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Class 12 Linear Programming · Very Long Answer5 Marksgraphical_optimization

Class 12 Maths · Linear Programming · Very Long Answer #65

Maximize Z=1000*x + 600*y subject to x + y ≤ 200; -x ≤ -20; 4*x - y ≤ 0; -x ≤ 0; -y ≤ 0. Solve graphically.
Coordinate-derived feasible-region graph for cbse12-math-lp-p1-practice_long-016
Coordinate-derived feasible-region graph for cbse12-math-lp-p1-practice_long-016

✅ Step-by-Step Solution

Final Answer: max Z=136000 at (40,160).
  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 = 200; -x = -20; 4*x - y = 0; -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: (20,80), (20,180), (40,160).
  4. Evaluate the objective at every vertex: (20,80) → 68000; (20,180) → 128000; (40,160) → 136000.
  5. The region is bounded. max Z=136000 at (40,160).

📋 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 = 200; -x = -20; 4*x - y = 0; -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: (20,80), (20,180), (40,160).
  • [1 mark] Evaluate the objective at every vertex: (20,80) → 68000; (20,180) → 128000; (40,160) → 136000.
  • [1 mark] The region is bounded. max Z=136000 at (40,160).

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