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प्रश्न
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
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उत्तर
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 A1(28, 0) and B1(0, 14). Join these points to obtain the line x + 2y = 28.
The line 3x + y = 24 meets meets the coordinate axes at A2(8, 0) and B2(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 A3(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 A3(2, 0), A2(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 | |
| A3(2, 0) | Z = 20 × 2 + 10 × 0 = 40 | |
| A2(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.
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