Class 12 Linear Programming · Very Long Answer5 Marksgraphical_optimization

Class 12 Maths · Linear Programming · Very Long Answer #41

Maximize Z=100*x + 120*y subject to 2*x + 3*y ≤ 30; 3*x + y ≤ 17; -x ≤ 0; -y ≤ 0. Solve graphically.

✅ Step-by-Step Solution

Final Answer: max Z=1260 at (3,8).

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 + 3*y = 30; 3*x + y = 17; -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), (0,10), (3,8), (17/3,0).
Evaluate the objective at every vertex: (0,0) → 0; (0,10) → 1200; (3,8) → 1260; (17/3,0) → 1700/3.
The region is bounded. max Z=1260 at (3,8).

[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 + 3*y = 30; 3*x + y = 17; -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), (0,10), (3,8), (17/3,0).
[1 mark] Evaluate the objective at every vertex: (0,0) → 0; (0,10) → 1200; (3,8) → 1260; (17/3,0) → 1700/3.
[1 mark] The region is bounded. max Z=1260 at (3,8).

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