Class 12 Linear Programming · Very Long Answer5 Marksgraphical_optimization

Class 12 Maths · Linear Programming · Very Long Answer #63

Maximize Z=5*x + 7*y subject to x + y ≤ 4; 3*x + 8*y ≤ 24; 10*x + 7*y ≤ 35; -x ≤ 0; -y ≤ 0. Solve graphically.

✅ Step-by-Step Solution

Final Answer: max Z=124/5 at (8/5,12/5).

Draw each boundary ax+by=c and retain the closed half-plane ax+by≤c; intersect all constraints, including x,y≥0.
Boundary equations: x + y = 4; 3*x + 8*y = 24; 10*x + 7*y = 35; -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), (0,3), (8/5,12/5), (7/3,5/3), (7/2,0).
Evaluate the objective at every vertex: (0,0) → 0; (0,3) → 21; (8/5,12/5) → 124/5; (7/3,5/3) → 70/3; (7/2,0) → 35/2.
The region is bounded. max Z=124/5 at (8/5,12/5).

[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: x + y = 4; 3*x + 8*y = 24; 10*x + 7*y = 35; -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), (0,3), (8/5,12/5), (7/3,5/3), (7/2,0).
[1 mark] Evaluate the objective at every vertex: (0,0) → 0; (0,3) → 21; (8/5,12/5) → 124/5; (7/3,5/3) → 70/3; (7/2,0) → 35/2.
[1 mark] The region is bounded. max Z=124/5 at (8/5,12/5).

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