Thread: Mixture Problem

I have no idea where even to start with this one, so anything will help!


A tank initially contains 60 gal of pure water. Brine containing 1 lb of salt per gallon enters the tank and 2 gal/min and the (perfectly mixed) solution leaves the tank at 3 gal/min; thus tthe tank is empty after exactly one hour. (a) Find the amount of salt in the tank after t minutes. (b) What is the maximum amount of salt ever in the tank?

simply let y be the amount of salt in the tank at time t

dy/dt = salt in -salt out

the volume of fluid in the tank is v(t) = 60 -t gals

dy/dt =(1lb/gal)(2gal/min) - (ylbs/60 -t gals)(3gals/min)

dy/dt = 1 - 3y/(60 -t) y(0) = 0

This is not separable so rewrite

dy/dt + 3y /(60-t) = 1 y(0) = 0

Now use an integrating factor to solve

How would one go about solving part b? I have a figured out, but part b is stumping me. I tried maximizing the function, but the solution is not in agreement with the solutions manual...ANy help from here would be appreciated!

See attachment

Thank You Very Much!!!

i think there's something wrong, i think it should be dy/dt +3y/(60-t) = 2

since 2(1) = 2 , not 1. make sense?