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Class 12 Linear Programming · Long Answer3 Marksgraphical_optimization

Class 12 Maths · Linear Programming · Long Answer #38

Minimize Z=10*x + 4*y subject to -4*x - y ≤ -80; -2*x - y ≤ -60; -x ≤ 0; -y ≤ 0. Solve graphically.
Coordinate-derived feasible-region graph for cbse12-math-lp-p1-short-012
Coordinate-derived feasible-region graph for cbse12-math-lp-p1-short-012

✅ Step-by-Step Solution

Final Answer: min Z=260 at (10,40).
  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: (0,80), (10,40), (30,0).
  3. Evaluate the objective at every vertex: (0,80) → 320; (10,40) → 260; (30,0) → 300.
  4. The region is unbounded. Its recession directions satisfy every homogeneous constraint; representatives: (1,0), (0,1), (0,1), (1,0). Their objective changes are 10, 4, 4, 10. This checks whether a finite optimum exists. min Z=260 at (10,40).

📋 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: (0,80), (10,40), (30,0).
  • [1 mark] Evaluate the objective at every vertex: (0,80) → 320; (10,40) → 260; (30,0) → 300.
  • [0.5 marks] The region is unbounded. Its recession directions satisfy every homogeneous constraint; representatives: (1,0), (0,1), (0,1), (1,0). Their objective changes are 10, 4, 4, 10. This checks whether a finite optimum exists. min Z=260 at (10,40).

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