Question

Sanderson Manufacturing produces ornate, decorative wood frame doors and windows. Each item produced goes through three manufacturing processes: cutting, sanding, and finishing. Each door produced requires 1 hour in cutting, 30 minutes in sanding, and 30 minutes in finishing. Each window requires 30 minutes in cutting, 45 minutes in sanding, and 1 hour in finishing. In the coming week Sanderson has 40 hours of cutting capacity available, 40 hours of sanding capacity, and 60 hours of finishing capacity. Assume all doors produced can be sold for a profit of $500 and all windows can be sold for a profit of $400.

Required:
a. Formulate an LP model for this problem.
b. Sketch the feasible region.
c. What is the optimal solution?

Answer 1

Let X1 be the number of decorative wood frame doors and X2 be the number of windows.

The profit earned from selling each door is $500 and the profit earned from selling of each window is $400.

The Sanderson Manufacturer wants to maximize their profit. So for this model, the objective function is

Max: 500X1 + 400X2

Now the total time available for cutting of door and window are 2400 minutes.

so the time taken in cutting should be less than or equal to 2400.

60X1 + 30X2 ≤ 2400

The total available time for sanding of door and window are 2400 minutes. Therefore, the time taken in sanding will be less than or equal to 2400. 30X1 + 45X2 ≤ 2400

The total time available for finishing of door and window is 3600 hours. Therefore, the time taken in finishing will be less than or equal to 3600. 30X1 + 60X2 ≤ 3600

As the number of decorative wood frame door and the number of windows cannot be negative.

Therefore, X1, X2 ≥ 0

so the question s

a)

The LP mode for this model is;

Max: 500X1 + 400X2

Subject to:

60X1 + 30X2 ≤ 2400

]30X1 +45X2 ≤ 2400

30X1 + 60X2 ≤ 3600

X1, X2 ≥ 0

b) Plot the graph of the LP

Max: 500X1+ 400X2

Subject to:

60X1 + 30X2 ≤ 2400

30X1 + 45X2 ≤ 2400

30X1 + 60X2 ≤ 3600

X1,X2

≥ 0

In the uploaded image of the graph, the shaded region in the graph is the feasible region.

c) Consider the following corner point's (0,0), (0, 53.33), (20, 40) and (40, 0) of the feasible region from the graph

At point (0, 0), the objective function,

500X1 + 400X2 = 500 × 0 + 400 × 0

= 0

At point (0, 53.33), the value of objective function,

500X1 + 400X2 = 500 × 0 + 400 × 53.33 = 21332

At point (40, 0), the value of objective function,

500X1 + 400X2 = 500 × 40 + 400 × 0 = 20000

At point (20, 40), the value of objective function

500X1 + 400X2 = 500 × 20 + 400 × 40 = 26000

The maximum value of the objective function is

26000 at corner point ( 20, 40 )

Hence, the optimal solution of this problem is

X1 = 20, X2 = 40 and the objective is 26000