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Tutorial 1.38 Solve applications involving linear programming
When we are solving real-world applications where linear programming might be useful, we will need to translate the written word into an appropriate (equivalent) set of inequalities that can then be solved as a system (as was explained in previous tutorials).
example 1.38a The Monmouth Bakery Company makes regular cake mixes as well as gourmet pancake mixes that are sold at various groceries stores in the area:
• Each pound of the regular cake mix uses 0.4 pound of flour and 0.1 pound of shortening.
• Each pound of the gourmet pancake mix uses 0.6 pound of flour, 0.1 pound of shortening and 0.4 pound of sugar.
• Suppliers can deliver at most 8400 pounds of flour, at least 800 pounds of shortening, and at most 2000 pounds of sugar.
If the profit per pound is 25¢ for regular cake mix and 35¢ for gourmet pancake mix, how many pounds of each mix should be made to earn maximum profit? What is that maximum profit?
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Sometimes, translation is made a bit easier by organizing the given information into a data table (as has been done in the table at the left below). We can then use the organized table to write the constraints for the given word problem. Notice that the supplier can deliver at most one quantity and at least another.
In mathematical translations, at most is translated into the ≤ (is less than or equal to) sign because the amount delivered cannot go over the stated amount. The phrase at least is translated into the ≥ (is greater than or equal to) sign because the amount delivered cannot be any lower than the stated amount. Also, remember that when completing a real-world business application, we want to use numbers that are greater than or equal to zero…thus: x ≥ 0 and y ≥ 0. The information given about the profit will aid us in creating the objective function for this scenario.
x y
cake pancake symbol totals
flour: 0.4 0.6 ≤ 8400
shortening: 0.1 0.1 ≥ 800
sugar: 0.0 0.4 ≤ 2000
profits: 0.25 0.35 = MAX
constraints (inequalities):
flour: 0.4x + 0.6y £ 8400 shortening: 0.1x + 0.1y ≥ 800 sugar: 0.4y £ 2000 x-values: x ≥ 0 y-values: y ≥ 0
objective: F = 0.25x + 0.35y
Now that we have the system of inequalities and the objective function translated from the given word problem, we can now go about the task of graphing the system on a coordinate grid, shading in the feasible solution region, determining the coordinates of any appropriate vertices and then determining the values of x and y that produce the optimal maximum profit for this given situation.
Note that one of the constraints (inequalities) has a different relationship sign than the others…thus, we have a system with mixed constraints. When shading, we will need to pay attention to this fact so that we arrive at the correct feasible solution region.
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The last two inequalities (constraints) guarantee that the graph will be in the first quadrant. We have graphed each of the first three inequalities (constraints). Now we just need to determine where the overlap exists among all of the given inequalities (constraints). Then we need to determine the vertex coordinates of the points where the graphed lines intersect and form the polygonal solution region.
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Once we have the vertex coordinates, we can evaluate the given objective function:
vertices objective
x y F = 0.25x + 0.35y
3000 5000 F = 0.25(3000) + 0.35(5000) = 2500 13,000 5000 F = 0.25(13000) + 0.35(5000) = 5000 21,000 0 F = 0.25(21000) + 0.35(0) = 5250
8000 0 F = 0.25(8000) + 0.35(0) = 2000
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Finally, just find the highest (maximum) value of these evaluations. Using the vertex coordinates, we obtain the evaluations of: 2500, 5000, 5250 and 2000…the highest of these evaluations being: 5250.
Therefore, the optimal maximum value for the given constraints is:
MAX = 5250 when x = 21000 and y = 0
(which might mean stopping production of the gourmet pancake mixes).
example 1.38b The Monmouth Drapery Company makes 3 types of draperies at 2 different locations:
• At location #1, it can make 20 pairs of deluxe drapes, 15 pairs of better drapes, and 6 pairs of standard drapes per day.
• At location #2, it can make 10 pairs of deluxe drapes, 20 pairs of better drapes, and 12 pairs of standard drapes per day.
The company has orders for 2000 pairs of deluxe drapes, 2250 pairs of better drapes, and 1200 pairs of standard drapes. If the daily costs are $900 per day at location #1 and $750 per day at location #2, how many days should be scheduled at each location in order to fill the orders at minimum cost?
What is that minimum cost?
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x y
location 1 location 2 symbol orders
deluxe 20 10 ≥ 2000
better 15 15 ≥ 2250
standard 6 12 ≥ 1200
daily costs 900 750 = MIN
constraints (inequalities):
deluxe: 20x + 10y ≥ 2000
better: 15x + 15y ≥ 2250
standard: 6x + 12y ≥ 1200
x-values: x ≥ 0
y-values: y ≥ 0
objective: F = 900x + 750y
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The last two inequalities (constraints) guarantee that the graph will be in the first quadrant. We have graphed each of the first three inequalities (constraints). Now we just need to determine where the overlap exists among all of the given inequalities (constraints). Then we need to determine the vertex coordinates of the points where the graphed lines intersect and form the polygonal solution region.
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Once we have the vertex coordinates, we can evaluate the given objective function:
vertices objective
x y F = 900x + 750y
0 200 F = 900(0) + 750(200) = 150,000 50 100 F = 900(50) + 750(100) = 120,000 100 50 F = 900(100) + 750(50) = 127,500 200 0 F = 900(200) + 750(0) = 180,000
Finally, just find the highest (maximum) value of these evaluations. Using the vertex coordinates, we obtain the evaluations of 150,000
120,000 127,500 180,000
The lowest of these evaluations is: 120,000. Therefore, the optimal minimum value for the given constraints is: MAX = $120,000 when x = 50 and y = 100 (which means that the company should schedule 50 days at location #1 and 100 days at location #2 to make enough draperies to satisfy the given orders).
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