Answer:
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
