Chapter 11 - Wisniewski
Essay by Adrine Michelle Krikorian • February 1, 2017 • Coursework • 1,702 Words (7 Pages) • 1,229 Views
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 A | Machine B | Finishing |
Truck | 2 hr | 3 hr | 5 hr |
Spinning Top | 1 hr | 1 hr | 1 hr |
The 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 week
y = number of spinning tops produced per week
Objective Function:
Maximize profit 7x + 2y
Subject to:
Machine A capacity limit (in hours): 2x + 1y ≤ 80 hours/week
Machine B capacity limit (in hours) 3x + 1y ≤ 50 hours/week
Finishing capacity limit (in hours) 5x + 1y ≤ 70 hours/week
Non-negativity x, y ≥ 0
Calculations:
Graphing the constraints:
The x- and y- intercepts for the of the constraints are
Constraint | x-intercept | y-intercept | |
Machine A capacity | 2x + 1y ≤ 80 | (40, 0) | (0, 80) |
Machine B capacity | 3x + 1y ≤ 50 | (16.67, 0) | (0, 50) |
Finishing capacity | 5x + 1y ≤ 70 | (14, 0) | (0, 70) |
Finding the corner points
The 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 x
Machine B 3x + 1y = 50
Finishing 5x + 1y = 70[pic 1]
-2x = -20
x = 10
Substitute the value for x into one of the constraints and solve for y
Machine 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 week
y = Number of bags of Blend II the grower should purchase per week
Objective Function:
Minimize cost $8x + $10y
Subject to:
Minimum weekly requirement for Nutrient A: 2x + 2y ≥ 80
Minimum weekly requirement for Nutrient B 6x + 2y ≥ 120
Minimum weekly requirement for Nutrient C 4x + 12y ≥ 240
Non-negativity x, y ≥ 0
Calculations:
Graphing the constraints:
The x- and y- intercepts for the of the constraints are
Constraint | x-intercept | y-intercept | |
Nutrient A | 2x + 2y ≥ 80 | (40, 0) | (0, 40) |
Nutrient B | 6x + 2y ≥ 120 | (20, 0) | (0, 60) |
Nutrient C | 4x + 12y ≥ 240 | (60, 0) | (0, 20) |
Finding the corner points
The 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 x
Nutrient A 2x + 2y = 80
Nutrient B 6x + 2y = 120
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