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 Paint My Love (Posted on 2003-05-12)
Your paint inventory consists of 60 gallons of blue, 40 gallons of red, and 30 gallons of yellow. To make purple paint you mix equals parts of blue and red. To make orange paint you mix equal parts of red and yellow. To make green paint you mix equal parts of blue and yellow. Purple paint sells for \$6 a gallon, orange for \$20, and green for \$9.

There is a fixed disposal charge for every unused gallon. How much of each of purple, orange, and green paint should you mix to maximize profits if the disposal cost is (a) \$4 per gallon, and (b) \$6 per gallon.

 See The Solution Submitted by Ravi Raja Rating: 2.6667 (3 votes)

Comments: ( Back to comment list | You must be logged in to post comments.)
 Linear maximization (Solution) | Comment 2 of 4 |
This problem is a linear maximization problem.

First define some variables:
Let a be the amount of red used to make purple.
Let b be the amount of red used to make orange.
Let c be the amount of yellow used to make orange.
Let d be the amount of yellow used to make green.
Let e be the amount of blue used to make green.
Let f be the amount of blue used to make purple.
Let k be the disposal cost per gallon (\$4 or \$6).

Now create some equations:
a = f (equal parts red and blue)
b = c (equal parts red and yellow)
d = e (equal parts yellow and blue)
a + b <= 40 (amount of red)
c + d <= 30 (amount of yellow)
e + f <= 60 (amount of blue)
a,b,c,d,e,f >= 0 (all variables are non negative)

Net profit is the amount of sales minus the amount of disposal. Net profit equation = 6*(a + f) + 20*(b + c) + 9*(d + e) - k*(130 - a - b - c - d - e - f)

After some substitution a linear system of inequalities in 3 variables emerges:

a + b <= 40
b + d <= 30
d + a <= 60
a >= 0, b >= 0, d >= 0

The maximum net profit occurs at one of the verticies of the region defined by the inequalities. The verticies are (0,0,0), (0,0,30), (0,30,0), (40,0,0), (30, 0, 30), (40, 0, 20), (10, 30, 0), (35,5,25).

If the disposal cost is \$4 then Net Profit = 20a + 48b + 26d - 520
This equation has a maximum value of \$1120 with vertex (10, 30, 0).

If the disposal cost is \$6 then Net Profit = 24a + 52b + 30d - 780
This equation has a maximum value \$1070 with vertecies (10, 30, 0) and (35, 5, 25).

With a \$4/gallon disposal cost, the most revenue \$1120 will be made by making 20 gallons orange and 60 gallons green.

With a \$6/gallon disposal cost, the most revenue \$1070 will be made by making 20 gallons purple and 60 gallons orange or by making 70 gallons purple, 10 gallons orange and 50 gallons green.
 Posted by Brian Smith on 2003-05-12 05:09:37
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