We are to maximise
Z = 4x + y
subject to the constraints
x + y ≤ 50
3x + y ≤ 90
x ≥ 0, 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.
Now we draw the graph of the line x + y = 50
For x = 0, y = 50
For y = 0, x = 50
∴ line meets OX in A(50, 0) and OY in L(0, 50)
Let us draw the graph of line 3 x + y = 90
For x = 0, y = 90
For y = 0, 3x = 90 or x = 30
∴ line meets OX in B(30, 0) and OY in M(0, 90).
Since feasible region is the region which satisfies all the constraints.
∴ OBCL is the feasible region, which is bounded.
The comer points are
O(0, 0), B(30, 0), C(20, 30), L(0, 50)
At O(0, 0), Z = 0 + 0 = 0
At B(30, 0), Z = 120 + 0 = 120
At C(20, 30), Z = 80 + 30 = 110
At L(0, 50), Z = 0 + 50 = 50
∴ maximum value = 120 at the point (30, 0).
Tips: -
Note: Coordinates of C can be found by two methods:
Method I: Draw the graph of inequalities on the graph paper. So coordinates of C can be determined.
Method II: Solve the two equation x + y = 50, 3x + y = 90 by any method to find coordinates of C.