Mixture Linear Equation Word Problem

Feb 2019
6
1
Alabama
A 200 gallon tank is filled with water in which 20 pounds of salt has been dissolved. A water solution containing 2 pounds of salt per gallon runs into the tank at the rate of 2 gallons per minute and the well-stirred mixture runs out of the tank at the rate of 4 gallons per minute.
(a) How much salt will be in the tank after 25 minutes?
(b) How long will it take for there to be 100 pounds of salt in the tank? (NOTE: there will be two answers.)
(c) How much salt will be in the tank after 100 minutes have passed?
 
Dec 2011
2,314
915
St. Augustine, FL.
Let's let \(\displaystyle A(t)\) represent the amount (in pounds) of salt in the tank at time \(\displaystyle t\) (in minutes). We are told \(\displaystyle A(0)=20\). Now, the time rate of change of salt in the tank is the salt coming in minus the salt going out. We are told the salt coming in is 4 lbs. per minute. And the salt leaving is 4 times the amount of salt per gallon present in the tank at time \(\displaystyle t\). So, we need to know how many gallons of brine are present in the tank at time \(\displaystyle t\), which is:

\(\displaystyle 200+(2-4)t=2(100-t)\)

Now the amount coming in is:

\(\displaystyle \left(2\frac{\text{gal}}{\text{min}}\right) \left(2\frac{\text{lb}}{\text{gal}}\right)= 4\frac{\text{lb}}{\text{min}}\)

And the amount going out is:

\(\displaystyle \left(4\frac{\text{gal}}{\text{min}}\right) \left(\frac{A}{2(100-t)}\frac{\text{lb}}{\text{gal}}\right)=\frac{2}{100-t}\frac{\text{lb}}{\text{min}}\)

And so this leads to the IVP:

\(\displaystyle \frac{dA}{dt}=4-2\frac{A}{100-t}\) where \(\displaystyle A(0)=20\)

Can you proceed?
 
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HallsofIvy

MHF Helper
Apr 2005
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Have you not at least tried this yourself? You have posted several problems where you just state the problem. That violates the rules of this forum.

Start by setting it up as a "differential equation". Let X(t) be the amount of salt, in pounds, in the tank at time t in minutes. The rate of change of X, dX/dt, is the rate at which salt is coming into the tank minus the rate at which salt is going out of the tank. The rate at which salt is going into the tank is 2 pounds per gallon times 2 gallons per minute= 4 pounds per minute.

The rate at which salt is going out of the tank is a little more complicated. It is the number of pounds per gallon times 4 gallons per minute.

One complication is that, because water is coming in at 2 gallons per minute but going out of the tank at 4 gallons per minute so the amount of water in the tank is reducing at 2- 4= -2 gallons per minute. The tank originally contains 200 gallons so after t minutes there are 200- 2t gallons of water in the tank. With X(t) pounds of salt in the tank there are \(\displaystyle \frac{X(t)}{200- 2t}\) "pounds per gallon". The salt is going out of the tank at the rate of \(\displaystyle -4\frac{X(t)}{200- 2t}\).

Putting those together, \(\displaystyle \frac{dX}{dt}= 4- \frac{X}{200- 2t}\). Solve that differential equation with the initial condition X(0)= 20.

Once you know X(t)
a) What is X(25)?
b) Solve X(t)= 100 for t.
c) What is X(100)?
 
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Dec 2011
2,314
915
St. Augustine, FL.
Have you not at least tried this yourself? You have posted several problems where you just state the problem. That violates the rules of this forum.

Start by setting it up as a "differential equation". Let X(t) be the amount of salt, in pounds, in the tank at time t in minutes. The rate of change of X, dX/dt, is the rate at which salt is coming into the tank minus the rate at which salt is going out of the tank. The rate at which salt is going into the tank is 2 pounds per gallon times 2 gallons per minute= 4 pounds per minute.

The rate at which salt is going out of the tank is a little more complicated. It is the number of pounds per gallon times 4 gallons per minute.

One complication is that, because water is coming in at 2 gallons per minute but going out of the tank at 4 gallons per minute so the amount of water in the tank is reducing at 2- 4= -2 gallons per minute. The tank originally contains 200 gallons so after t minutes there are 200- 2t gallons of water in the tank. With X(t) pounds of salt in the tank there are \(\displaystyle \frac{X(t)}{200- 2t}\) "pounds per gallon". The salt is going out of the tank at the rate of \(\displaystyle -4\frac{X(t)}{200- 2t}\).

Putting those together, \(\displaystyle \frac{dX}{dt}= 4- \frac{X}{200- 2t}\). Solve that differential equation with the initial condition X(0)= 20.

Once you know X(t)
a) What is X(25)?
b) Solve X(t)= 100 for t.
c) What is X(100)?
I think your ODE should be:

\(\displaystyle \frac{dX}{dt}= 4-4\frac{X}{200- 2t}\)

And this is equivalent to what I posted. :)
 
Feb 2019
6
1
Alabama
I have tired and yes I did post 2 problems that I had questions on. I think I have the problems figured out.

Thank you for your response and it appears that I was correct in my thought process.
 
Dec 2011
2,314
915
St. Augustine, FL.
I have tired and yes I did post 2 problems that I had questions on. I think I have the problems figured out.

Thank you for your response and it appears that I was correct in my thought process.
Hello, Amy!

What most math helpers want from people posting questions is to see what they've tried so they can see where the issue preventing them from completing the exercise might be. Unfortunately, there are so many people who just want someone else to do their homework, that when someone posts a question without any effort shown, that's what it might look like. A student who posts the question along with what they've tried, or what their thoughts are on how to begin is much more likely to receive timely help.
 
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Feb 2019
6
1
Alabama
Thank you Mark...that is not a problem! I have worked out the problems and my teacher stated to get any and all help I could get on the 6 questions he gave out. I was merely trying to make sure I was on the right track. I appreciate your kind response and explanation!
 
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