The Gutchi Company manufactures purses, shaving bags, and backpacks. The construction of the three products requires leather and synthetic material, with leather being the limiting raw material.
The production process uses two types of skilled labor, sewing and finishing. The following table gives the availability of the resources, their usage by the three products, and the profits per unit

Resource requirements per unit 

Resource 
Purse 
Bag 
Backpack 
Daily availability 
Leather (ft^{2}) 
2 
1 
3 
42 
Sewing (hr) 
2 
1 
2 
40 
Finishing (hr) 
1 
0.5 
1 
45 
Profit ($) 
24 
22 
45 

 Formulate the problem as a linear program, and find the optimal solution.
 From the optimal solution, determine the state of each resource.
We got this problem from the book “Operations Research” by HAMDY A. TAHA. Chapter 3: Simplex Method and Sensitivity Analysis, Exercise 3.3B – 11. The Gutchi Company.
Solution
a. We define the variables:
 x_{1} = number of purses per day.
 x_{2}=number of bags per day.
 x_{3}=number of backpacks per day.
The statement of the linear programming problem is:
Maximize Z=24x_{1}+22x_{2}+45x_{3}
Subject to:
2x_{1}+x_{2}+3x_{3}β€42
2x_{1}+x_{2}+2x_{3}β€40
x_{1}+0.5x_{2}+x_{3}β€45
x_{1}, x_{2}, x_{3}β₯0
To solve the problem, we use our online calculator of the simplex method:
Initial Table
Table 1 
C_{j} 
24 
22 
45 
0 
0 
0 

C_{b} 
Base 
X_{1} 
X_{2} 
X_{3} 
S_{1} 
S_{2} 
S_{3} 
R 
0 
S_{1} 
2 
1 
3 
1 
0 
0 
42 
0 
S_{2} 
2 
1 
2 
0 
1 
0 
40 
0 
S_{3} 
1 
1/2 
1 
0 
0 
1 
45 

Z 
24 
22 
45 
0 
0 
0 
0 
Enter the variable X_{3} and the variable S_{1} leaves the base. The pivot element is 3
Iteration 1
Table 2 
C_{j} 
24 
22 
45 
0 
0 
0 

C_{b} 
Base 
X_{1} 
X_{2} 
X_{3} 
S_{1} 
S_{2} 
S_{3} 
R 
45 
X_{3} 
2/3 
1/3 
1 
1/3 
0 
0 
14 
0 
S_{2} 
2/3 
1/3 
0 
2/3 
1 
0 
12 
0 
S_{3} 
1/3 
1/6 
0 
1/3 
0 
1 
31 

Z 
6 
7 
0 
15 
0 
0 
630 
Enter the variable X_{2} and the variable S_{2} leaves the base. The pivot element is 1/3
Iteration 2
Table 3 
C_{j} 
24 
22 
45 
0 
0 
0 

C_{b} 
Base 
X_{1} 
X_{2} 
X_{3} 
S_{1} 
S_{2} 
S_{3} 
R 
45 
X_{3} 
0 
0 
1 
1 
1 
0 
2 
22 
X_{2} 
2 
1 
0 
2 
3 
0 
36 
0 
S_{3} 
0 
0 
0 
0 
1/2 
1 
25 

Z 
20 
0 
0 
1 
21 
0 
882 
The optimal solution is Z = 882
X_{1}= 0, X_{2}= 36, X_{3}= 2, S_{1}= 0, S_{2}= 0, S_{3}= 25
b. Status of Resources:
Resource 
Slack 
Status 
Leather 
0 
Scarce 
Sewing 
0 
Scarce 
Finishing 
25 
Abundant 