while losing 4y/30=2y/15 through drainage and 2y/30=y/15 pumped back to A per minute. Total loss=3y/15=y/5. Net change dy/dt=3x/40-y/5.

Initial conditions show the amount of salt in the tanks. x(0)=20, y(0)=5.