# A Simple Salt Solution Mixing Problem - yet stuck

1. Jun 9, 2013

### Rokas_P

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.

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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 ...

2. Jun 9, 2013

### SteamKing

Staff Emeritus
Well, 2 + x must be less than or equal to 5 liters, since that is the maximum capacity of the vessel.

3. Jun 9, 2013

### Staff: Mentor

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?

4. Jun 9, 2013

### etudiant

Alternatively, the answer is to toss out the 15% solution. Just use the full 20% salt solution.

5. Jun 10, 2013

### Rokas_P

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.

6. Jun 10, 2013