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Unformatted text preview: Objection Function: Maximize Profit using: 60X 1 + 80X 2 = Z[Profit] Constraints Melting: X 1 + X 2 ≤ 45 Extruder A: 3X 1 ≤ 90 Extruder B: 1X 2 ≤ 160 Raw Material: 5X 1 + 4X 2 ≤ 200 B) Graph Below. Optimal Solution at point D. [0,45] C) 60(0) + 80(45) = $3600 = MAX Profit 8) Considering Problem 7 A) Looking at the shadow price on the chart I got these answers: Cutting Time: If costs were less than $4.75, then we’d be willing to pay for an extra hour of cutting Sewing Time: I would not be willing to pay anything for an extra hour of sewing because there is a lot of slack already Yard of Material: I would be willing to pay less than $2.75 for an extra yard of material. It is because the shadow price of an extra yard is $2.75. B) Range of feasibility for each: Cutting Range is [60, 132] Material is [33.33, 100] Shadow Price is valid for those ranges....
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 Spring '11
 christychen
 Optimization, #, optimal solution, $3000, $2.75, $4.75

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