Coordinate-derived feasible-region graph for cbse12-math-lp-p1-long-006
✅ Step-by-Step Solution
Final Answer: min Z=38 at (2,4).
Draw each boundary ax+by=c and retain the closed half-plane ax+by≤c; intersect all constraints, including x,y≥0.
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.
Solving pairs of boundary equations and rejecting points violating any inequality gives vertices: (0,8), (2,4), (10,0).
Evaluate the objective at every vertex: (0,8) → 56; (2,4) → 38; (10,0) → 50.
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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