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Class 12 Linear Programming · Very Long Answer5 Marksgraphical_optimization

Class 12 Maths · Linear Programming · Very Long Answer #67

Maximize Z=7*x + 10*y subject to 4*x + 6*y ≤ 240; 6*x + 3*y ≤ 240; -x ≤ -10; -x ≤ 0; -y ≤ 0. Solve graphically.
Coordinate-derived feasible-region graph for cbse12-math-lp-p1-practice_long-018
Coordinate-derived feasible-region graph for cbse12-math-lp-p1-practice_long-018

✅ Step-by-Step Solution

Final Answer: max Z=410 at (30,20).
  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: 4*x + 6*y = 240; 6*x + 3*y = 240; -x = -10; -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: (10,0), (10,100/3), (30,20), (40,0).
  4. Evaluate the objective at every vertex: (10,0) → 70; (10,100/3) → 1210/3; (30,20) → 410; (40,0) → 280.
  5. The region is bounded. max Z=410 at (30,20).

📋 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: 4*x + 6*y = 240; 6*x + 3*y = 240; -x = -10; -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: (10,0), (10,100/3), (30,20), (40,0).
  • [1 mark] Evaluate the objective at every vertex: (10,0) → 70; (10,100/3) → 1210/3; (30,20) → 410; (40,0) → 280.
  • [1 mark] The region is bounded. max Z=410 at (30,20).

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