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Question
Find graphically, the maximum value of Z = 2x + 5y, subject to constraints given below:
2x + 4y ≤ 8
3x + y ≤ 6
x + y ≤ 4
x ≥ 0, y ≥ 0
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Solution
The given linear programming problem is Maximize Z = 2x + 5y
subject to the constraints
2x + 4y ≤ 8
3x + y ≤ 6
x + y ≤ 4
x ≥ 0, y ≥ 0
Converting the inequations into equations, we obtain the following equations of straight lines:
2x + 4y = 8, 3x + y = 6, x + y = 4
The line 2x + 4y = 8 meets the coordinate axes at (4, 0) and (0, 2).
The line 3x + y = 6 meets the coordinate axes at (2, 0) and (0, 6).
The line x + y = 4 meets the coordinate axes at (4, 0) and (0, 4).
The feasible region determined by the given constraints can be diagrammatically represented as,
2x + 4y = 8 and 3x + y = 6 gives (8/5, 6/5)
The coordinates of the corner points of the feasible region are O(0, 0), A(0, 2), B \[\left(\frac{8}{5}, \frac{6}{5} \right)\] and C(2, 0).
The value of the objective function at these points are given in the following table.
| Corner Point | Z = 2x + 5y |
| (0, 0) | 2 × 0 + 5 × 0 = 0 |
| (2, 0) | 2 × 2 + 5 × 0 = 4 |
| (0, 2) | 2 × 0 + 5 × 2 = 10 → Maximum |
|
\[\left( \frac{8}{5}, \frac{6}{5} \right)\]
|
\[2 \times \frac{8}{5} + 5 \times \frac{6}{5} = \frac{46}{5}\] = 9.2
|
Thus, the maximum value of Z is 10 at x = 0, y = 2.
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