Problem 2.7
Problem Details:
Wilson manufacturing produces both baseballs and softballs, which it wholesales to vendors around the country. Its facilities permit the manufacture of a maximum of 500 dozen baseballs and a maximum of 500 dozen softballs each day. The cowhide covers for each ball are cut from the same processed cowhide sheets. Each dozen baseballs require 5 square feet of cowhide, including waste, whereas each dozen softballs require 6 square feet. Wilson has 3600 square feet of cowhide sheets available every day.
Production of baseballs and softballs includes making the inside core, cutting and sewing the cover, and packaging. It takes about 1 minute to manufacture a dozen baseballs and 2 minutes to manufacture a dozen softballs. A total of 960 minutes is available for production daily.
A) Formulate the Problem & Graph the Feasible Region
Decision Variables:
X1 = Number of dozens of baseballs to produce
X2 = Number of dozens of softballs to produce
Objective Function:
The objective is total profit maximization for one day to be achieved by selecting the optimal number of baseballs and softballs to produce given all applicable explicit and implicit constraints.
(Profit per dozen units produced) x (Number of Dozen Units Produced Daily)
Maximize $7X1 + $10X2
Constraints:
- Total number of dozen baseballs produced cannot exceed manufacturing capacity of 500 dozen baseballs per day.
- X1 ≤ 500 (production constraint)
- Total number of dozen softballs produced cannot exceed manufacturing capacity of 500 dozen softballs per day.
- X2 ≤ 500 (production constraint)
- Total number of square feet of cowhide used to manufacture both baseballs and softballs combined cannot exceed 3,600 sqft per day.
- 5X1 + 6X2 ≤ 3600 (Material constraint)
- Total number of minutes used to manufacture to manufacture both baseballs and softballs combined cannot exceed 960 minutes per day.
- X1 + 2X2 ≤ 960 (Manufacturing time constraint)
- The number of dozens of baseballs as well as softballs produced may not be negative.
- X1≥ 0, X2 ≥ 0 (non-negativity)
Linear Programming Model:
Maximize $7X1 + $10X2(Total daily profit)
Subject to:
C1 = X1≤ 500 (Production constraint)
C2 = X2 ≤ 500 (Production constraint)
C3 = 5X1+ 6X2≤ 3600 (Material constraint)
C4 = X1+ 2X2≤ 960 (Manufacturing time constraint)
X1≥ 0, X2≥ 0 (non-negativity constraint)
Graph of the Feasible Region
Graph Description:
Decision variable X1 is plotted on the horizontal axis while X2 on the vertical axis. Due to the non-negativity constraint, the graph lies within the first quadrant. Next, each constraint line is plotted on the graph. The feasible area, shaded in gray, is the area containing all points which satisfy all constraints simultaneous. Any point outside this region, is infeasible, meaning that it violates at least one of the constraints of the model.
B)Point X (300,300) is a interior point as it satisfies all constraints, lies within the feasible region and not on the boundary of a constraint line, nor is it an extreme or corner point. Point Y (350,350) is an infeasible point as it lies outside the feasible region and does not satisfy all constraints (violates: 5X1 + 6X2 ≤ 3600 (Material constraint) 5(350) + 6(350)=3,850 ≤ 3600 [False] and X1 + 2X2 ≤ 960 (Manufacturing time constraint) 350 + 2(350)= 1050 ≤ 960=false ). The optimal solution to a Linear Programming Model is always an extreme/corner point. Since neither point is acorner point of the feasible region neither could possibly be the optimal solution.
C)The optimal solution to a linear program model with a bounded feasible region is an extreme/corner point. By inputting the coordinates of each corner point into the objective function $7X1 + $10X2 one can determine the optimal solution which maximizes the objective function. The results of this process, as indicated in the chart below, denote that CornerPointD (360, 300), maximizes profit and is therefore the optimal solution. Thus, 360 baseballs and 300 softballs should be produced weekly in order to maximize profit and the optimal value of the objective function or maximized profit is $5,520.
Corner Points / X1 / X2 / $7X1 + $10X2A / 0 / 0 / $0.00
B / 500 / 0 / $3,500.00
C / 500 / 550/3 / $5,333.33
D / 360 / 300 / $5,520.00
E / 0 / 480 / $4,800.00
* Rounded down to 183
All corner points were readily known from the graph of the feasible region with the exception of C & D which had to be calculated by solving the system of equations for the two constraint lines which intersect at each of those respective points.
Calculation for CornerPointC
C1 = X1 <= 500
C3 = 5X1+ 6X2≤ 3,600
Solve by system via substitution
5X1+ 6X2= 3,600;X1=500
5(500) +6x2 =3,600
6X2=1,100
X2= 550/3 = 183.33 (X2 = 183,rounded down, or up 184 may cause infeasibility)
(Notice that one should not round the LP solution, because since the Wilson is not going out of business, he can produce 183 plus 1/3 of work in progress. Rounding may also cause infeasibility.Point C = (500, 550/3) soon Friday he make 183 + 1/3 then on Monday he complete the work in progress first.
Calculation for CornerPointD
C3 = 5X1+ 6X2≤ 3,600
C4 = X1+ 2X2≤ 960
Solve system via elimination method
5X1+ 6X2= 3,600
3(X1+ 2X2)= 3(960)
5X1+ 6X2= 3,600
3x1 + 6x2 = 2880
2x1 = 720
X1 = 360
X1+ 2X2=960;X1=360
360 + 2X2=960
2X2=600
X2=300
Point D = (360,300)