Question

A jean manufacturer makes three types of jeans, each of which goes through three manufacturing phases: cutting, sewing, and finishing. The number of minutes each type of product requires in each of the three phases is given below:

Jean Cutting Sewing Finishing
I 8 12 4
II 12 18 8
III 18 24 12

There are 5200 minutes of cutting time, 6000 minutes of sewing time, and 2200 minutes of finishing time each day. The company can sell all the jeans it makes and makes a profit of $4 on each Jean I, $4.50 on each Jean II, and 6 on each Jean III.

Required:
a. What number of jeans in each category should be made each day to maximize profits?
b. Formulate a linear programming problem that models the problem given above. Be sure to identify all variables used.
c. Analyze the solution.

Answer 1

z ( max) = 2000 $

x₁ = 500 x₂ = 0 x₃ = 0

Explanation:

cutting sewing finishing Profit $

Jean1 (x₁ ) 8 12 4 4

Jean2 (x₂ ) 12 18 8 4.5

Jean3 (x₃ ) 18 24 12 6

Time available 5200 6000 2200

b) Formulation of a linear programming problem:

Objective Function z

z = 4*x₁ + 4.5*x₂ + 6*x₃ to maximize

Constrains:

Constrain 1

Available cutting time 5200 minutes

8*x₁ + 12*x₂ + 18*x₃ ≤ 5200

Constrain 2:

Available sewing time 6000 minutes

12*x₁ + 18*x₂ + 24*x₃ ≤ 6000

Constrain 3:

Available finishing time 2200 minutes

4*x₁ + 8*x₂ + 12*x₃ ≤ 2200

General constraints:

x₁ ≥ 0 x₂ ≥ 0 x₃ ≥ 0 all integers

After 6 iterations the optimal solution is:

z ( max) = 2000 $

x₁ = 500 x₂ = 0 x₃ = 0

We can see in the model that times required to make jean 2 and jean 3

are much bigger than jean 1 and but the price of jean 2 is only 0.5 $ above the price of jean 1 and jean 3 is only 2 $.