9. Maximise Z = – x + 2y, subject to the constraints x ≥ 3, x + y ≥ 5, x + 2y ≥ 6, y ≥ 0.
9. Maximise Z = – x + 2y, subject to the constraints x ≥ 3, x + y ≥ 5, x + 2y ≥ 6, y ≥ 0.
Maximise , subject to the constraints
The corresponding equation of the given inequalities are
The graph of the given inequalities is shown

The feasible region unbounded.
The values of Z at corner points A (6,0), B (4,1), and C (3,2) are as follows

As the feasible region is unbounded z=1 may or may
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Maths Ncert Solutions class 12th 2026
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