Solve the following linear programming problem graphically:
Minimise Z = x + y
subject to the constraints x - y ≤ - 1, - x + y ≤ 0, x, y ≥ 0.
We are to maximise
Z = x + y subject to the constraints x - y ≤ - 1, - x + y ≤ 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 - y = - 1
For x = 0, - y = - 1 ⇒ y = 1
For y = 0, x = - 1
∴ line x - y = 1 meets OX in A (- 1, 0) and OY in B (0, 1)
Again we draw the graph of - x + y = 0
For x = 0, y = 0
For y = 0, x = 0
∴ line - x + y = 0 passes through O (0, 0).
Since feasible region satisfies all the constraints
∴ in this case, feasible region is empty
∴ there exists no solution to the given linear programming problem.
