Suppose that a large mixing tank initially holds 500 gallons of water in which 50 pounds of salt have been dissolved. Another brine solution is pumped into the tank at a rate of 3 gal/min, and when the solution is well stirred, it is then pumped out at a slower rate of 2 gal/min. If the concentration of the solution entering is 2 lb/gal, determine a differential equation (in lb/min) for the amount of salt A(t) (in lb) in the tank at time t > 0. (Use A for A(t).)

Step-by-step explanation:

The large mixing tank initially holds 500 gallons of water in which 50 pounds of salt have been dissolved.

Volume = 500 gallons

Initial Amount of Salt, A(0)=50 pounds

Brine solution with concentration of 2 lb/gal is pumped into the tank at a rate of 3 gal/min

=(concentration of salt in inflow)(input rate of brine)

When the solution is well stirred, it is then pumped out at a slower rate of 2 gal/min.

Concentration c(t) of the salt in the tank at time t

Concentration,

=(concentration of salt in outflow)(output rate of brine)

Now, the rate of change of the amount of salt in the tank

We solve the resulting differential equation by separation of variables.  

Taking the integral of both sides

Recall that when t=0, A(t)=50 (our initial condition)

a is the correct

my guess would be b and d, but i didn’t actually get out the pen and paper. lol