Dwight is an elementary school teacher who also raises pigs for supplemental income. He is trying to decide what to feed his pigs. He is considering using a combination of pig feeds available from local suppliers. He would like to feed the pigs at minimum cost while also making sure each pig receives an adequate supply of calories and vitamins. Dwight has formulated the following Linear Programming (LP) model:
Let A = pounds of Feed “Type A” in diet
B = pounds of Feed “Type B” in diet
Minimize Z = $0.50A + $0.75B (Cost)
subject to 600A + 1,100B ≥ 10,000 (requirement of calories per day)
110A + 90B ≥ 700 (requirement of units of vitamins)
A ≤ (1/4)(A + B) (constraint 3)
and A ≥ 0, B ≥ 0.
To graph the constraints, we first need to determine the coordinates of the points that satisfy each constraint. We can do this by setting one of the variables to zero and solving for the other. Then, we can plot these points and connect them to form the lines that represent each constraint.
First, let’s find the points for constraint 1 (600A + 1,100B ≥ 10,000). Setting A = 0, we have: 1,100B ≥ 10,000 B ≥ 9.09
Setting B = 0, we have: 600A ≥ 10,000 A ≥ 16.67
We now have two points: (0, 9.09) and (16.67, 0).
Next, let’s find the points for constraint 2 (110A + 90B ≥ 700). Setting A = 0, we have: 90B ≥ 700 B ≥ 7.78
Setting B = 0, we have: 110A ≥ 700 A ≥ 6.36
We now have two points: (0, 7.78) and (6.36, 0).
Finally, let’s find the points for constraint 3 (A ≤ (1/4)(A + B)). Setting A = 0, we have: 0 ≤ (1/4)(0 + B) B ≥ 0
Setting B = 0, we have: A ≤ (1/4)(A + 0) A ≥ 0
We now have two points: (0, 0) and (0, Infinity).
We can plot these points and connect the lines to form the feasible region. The feasible region is the area that lies within or on all of the constraint lines, and satisfies all of the constraints. In this case, the feasible region is the area that lies on or below the line for constraint 1, on or to the right of the line for constraint 2, and on or below the line for constraint 3. The feasible region can be shaded or darkened to clearly distinguish it from the rest of the graph.
The objective function, Z = $0.50A + $0.75B, can then be evaluated at several points within the feasible region to find the minimum cost. The optimal solution occurs at the point on the feasible region where the objective function takes on its minimum value.
Answer the following 2 Questions:
1. Determine the optimal solution(s) and the minimum cost (show your calculations). Include “managerial statements” that communicate the results of the analyses (i.e. describe verbally the results).
2. Determine the amount of slack for each of the constraints.
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