Question & Answer: Maximize Z= 15X1 +15X2 Subject​ to: C1 3×1 + 1×2…..

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

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