Step 1: Conceptual Overview: This problem involves minimizing an objective function within a feasible region defined by linear constraints. For a bounded feasible region, the optimal solution is found at a corner point. For an unbounded region, corner points are evaluated, and then the region is checked to confirm if a lower objective function value is achievable.
Step 2: Methodological Outline:
1. Define the feasible region based on the given constraints.
2. Determine the coordinates of all corner points of this region.
3. Calculate the objective function's value at each corner point.
4. The corner point yielding the minimum value is the optimal solution. If multiple corner points result in the same minimum value, the line segment connecting them represents the set of optimal solutions.
Step 3: Detailed Analysis:
The Linear Programming Problem (LPP) is:
Minimize Z = x + 2y
Subject to:
1) \(2x + y \ge 3\)
2) \(x + 2y \ge 6\)
3) \(x \ge 0, y \ge 0\)
The corner points are derived from the intersections of the boundary lines of the constraints.
Point A (y-axis intercept):
Set x = 0. Constraints become \(y \ge 3\) and \(2y \ge 6 \implies y \ge 3\). The intersection is (0, 3).
Point B (x-axis intercept):
Set y = 0. Constraints become \(2x \ge 3 \implies x \ge 1.5\) and \(x \ge 6\). The intersection is (6, 0).
Point C (Intersection of \(2x + y = 3\) and \(x + 2y = 6\)):
From \(2x + y = 3\), derive \(y = 3 - 2x\). Substitute into the second equation:
\(x + 2(3 - 2x) = 6\)
\(x + 6 - 4x = 6\)
\(-3x = 0 \implies x = 0\).
Substituting x = 0 into \(y = 3 - 2x\) yields \(y = 3\). The intersection is (0, 3), which is Point A.
The corner points of the unbounded feasible region are (0, 3) and (6, 0).
Evaluate the objective function Z = x + 2y at these points:
At point (0, 3): \(Z = 0 + 2(3) = 6\).
At point (6, 0): \(Z = 6 + 2(0) = 6\).
The minimum value of Z at the corner points is 6, occurring at both (0, 3) and (6, 0).
For an unbounded region, it's necessary to verify if Z can be less than 6. The region \(x + 2y<6\) is examined. This half-plane does not intersect the feasible region because one of the constraints is \(x + 2y \ge 6\). Thus, the minimum value of Z is confirmed to be 6.
Step 4: Conclusion:
The minimum value of Z is 6. This minimum is attained at both corner points (0, 3) and (6, 0). Therefore, the optimal feasible solutions are located at (6, 0) and (0, 3).