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Class 12 Linear Programming · Very Long Answer5 Marksgraphical_optimization

Class 12 Maths · Linear Programming · Very Long Answer #71

Maximize Z=45000*x + 5000*y subject to x + y ≤ 250; 5*x + 8*y ≤ 1400; -x ≤ 0; -y ≤ 0. Solve graphically.
Coordinate-derived feasible-region graph for cbse12-math-lp-p1-practice_long-022
Coordinate-derived feasible-region graph for cbse12-math-lp-p1-practice_long-022

✅ Step-by-Step Solution

Final Answer: max Z=11250000 at (250,0).
  1. Draw each boundary ax+by=c and retain the closed half-plane ax+by≤c; intersect all constraints, including x,y≥0.
  2. Boundary equations: x + y = 250; 5*x + 8*y = 1400; -x = 0; -y = 0. Use an off-line test point to choose each half-plane.
  3. Solving pairs of boundary equations and rejecting points violating any inequality gives vertices: (0,0), (0,175), (200,50), (250,0).
  4. Evaluate the objective at every vertex: (0,0) → 0; (0,175) → 875000; (200,50) → 9250000; (250,0) → 11250000.
  5. The region is bounded. max Z=11250000 at (250,0).

📋 CBSE Stepwise Marking Scheme

  • [1 mark] Draw each boundary ax+by=c and retain the closed half-plane ax+by≤c; intersect all constraints, including x,y≥0.
  • [1 mark] Boundary equations: x + y = 250; 5*x + 8*y = 1400; -x = 0; -y = 0. Use an off-line test point to choose each half-plane.
  • [1 mark] Solving pairs of boundary equations and rejecting points violating any inequality gives vertices: (0,0), (0,175), (200,50), (250,0).
  • [1 mark] Evaluate the objective at every vertex: (0,0) → 0; (0,175) → 875000; (200,50) → 9250000; (250,0) → 11250000.
  • [1 mark] The region is bounded. max Z=11250000 at (250,0).

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