We are to minimise and maximise Z = 5x + 10y subject to constraints x + 2 y ≤ 120
x + y ≥ 60
x - 2y ≥ 0
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 = 120
For x = 0, 2 y = 120 or y = 60
For y = 0, x = 120
∴ line meets OX in A( 120, 0) and OY in L(0, 60).
Also we draw the graph of
x + y = 60.
For r = 0, y = 60
For y = 0, x = 60
∴ line meets OX in B(60, 0) and OY in L(0, 60).
Again we draw the graph of
x - 2y = 0
This is a line through the origin and C(40, 0), which is point of intersection of x - 2 y = 0 and x + y = 60

Since feasible region satisfies all the constraints.
∴ BADC is the feasible region.
Comer points are B(60, 0), A(120, 0), D(60, 30), C(40, 20).
At B(60, 0), Z = 300 + 0 = 300
At A(120, 0), Z = 600+ 0 = 600
At D(60, 30), Z = 300 + 300 = 600
At C(40, 20), Z = 200 + 200 = 400
∴ minimum value = 300 at (60, 0) and maximum value = 600 at (120, 0) and (60, 30).