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January 29, 2015

January 29, 2015

Posted by **sandy** on Saturday, April 26, 2008 at 8:21pm.

a use as little as possible of the 50% solution

b use as much as possible of the 50% solution

c use 2 liters of the 50% solution

- algebra -
**Reiny**, Saturday, April 26, 2008 at 10:08pmlet the amount of 50% solution be x L

let the amount of 20% solution by y L

let the amount of 10% solution be z or (10-x-y) L

then .1(10-x-y) + .2y + .5x = .25(10)

10-x-y + 2y + 5x = 25

y = -4x + 15 , relation (A)

also x+y ≤ 10 , relation (B)

so any (x,y) which satisfies both relations A and B will work

where do they intersect?

y = -4x + 15 with y = -x + 10

x = 5/3, then y = 25/3 , z = 0

notice the x intercept falls within (B)

let y=0, then x= 3.75 , z = 6.25

if x = 2 , then y = 7, and z = 1

50% . 20% . 10%

0 .. 15 .. not possible

5/3 . 25/3 ..0

15/4 . 0 ...25/4

2 ... 7 ... 1

a) for which of these is x as small as possible

b) for which of these is x as large as possible

c) when x = 2 , y = 7, z = 1

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