Class 12 Linear Programming · Very Long Answer5 Marksgraphical_optimization

Class 12 Maths · Linear Programming · Very Long Answer #42

Maximize Z=60*x + 40*y subject to 5*x + 6*y ≤ 45; 3*x + 2*y ≤ 18; -x ≤ 0; -y ≤ 0. Solve graphically.

✅ Step-by-Step Solution

Final Answer: max Z=360 at (9/4,45/8), (6,0) and every point on their joining segment.

Draw each boundary ax+by=c and retain the closed half-plane ax+by≤c; intersect all constraints, including x,y≥0.
Boundary equations: 5*x + 6*y = 45; 3*x + 2*y = 18; -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: (0,0), (0,15/2), (9/4,45/8), (6,0).
Evaluate the objective at every vertex: (0,0) → 0; (0,15/2) → 300; (9/4,45/8) → 360; (6,0) → 360.
The region is bounded. max Z=360 at (9/4,45/8), (6,0) and every point on their joining segment.

[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: 5*x + 6*y = 45; 3*x + 2*y = 18; -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: (0,0), (0,15/2), (9/4,45/8), (6,0).
[1 mark] Evaluate the objective at every vertex: (0,0) → 0; (0,15/2) → 300; (9/4,45/8) → 360; (6,0) → 360.
[1 mark] The region is bounded. max Z=360 at (9/4,45/8), (6,0) and every point on their joining segment.

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