Beer containing 6% alcohol per gallon is pumped into a vat that initially contains 400 gallons of beer at 3% alcohol.

The rate at which the beer is pumped in is 3 gallons per minute, whereas the mixed liquid is pumped out at a rate of

4 gallons per minute. (a) Find the number of gallons of alcohol A(t) in the tank at any time t. (b) What is the

percentage of alcohol in the tank after one hour? (c) When is the tank empty?

dy/dt = .06*3 - y(t)/(400-t)

Is this the correct way to set up the problem? What is the relevance of the initial concentration of the beer in the vat? Does that not factor into the problem?