Respuesta :
Answer:
Let X1 be the number of decorative wood frame doors and X2 be the number of windows. Â
The profit earned from selling each door is $500 and the profit earned from selling of each window is $400. Â
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. Â
so the time taken in cutting should be less than or equal to 2400. Â
60X1 + 30X2 ≤ 2400 Â
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 Â
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 Â
As the number of decorative wood frame door and the number of windows cannot be negative. Â
Therefore, X1, X2 ≥ 0
so the question s
a)
The LP mode for this model is;
Max: 500X1 + 400X2 Â
Subject to: Â
60X1 + 30X2 ≤ 2400 Â
]30X1 +45X2 ≤ 2400 Â
30X1 + 60X2 ≤ 3600 Â
X1, X2 ≥ 0 Â
b) Plot the graph of the LP Â
Max: 500X1+ 400X2 Â
Subject to: Â
60X1 + 30X2 ≤ 2400 Â
30X1 + 45X2 ≤ 2400 Â
30X1 + 60X2 ≤ 3600
X1,X2 Â
≥ 0
In the uploaded image of the graph, the shaded region in the graph is the feasible region. Â
c) Consider the following corner point's (0,0), (0, 53.33), (20, 40) and (40, 0) of the feasible region from the graph Â
At point (0, 0), the objective function, Â
500X1 + 400X2 = 500 × 0 + 400 × 0 Â
= 0
At point (0, 53.33), the value of objective function,
500X1 + 400X2 = 500 × 0 + 400 × 53.33 = 21332 Â
At point (40, 0), the value of objective function, Â
500X1 + 400X2 = 500 × 40 + 400 × 0 = 20000 Â
At point (20, 40), the value of objective function
500X1 + 400X2 = 500 × 20 + 400 × 40 = 26000 Â
The maximum value of the objective function is Â
26000 at corner point ( 20, 40 )
Hence, the optimal solution of this problem is Â
X1 = 20, X2 = 40 and the objective is 26000
