1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

A Simple Salt Solution Mixing Problem - yet stuck

  1. Jun 9, 2013 #1
    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:

    [tex]f(x)=\frac{0.3+0.2x}{2+x}[/tex]

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

    I obtained the derivative of the above function:

    [tex]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}[/tex]

    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 ...
     
  2. jcsd
  3. Jun 9, 2013 #2

    SteamKing

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper

    Well, 2 + x must be less than or equal to 5 liters, since that is the maximum capacity of the vessel.
     
  4. Jun 9, 2013 #3

    Borek

    User Avatar

    Staff: Mentor

    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?
     
  5. Jun 9, 2013 #4

    etudiant

    User Avatar
    Gold Member

    Alternatively, the answer is to toss out the 15% solution. Just use the full 20% salt solution.
     
  6. Jun 10, 2013 #5
    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.
     
  7. Jun 10, 2013 #6

    Borek

    User Avatar

    Staff: Mentor

Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: A Simple Salt Solution Mixing Problem - yet stuck
  1. Water mixing problem (Replies: 2)

Loading...