We are to minimise and maximise
Z = x + 2y subject to constraints x + 2y ≥ 100, 2x - y ≤ 0, 2x + y ≤ 200, x, y ≥ 0
Consider a set of rectangular cartesian axes OXY in the plane.
It is clear that any point which satisfies x ≥ 0, y ≥ 0 lies in the first quadrant.
Let us draw the graph of x + 2 y = 100
For x = 0, 2 y = 100 or y = 50
For y = 0, x = 100
∴  line meets OX in A(100, 0) and OY in L(0, 50).
2x - y = 0 is a straight line passing through origin and C(20, 4), which is point of intersection of 2x - y = 0 and x + 2y = 100.
Again we draw the graph of 2x + y = 200.
For x = 0, y = 200
For y = 0, 2x = 200 or x = 100
∴ line meets OX in A(100, 0) and OY in M(0, 200).
Since feasible region satisfies all the constraints.
∴ CDML is the feasible region.
The comer points are C(20, 40), D(50. 100), M(0, 200), L(0, 50).
At C(20, 40), Z = 20 + 80 = 100
At D(50, 100), Z = 50 + 200 = 250
At M(0, 200), Z = 0 + 400 = 400
At L(0, 50), Z = 0 + 100 = 100

∴  maximum value = 400 at (0, 200)
and minimum value = 100 at (20, 40) and (0, 50) i.e. along the segment joining (20, 40) and (0, 50).