Solve the following L.P.P. graphically Maximise Z = 4x + y
Subject to following constraints x + y ≤ 50
3x + y ≤ 90,
x ≥ 10
x, y ≥ 0
Solution
The given constraints are x + y ≤ 50, 3x + y ≤ 90, x ≥ 10 and x, y ≥ 0.
Converting the inequations into equations, we obtain the following equations:
x + y = 50, 3x + y = 90, x = 10, x = 0 and y = 0
These equations represents straight lines in XOY plane.
The line x + y =50 meets meets the coordinate axes at A_{1}(50, 0) and B_{1}(0, 50). Join these points to obtain the line x + y = 50.
The line 3x + y = 90 meets meets the coordinate axes at A_{2}(30, 0) and B_{2}(0, 90). Join these points to obtain the line 3x + y = 90.
The line x = 10, is parallel to y-axis, passes through the point A_{3}(10, 0).
Also, x = 0 is the y-axis and y = 0 is the x-axis.
The feasible region of the LPP is shaded below.
Point of intersection of lines x + y = 50 and 3x + y = 90 is Q(20, 30).
Point of intersection of lines x = 10 and x + y = 50 is R(10, 40).
The coordinates of the corner points of the feasible region are A3(10, 0), A2(30, 0), Q(20, 30) and R(10, 40).
The values of the objective function at these points are given in the following table:
Point | Value of the objective function Z = 4x + y | |
A_{3}(10, 0) | Z = 4 × 10 + 0 = 40 | |
A_{2}(30, 0) | Z = 4 × 30 + 0 = 120 | Maximum |
Q(20, 30) | Z = 4 × 20 + 30 = 110 | |
R(10, 40) | Z = 4 × 10 + 40 = 80 |
Thus, Z is maximum at A_{2}(30, 0). The maximum value of Z is 120