Chapter 11 – Wisniewski
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A toy manufacturer preparing a production schedule for two new toys, trucks and spinning tops, must use the information concerning their manufacturing times, given in the following table. Machine AMachine BFinishingTruck2 hr3 hr5 hrSpinning Top1 hr1 hr1 hrThe available hours per week per machine are: Machine A, 80 hours; Machine B, 50 hours; Finishing, 70 hours. If the profits on each truck and spinning top are $7 and $2 respectively, how many of each toy should be made per week in order to maximize profit? What is the maximum profit?Decision Variables: x = number of trucks produced per weeky = number of spinning tops produced per weekObjective Function:Maximize profit 7x + 2ySubject to:Machine A capacity limit (in hours): 2x + 1y ≤ 80 hours/weekMachine B capacity limit (in hours) 3x + 1y ≤ 50 hours/weekFinishing capacity limit (in hours) 5x + 1y ≤ 70 hours/weekNon-negativity x, y ≥ 0Calculations:Graphing the constraints:The x- and y- intercepts for the of the constraints areConstraintx-intercepty-interceptMachine A capacity2x + 1y ≤ 80(40, 0)(0, 80)Machine B capacity3x + 1y ≤ 50(16.67, 0)(0, 50)Finishing capacity5x + 1y ≤ 70(14, 0)(0, 70)Finding the corner pointsThe intersection of the Machine B constraint and the Finishing constraint forms one of the corners of the feasible region so the coordinates must be calculated.Subtract the equation for Finishing from the equation for Machine B and solve for xMachine B 3x + 1y = 50Finishing 5x + 1y = 70[pic 1]
-2x = -20 x = 10Substitute the value for x into one of the constraints and solve for yMachine B 3(10)+ 1y = 50 30+ 1y = 50 y = 20 Determine which corner point maximizes weekly profit[pic 2](0,0) 7(0) + 2(0) = $0/week (10, 20) 7(10) + 2(20) = $110/week(0,50) 7(0) + 2(50) = $100/week (14, 0) 7(14) + 2(0) = $98/week[pic 3][pic 4][pic 5]A produce grower is purchasing fertilizer containing three nutrients: A, B and C. The minimum weekly requirements are 80 units of A, 120 of B, and 240 of C. There are two popular blends of fertilizer on the market. Blend I costing $8 per bag, contains 2 units of A, 6 of B and 4 of C. Blend II costing $10 per bag, contains 2 units of A, 2 of B and 12 of C. How many bags of each blend should the grower buy each week to minimize the cost of meeting the nutrient requirements?Decision Variables: x = Number of bags of Blend I the grower should purchase per weeky = Number of bags of Blend II the grower should purchase per weekObjective Function: Minimize cost $8x + $10ySubject to:Minimum weekly requirement for Nutrient A: 2x + 2y ≥ 80Minimum weekly requirement for Nutrient B 6x + 2y ≥ 120Minimum weekly requirement for Nutrient C 4x + 12y ≥ 240Non-negativity x, y ≥ 0Calculations:Graphing the constraints:The x- and y- intercepts for the of the constraints areConstraintx-intercepty-interceptNutrient A2x + 2y ≥ 80(40, 0)(0, 40)Nutrient B6x + 2y ≥ 120(20, 0)(0, 60)Nutrient C4x + 12y ≥ 240(60, 0)(0, 20)Finding the corner pointsThe intersection of the Nutrient A and Nutrient B constraints forms one of the corners of the feasible region so the coordinates must be calculated.Subtract the equation for Nutrient B from the equation for Nutrient A and solve for xNutrient A 2x + 2y = 80Nutrient B 6x + 2y = 120