A Simple Salt Solution Mixing Problem - yet stuck

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    Mixing Salt Stuck
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Discussion Overview

The discussion revolves around a mixing problem involving salt solutions, specifically how to determine the optimal amount of a 20% salt solution to mix with a 15% salt solution in a vessel with a limited capacity. The problem is approached through mathematical reasoning and optimization techniques.

Discussion Character

  • Mathematical reasoning
  • Exploratory
  • Homework-related

Main Points Raised

  • The original poster presents a function for salt concentration and attempts to find its maximum using derivatives.
  • Some participants note the vessel's capacity constraint, indicating that the total volume of solutions must not exceed 5 liters.
  • One participant suggests that the optimal strategy may be to discard the 15% solution entirely and use only the 20% solution.
  • The original poster expresses confusion regarding the lack of a local maximum in their function and reflects on the implications of this observation.

Areas of Agreement / Disagreement

Participants do not reach a consensus on the best approach to the problem. While some suggest maximizing the amount of 20% solution added, others propose an alternative method of discarding the 15% solution entirely.

Contextual Notes

The discussion highlights the importance of considering the vessel's capacity and the nature of the solutions being mixed. The original poster's expectation of a local maximum in the function is challenged by the nature of the problem.

Rokas_P
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Hi, here is a very simple solution mixing problem that I can't solve which I am really ashamed of.

Problem. A vessel whose capacity is 5 liters contains 2 liters of 15% salt solution. How many liters of 20% salt solution have to be mixed to the 15% solution to produce a solution with as high salt concentration as possible?

My solution. I constructed this function for salt concentration in the new solution:

f(x)=\frac{0.3+0.2x}{2+x}

(The amount of salt in the first solution is 2*0.15 = 0.3)

I obtained the derivative of the above function:

f'(x)=\frac{0.2(2+x)-(0.3+0.2x)}{\left(2+x\right)^2}=\frac{0.4+0.2x-0.3-0.2x}{\left(2+x\right)^2}=\frac{0.1}{\left(2+x\right)^2}

I'm stuck at this point. If I try to find points of extremum, there are none. It means that I cannot increase salt concentration in the original solution by adding the said 20% salt solution. But I know the answer is 3 liters, so I must be doing something wrong.

--

As a sidenote, this problem is taken from a national maths exam. As such, you are not expected to think of mass, volume and density of water and salt. Also, I didn't make use of the fact that the vessel can contain up to 5 liters of liquid but just don't see where it fits in. Finally, I am not sure if this problem is even supposed to be solved using optimization techniques (derivatives and minima/maxima) but that was the first idea I got ...
 
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Well, 2 + x must be less than or equal to 5 liters, since that is the maximum capacity of the vessel.
 
Rokas_P said:
Also, I didn't make use of the fact that the vessel can contain up to 5 liters of liquid but just don't see where it fits in.

And that was your mistake...

Think about it - the more 20% solution you add, the closer to 20% is the concentration of the salt in the mixture. So you have to add as much as possible. How much can you add?
 
Alternatively, the answer is to toss out the 15% solution. Just use the full 20% salt solution.
 
Thanks everyone for the explanations.

The problem here was that I expected the function to have a local maximum and when the derivative showed me it had none, I thought I must have done something wrong but now I realize that there is no reason for the function to have a maximum.

We can consider this problem solved.
 

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