Quantative Methods Essay

582 WordsOct 29, 20123 Pages
Question 1: Maximize 1.25s + 2L | Small bag | Large bag | Available | Bat-Bars | 2 | 6 | 200 | Crazee-Crunches | 3 | 4 | 150 | Ghostly-Gobs | 5 | 8 | 300 | Selling price | $ 1.25 | $ 2 | * 2s + 6L < 200 3s + 4L < 150 5s + 8L < 300 s, L > 0 Question 2: Minimize 0.6G1 + 0.8G2 | Gain I | Gain II | Requirements | Protein | 2 | 4 | at least 20 | Iron | 5 | 1 | at least 16 | Carbohydrates | 5 | 6 | at least 46 | Selling price | $ 0.60 | $ 0.80 | 2G1 + 4G2 > 20 5G1 + G2 > 16 5G1 + 6G2 > 46 G1, G2 > 0 Question 3: Maximize Z = 2 x + 4 y Graph 1: 2x + y ≤ 40 2x + y = 40 Using a table of values: x | y | 0 | 40 | 10 | 20 | 20 | 0 | Trying (0,0) 2 (0) + 1(0) < 40 0 < 40 TRUE Graph 2: x + y ≤ 25 x + y = 25 Using a table of values: x | y | 0 | 25 | 10 | 15 | 20 | 30 | Trying (0,0) 1 (0) + 1(0) < 25 0 < 25 TRUE Graph 3: 3x - 2y > 0 3x - 2y = 0 Using a table of values: x | y | 0 | 0 | 10 | 15 | 20 | 30 | Trying (30,0) 3 (30) - 2(0) > 0 90 > 0 TRUE The feasible region has four corner points A, B, C, D | x | y | 2 x + 4 y | A | 0 | 0 | 2(0) + 4(0) = 0 | B | 20 | 0 | 2(20) + 4(0) = 40 | C | 15 | 10 | 2(15) + 4(10) = 70 | D | 10 | 15 | 2(10) + 4(15) = 80 | * To solve point C ( intersection of line 1 and line 2 ): 2x + y = 40 y = -2x + 40 x + y = 25 y = -x + 25 -2x + 40 = -x + 25 -2x + x = 25 – 40 x = 15 to find y: 2(15) + y = 40 y = 40 – 30 y = 10 * To solve point D ( intersection of line 2 and line 3 ): x + y = 25 y = -x + 25 3x - 2y = 0 y = 1.5x -x + 25 = 1.5x

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