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Question
Maximise z = 8x + 9y subject to the constraints given below :
2x + 3y ≤ 6
3x − 2y ≤6
y ≤ 1
x, y ≥ 0
Solution
The given constraints are
2x + 3y ≤ 6
3x − 2y ≤ 6
y ≤ 1
x, y ≥ 0
Converting the given inequations into equations, we get 2x + 3y = 6, 3x − 2y = 6, y = 1, x = 0 and y = 0. These lines are drawn on the graph and the shaded region OABCD represents the feasible region of the given LPP.
It can be observed that the feasible region is bounded. The coordinates of the corner points of the feasible region are O(0, 0), A(1, 0), B
\[\left( \frac{3}{2}, 1 \right)\], C \[\left( \frac{30}{13}, \frac{6}{13} \right)\] and D(2, 0).
The values of the objective function, z at these corner points are given in the following table:
| Corner Point | Value of the Objective Function z = 8x + 9y |
| O(0, 0) | z = 8 × 0 + 9 × 0 = 0 |
| A(1, 0) | z = 8 × 1 + 9 × 0 = 8 |
| B \[\left( \frac{3}{2}, 1 \right)\] | z = 8 × \[\frac{3}{2}\] + 9 ×1 = 21 |
| C \[\left( \frac{30}{13}, \frac{6}{13} \right)\] | z = 8 × \[\frac{30}{13}\] + 9 × \[\frac{6}{13}\] = \[\frac{294}{13}\] |
| D (2, 0) | z = 8 × 2 + 9 × 0 = 16 |
From the table, z is maximum at \[x = \frac{30}{13}\] and \[y = \frac{6}{13}\] and the maximum value of z is \[\frac{294}{13}\].
Thus, the maximum value of z is \[\frac{294}{13}\].
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