I only know the correct answer for part c which is 2986 lbs

3. A 300 gallon tank initially holds 150 gal of water containing 60 lbs of salt. Brine containing 1 lb of salt per gallon enters the tank at 7 gal/mm, and the (perfectly mixed) solution is being pumped out at a rate of 6 gal/mm. (Notice that the volume is a variable function of time). a) Define the volume of the tank as a function of t. b) Find the amount of salt in the tank after t minutes. c) When the tank is full how much salt will it contain?

Solution

a) The initial volume in the container at t=0 is 150 gal of water

Incoming rate = 7 gal/min

Outgoing rate = 6 gal/min

Net incoming rate = 1 gal/min

Hence volume of brine in the tank at any time t is 150 + t(7-6) = 150 + t

b) given initial condition y(0) = 60

After solving the initial value problem

Integrating factor = e^{6/(150+t)dt} = e^{6ln(150+t)} = (150+t)^{6}

(150+t)^6 dy/dt + 6y(150+t)^5 = 7(150+t)^{6}

y(150+t)^6 = Integral(7(150+t)^6} + C

y(150+t)^6 = (150+t)^7 + C

y = (150+t) + C(150+t)^{-6}

y(0) = 60 = 150 + C(150)^{-6}

Hence we get y(t) = 150 + t + (-90*150^{6})/(150+t)^{6}

Now tank will be full at t=150 min

Hence y(t) = 150 + 150 + (-90)/64 = 300 - 90/64 = 298.6 lbs