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Class 12 Linear Programming · Very Long Answer5 Marksgraphical_optimization

Class 12 Maths · Linear Programming · Very Long Answer #45

Minimize Z=5*x + 7*y subject to -2*x - y ≤ -8; -x - 2*y ≤ -10; -x ≤ 0; -y ≤ 0. Solve graphically.
Coordinate-derived feasible-region graph for cbse12-math-lp-p1-long-006
Coordinate-derived feasible-region graph for cbse12-math-lp-p1-long-006

✅ Step-by-Step Solution

Final Answer: min Z=38 at (2,4).
  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: -2*x - y = -8; -x - 2*y = -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: (0,8), (2,4), (10,0).
  4. Evaluate the objective at every vertex: (0,8) → 56; (2,4) → 38; (10,0) → 50.
  5. The region is unbounded. Its recession directions satisfy every homogeneous constraint; representatives: (1,0), (0,1), (0,1), (1,0). Their objective changes are 5, 7, 7, 5. This checks whether a finite optimum exists. min Z=38 at (2,4).

📋 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: -2*x - y = -8; -x - 2*y = -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: (0,8), (2,4), (10,0).
  • [1 mark] Evaluate the objective at every vertex: (0,8) → 56; (2,4) → 38; (10,0) → 50.
  • [1 mark] The region is unbounded. Its recession directions satisfy every homogeneous constraint; representatives: (1,0), (0,1), (0,1), (1,0). Their objective changes are 5, 7, 7, 5. This checks whether a finite optimum exists. min Z=38 at (2,4).

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