A textbook example asks me:
A large tank is filled to capacity with 100 gallons of pure water. Brine
containing 3 pounds of salt per gallon is pumped into the tank at a rate of 4 gal/min.
The well-mixed solution is pumped out of the tank at the rate of 5 gal/min. Find the amount of salt after 30 minutes.
I took the following apparently incorrect approach to solving it:
I first calculated the total amount of liquid in the tank to be $100+4t-5t$ with $t$ denoting time in minutes.
I then observed that the amount of salt coming in was $\frac{3 \ \text{lbs}}{\text{min}}$. I determined that the amount of salt going out was $ \frac{\text{Salt}\frac{\text{lbs}}{\text{gal}}}{100-t}$.
To calculate the total amount of salt, I simply took the amount of incoming salt and subtracted the outgoing salt, plugged in a value of 30 for $t$ and arrived at 87.85 pounds of salt at 30 minutes.
However, the correct answer in the answer section is given as 209.97 lbs.
Can anyone suggest what I did wrong?
Best Answer
I misread the question and misinterpreted the amount of incoming salt.
There are 3 pounds per gallon of incoming brine, and 4 gallons being pumped in per minute, in other words, 12 pounds incoming.
Solving the differential equation gave me $$A(t) = [3(100-t)^{-4}+C](100-t)^5 $$ and then I solved for C by using $A(0) = 0$ as an initial condition (since the tank has no brine at the beginning). My final function for $$A(t)= 3(100-t)-\frac{3}{100^4}(100-t)^5$$ and $$A(30) = 159.57$$
:)