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Class 12 Linear Programming · Long Answer3 Marksgraphical_optimization

Class 12 Maths · Linear Programming · Long Answer #34

Maximize Z=300*x + 190*y subject to x + y ≤ 24; 2*x + y ≤ 32; -x ≤ 0; -y ≤ 0. Solve graphically.
Coordinate-derived feasible-region graph for cbse12-math-lp-p1-short-008
Coordinate-derived feasible-region graph for cbse12-math-lp-p1-short-008

✅ Step-by-Step Solution

Final Answer: max Z=5440 at (8,16).
  1. Draw each boundary ax+by=c and retain the closed half-plane ax+by≤c; intersect all constraints, including x,y≥0.
  2. Solving pairs of boundary equations and rejecting points violating any inequality gives vertices: (0,0), (0,24), (8,16), (16,0).
  3. Evaluate the objective at every vertex: (0,0) → 0; (0,24) → 4560; (8,16) → 5440; (16,0) → 4800.
  4. The region is bounded. max Z=5440 at (8,16).

📋 CBSE Stepwise Marking Scheme

  • [0.5 marks] Draw each boundary ax+by=c and retain the closed half-plane ax+by≤c; intersect all constraints, including x,y≥0.
  • [1 mark] Solving pairs of boundary equations and rejecting points violating any inequality gives vertices: (0,0), (0,24), (8,16), (16,0).
  • [1 mark] Evaluate the objective at every vertex: (0,0) → 0; (0,24) → 4560; (8,16) → 5440; (16,0) → 4800.
  • [0.5 marks] The region is bounded. max Z=5440 at (8,16).

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