I worked through this an am not confident with my answer:

Question;

An aquarium pool has volume 2x10^6 litres. The pool initially contains fresh water. At t=0 minuts, water containing 10 gms/litre of salt is poured into the pool at a rate of 60 litre/min. The salt water instantly mixes with the fresh water and excess mixture is drained out of the bottom of the pool at the same rate (60 litres/min). Let S(t) = mass of salt in pool at time t.

a. solve for S(t)

b. What happens to S(t) as t --> infinity?

My answer;

2x10^6 - 2x10^6 e^(-3t/100,000) = S(t)

therefore as t approaches infinity, S(t) approaches 2x10^6, but I don't think this makes sense with the story problem...

HELP please!!!