# Dilution problem. A very large tank initially contains 15 gallons of saltwater.

A very large tank initially contains 15 gallons of saltwater containing 6 pounds of salt. Saltwater containing 1 pound of salt per gallon is pumped in to the top of the tank a rate of 2 gallons per minute while a well mixed solution leaves the bottom of the tank at a rate of 1 gallon per minute. What does the solution to the differential equation predict about the concentration of salt in the tank in the long run.

I assumed this was an infinite tank since we are talking about an infinite time period.

Here is my differential equation where $$V$$ is the volume and $$S$$ is the weight of salt. $$V=15+2t-t=15+t$$

$$\frac{dS}{dt}=2-\frac{S}{15+t}$$

Solving it gives $$S=\frac{t^2+30t+C}{t+15}$$

We know concentration is going to be the ratio of amount of salt $$S$$ and the volume $$V$$.

$$\lim_{t \rightarrow\infty}{\frac{S}{V}}= \lim_{t \rightarrow\infty}\frac{t^2+30t+90}{(t+15)^2} = 1$$ So I believe that the concentration will go to $$1\frac{lbs}{gal}$$ in the long run. This seems intuitive to me since the concentration coming in is $$1$$ and will eventually dilute the rest.

However, my teacher says that eventually, the tank will be full of salt. She says the limit of $$1$$ corresponds to the percentage of the tank full of salt?? Who is wrong, and why?

• You have not defined $S$, and have not explained the ODE ${dS\over dt}=\ldots$. But otherwise your explanations are correct. In particular your limit $1$ means that in the end we shall have $1$ pound of salt per gallon of fluid. – Christian Blatter Feb 27 at 19:40
• @ChristianBlatter I should clarify. $S$ is the lbs of salt – Jac Frall Feb 27 at 19:45

$$\lim_{t \rightarrow\infty}{\frac{S}{V}}= 1$$
means that in the long run the concentration approaches 1 lb per gallon. The units of $$\frac{S}{V}$$ are $$[lb/gal]$$ and not, as your teacher proposed, a fraction.