Example 5 - Chapter 12 Class 12 Linear Programming

Last updated at August 14, 2026 by

Example 5 - Minimise Z = 3x + 2y subject constraints:x + y< 8 - Linear equations given - Not feasible

Example 5 - Chapter 12 Class 12 Linear Programming - Part 2

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Example 5 - Chapter 2 Class 12 Linear Programming - NCERT Minimise Z = 3x + 2y subject to the constraints: x + y ≥ 8 3x + 5y ≤ 15 x ≥ 0, y ≥ 0 Plotting points For x + y ≥ 8 Points are (0, 8) and (8, 0) For 3x + 5y ≤ 15 Points are (0, 3) and (5, 0) Plotting graphs of both lines And shading the relevant parts We notice that There is no common region, thus, there is no feasible region. Hence, there is no maxnimum value of Z

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