Given : Z = 5x + 10y ……………. (1)
subject to constraints x + 2y ≤ 120 …………….(2)
x + y > 60 …………….(3)
x – 2y ≥ 0 …………….(4)
x ≥ 0, y ≥ 0 …………….(5)
First we locate the region represented by (2), (3), (4) & (5) for this consides the lines ;
x + 2y = 120 which passes through A(120, 0) & B(0, 60)
x + y = 60 which passes through C (60, 0) & B (0, 60) and x – 2y
= 0 which passes through 0 (0, 0) & P (40, 20)
Now that P(40, 20) lies on BC also. Also x – 2y = 0&x + 2y= 120 meet in Q (60, 30)
The feasible region is shown shaded in the figure (indicated as feasible region) note that (0,0) does not lie in this region as it does not satisfy (3) The point C(60, 0) lie this region as it satisfy all the constraints (2) (3) (4) and (5).
The corner points of the feasible region which are to be examined for optimum solution are C(60,0) A( 120,0) Q(60, 30) and P(40,20)
At C(60, 0), Z = 5 × 60 + 10 × 0 = 300
At A(120, 0), Z = 5 × 120 + 10 × 0 = 600
At Q(60, 30), Z = 5 × 60 + 10 × 30 = 600
At P(40, 20), Z = 5 × 40 + 10 × 20 = 400
Hence Z is minimum at C(60,0) and min Z = 300 and max Z = 600 at A and also at Q ⇒ max Z will be attained at every point of segment AQ.
