Systems of equation w/3 stock solutions

[melby2188]

one must create 250ml of a 17% salt solution. You have three stock solutions. One liter container of a 5% salt, a 500ml of a 28% salt solution and a 400ml of 40% salt solution. Calculate the cheapest method of preparing the 17% salt sol. if the 5% salt solution costs $28 per liter, the 28% solution costs $38 per liter, and 40% sol. costs $50 per liter.

 

[Mark44]

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[melby2188]

Sorry,This is what i have so far: x+y+z=250 and .05x+.28y+.40z=.17(250) and i think the cost function should be something like c(x,y,z)=1/1000(28x+38y+50z) I know i need to use systems of equations I am just not sure how to solve it with the cost equation or if the cost equation is correct.

 

[Mark44]

With just a casual perusal of the problem, it seems to me that you want to use only the two lower concentration solutions and none of the high concentration solution. Since the desired solution should have a concentration of 17% salt, and since there is enough of the two cheaper solutions to make 250 ml, you should be able to find out how much of each of the two cheaper solutions gives you
a) a total of 250 ml. of solution
b) the right amount of salt in the solution

So your two equations should keep track of the total liquid amount, and the total amount of salt.

Notice that if the desired solution had a concentration higher than 28%, you would have to use some of the more expensive mix, or if the amount of the cheapest mix happened to be too small.

 

[melby2188]

ok thanks this is what I am getting

using x+y=250 and .05x+.28y=.17(250)
x=119.565 and y=130.435
putting this back into the cost function 1/1000(28x+38y) i get $8.30
does this sound right?

 

[Mark44]

Yes, this looks fine.