Solve the following L.P.P. graphically: Minimize Z=5x+10y subject to the constraints x+2y≤120 x+y≥60 x−2y≥0 x≥0,y≥0
Answer: Min Z = 300 at (60, 0)
Solution
Boundary lines: x+2y=120, x+y=60 and x−2y=0. Corner points of the feasible region: A(60,0), B(120,0), C(60,30) (intersection of x+2y=120 and x−2y=0) and D(40,20) (intersection of x+y=60 and x−2y=0). Evaluate Z=5x+10y: A(60,0): Z=300 ← minimum B(120,0): Z=600 C(60,30): Z=300+300=600 D(40,20): Z=200+200=400
∴Zmin=300 at (60,0)
Feasible region for x + 2y = 120, x + y = 60 and x - 2y = 0, with corner points A(60, 0), B(120, 0), C(60, 30) and D(40, 20).