Ex 12.1, 9 - Maximise Z = -x + 2y subject to x > 3, x + y > 5

Ex 12.1, 9 - Chapter 12 Class 12 Linear Programming - Part 2
Ex 12.1, 9 - Chapter 12 Class 12 Linear Programming - Part 3 Ex 12.1, 9 - Chapter 12 Class 12 Linear Programming - Part 4

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Transcript

Ex 12.1, 9 Maximise Z = – x + 2y, subject to the constraints: x ≥ 3, x + y ≥ 5, x + 2y ≥ 6, y ≥ 0 Maximize Z = –x + 2y Subject to, x ≥ 3 x + y ≥ 5 x + 2y ≥ 6 y ≥ 0 But as the feasible region is unbounded Hence 1 can or cannot be the maximum value of z So, we need to graph Inequality –x + 2y > 1 Since feasible region of –x + 2y > 1 has some points in common. So there is no maximum value for Z subject to given constraints.

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