Question
Solve the following L.P.P graphically: Maximise Z = 20x + 10y
Subject to the following constraints x + 2y ≤ 28,
3x + y ≤ 24,
x ≥ 2,
x, y ≥ 0
Solution
The given constraints are x + 2y ≤ 28, 3x + y ≤ 24, x ≥ 2 and x, y ≥ 0.
Converting the inequations into equations, we obtain the following equations:
x + 2y = 28, 3x + y = 24, x = 2, x = 0 and y = 0
These equations represents straight lines in XOY plane.
The line x + 2y = 28 meets meets the coordinate axes at A_{1}(28, 0) and B_{1}(0, 14). Join these points to obtain the line x + 2y = 28.
The line 3x + y = 24 meets meets the coordinate axes at A_{2}(8, 0) and B_{2}(0, 24). Join these points to obtain the line 3x + y = 24.
The line x = 2, is parallel to y-axis, passes through the point A_{3}(2, 0).
Also, x = 0 is the y-axis and y = 0 is the x-axis.
The feasible region of the LPP is shaded below.
The point of intersection of lines x + 2y = 28 and 3x + y = 24 is Q(4, 12).
The point of intersection of lines x = 2 and x + 2y = 28 is R(2, 13).
The coordinates of the corner points of the feasible region are A_{3}(2, 0), A_{2}(8, 0), Q(4, 12) and R(2, 13).
The values of the objective function at these points are given in the following table:
Point | Value of the objective function Z = 20x + 10y | |
A_{3}(2, 0) | Z = 20 × 2 + 10 × 0 = 40 | |
A_{2}(8, 0) | Z = 20 × 8 + 10 × 0 = 160 | |
Q(4, 12) | Z = 20 × 4 + 10 × 12 = 200 | Maximum |
R(2, 13 | Z = 20 × 2 + 10 × 13 = 170 |
Clearly, Z is maximum at Q(4, 12) and the maximum value of Z is 200.