Class 12 Linear Programming · Very Long Answer5 Marksgraphical_optimization

Class 12 Maths · Linear Programming · Very Long Answer #47

Maximize and minimize Z=5*x + 2*y subject to x - 2*y ≤ 2; 3*x + 2*y ≤ 12; -3*x + 2*y ≤ 3; -x ≤ 0; -y ≤ 0. Solve graphically.

✅ Step-by-Step Solution

Final Answer: min Z=0 at (0,0). max Z=19 at (7/2,3/4).

Draw each boundary ax+by=c and retain the closed half-plane ax+by≤c; intersect all constraints, including x,y≥0.
Boundary equations: x - 2*y = 2; 3*x + 2*y = 12; -3*x + 2*y = 3; -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), (0,3/2), (3/2,15/4), (2,0), (7/2,3/4).
Evaluate the objective at every vertex: (0,0) → 0; (0,3/2) → 3; (3/2,15/4) → 15; (2,0) → 10; (7/2,3/4) → 19.
The region is bounded. min Z=0 at (0,0). max Z=19 at (7/2,3/4).

[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 - 2*y = 2; 3*x + 2*y = 12; -3*x + 2*y = 3; -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), (0,3/2), (3/2,15/4), (2,0), (7/2,3/4).
[1 mark] Evaluate the objective at every vertex: (0,0) → 0; (0,3/2) → 3; (3/2,15/4) → 15; (2,0) → 10; (7/2,3/4) → 19.
[1 mark] The region is bounded. min Z=0 at (0,0). max Z=19 at (7/2,3/4).

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