![Ex 12.1, 9 - Chapter 12 Class 12 Linear Programming - Part 2](https://d1avenlh0i1xmr.cloudfront.net/7d4f1b75-4876-4d17-8627-113de468c31b/slide26.jpg)
![Ex 12.1, 9 - Chapter 12 Class 12 Linear Programming - Part 3](https://d1avenlh0i1xmr.cloudfront.net/26f0f3f4-b9f5-471b-97eb-d8fde0b55cf0/slide27.jpg)
![Ex 12.1, 9 - Chapter 12 Class 12 Linear Programming - Part 4](https://d1avenlh0i1xmr.cloudfront.net/bd2f57ef-63a7-4f90-a4d5-36afb76ddfb8/slide28.jpg)
Last updated at April 16, 2024 by Teachoo
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.