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MTH 279
Your 'question form' report has been received. Scroll down through the document to see any comments I might have inserted, and my final comment at the end.
** Question Form_labelMessages **
Query 4 Question 2 Question
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On Query 4, problem 2, I'm looking at what constant rate of flow is necessary to get 35 lbs of salt in a 1000 gal tank at the end of 8 hr. Initially, it contains 500 gal of a 5% solution and a 3% solution is pouring in.
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So, I understand that the problem has to be split up into two parts. First, you have to find out how much salt is in the tank when it is at full capacity. Initially, there are 25 lbs of salt (500 gal * .05 lbs/gal = 25 lb). That leaves 500 gal to be filled with the 3% solution to get the tank filled up. So 500 * .03 = 15 gal of salt. This gives a total of 40 lbs of salt in the tank when it is full. Now, I must find the time at which the tank will be at full capacity, and this is what I'm not sure if I'm doing right. I don't know the rate of flow, which I feel like I need to find the time. So, the closest time quantity I can come up with is t_full=500r. I came up with this by saying Q = r(.03)t + 25, using the initial condition Q(0) = 25. I know Q at the particular time is 40, so that is how I came up with t.
For the second part of the problem, I then knew that the time it would take for the tank to have a solution strength of 3.5% would have to be at 8 - 500r. So I ended up with the equation Q = 30+10e^(-r/1000*t) and plugging in this particular value for time, I ended up with a nasty quadratic equation. So, I feel like I maybe missed a detail that would've given me an actual numerical value for time in the first scenario, which would simplify the second scenario's equation.
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Have I solved for time correctly in the first part of the problem, and if not, how could I go about doing so?
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You are on the right track.
However if r is the flow rate then the time until the tank is first full is 500 gal / r (the time required to reach the full tank is inversely proportional to the rate, not directly proportional, as you expression 500 r would imply).
This leaves 8 hours - 500 gal / r to achieve the 3.5% solution.
So the problem becomes this:
Starting with a 4% solution, what flow rate r for a 3% solution, with the mixed solution flowing out at the same rate r, will achieve a 3.5% solution in 8 hours - 500 gal / r?
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