Maximize | Z= | 15X1 +15X2 | ||
Subject to: | C1 3×1 + 1×2 <= 300
C2 1×1 + 1×2 <= 200 C3 1×1<=100 C4 1×2>=50 C5 1×1 – 1×2 <= 0 x1, x2 >= 0 The optimum solution: x1: x2: Optimal solution for Z: |
Expert Answer
Using graphical method:
The 5 points and Z value is:
A = 0,200
Z = 15*0 + 15*200 = 3000
B = 0,50
Z = 15*0 + 15*50 = 750
C = Intersection of 3×1 + x2 <= 300 & x1 + x2 <= 200
C = 50,150
Z = 15*50 + 15*150 = 3000
D = Intersection of 3×1 + x2 <= 300 & x1 – x2 <= 0
D = 75,75
Z = 15*75 + 15*75 = 2250
E = 50,50
Z = 15*50 + 15*50 = 1500
Thus, both A & C points are optimum solution