Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?

share|improve this question
    
For starters, how did you arrive at three pounds per four gallons? The text says 3 pounds per gallon. –  joriki Oct 4 '12 at 16:00
    
Sorry, that should say $\frac{3 \ gallons}{4 \ minute}$. I'll fix that now... –  Imray Oct 4 '12 at 16:06
2  
Now I'm really confused. You announced that you'd change it to 3 gallons/4 minutes, and now you've changed it to 3 pounds/4 minutes. I think the most important thing you probably need to solve the problem is to be more careful. –  joriki Oct 4 '12 at 16:19
    
Sorry Sorry - silly mistake!! That might have been my issue, I'll try it again and let you know if I got it... –  Imray Oct 4 '12 at 16:25
    
I fixed it and solved it properly. Thank you for pointing out my mistake. I'll reread a few more times before posting it up next time :) –  Imray Oct 4 '12 at 16:42
add comment

1 Answer 1

up vote 0 down vote accepted

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.

If $A(t)$ is the function describing the total amount of salt as a function of time, then $\frac{dA}{dt} = 12 - \frac{5A(t)}{100-t}$.

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 $A(t)$ is $ = 3(100-t)-\frac{3}{100^4}(100-t)^5$ and $A(30) = 209.97$

:)

share|improve this answer
2  
Finally you got a differential equation into the solution of your differential equation problem :) –  rschwieb Oct 4 '12 at 20:03
    
lol that should've given it away :) –  Imray Oct 4 '12 at 20:09
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.