Quantative Methods

Question 1: Maximize 1.25s + 2L | Small bag| macroscopic bag| Available| Bat-Bars| 2| 6| 200| Crazee-Crunches| 3| 4| 150| Ghostly-Gobs| 5| 8| 300| dish out price| $ 1.25| $ 2| * 2s + 6L < 200 3s + 4L < 150 5s + 8L < 300 s, L > 0 Question 2: smear 0.6G1 + 0.8G2 | Gain I| Gain II| Requirements| Protein| 2| 4| at to the lowest degree(prenominal) 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 chart 1: 2x + y ? 40 2x + y = 40 utilise a lock up of values: x| y| 0| 40| 10| 20| 20| 0| hard (0,0) 2 (0) + 1(0) < 40 0 < 40 TRUE graphical record 2: x + y ? 25 x + y = 25 Using a get across 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 quaternary respite occlusions 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 menstruation C ( intersection of rootage 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 move up 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 -x 1.5x = -25 x = 10 to find y: 3(10) - 2y = 0 y = 15 The maximum...If you want to get a full essay, order it on our website: OrderCustomPaper.com

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