Step 1: Conceptual Understanding:
This is a Linear Programming minimization problem. For a bounded feasible region, the minimum value is found at a corner point. For an unbounded region, evaluate Z at corner points and verify if a smaller value exists within the region.Step 2: Methodology:
1. Define the feasible region using the given constraints.2. Determine the coordinates of all corner points of the feasible region.3. Calculate the objective function Z at each corner point.4. The point yielding the minimum Z value is the optimal solution. If multiple corner points result in the same minimum value, all points on the line segment connecting them are also optimal.Step 3: Detailed Procedure:
Linear Programming Problem (LPP):Minimize Z = x + 2ySubject to:1) \(2x + y \ge 3\)2) \(x + 2y \ge 6\)3) \(x \ge 0, y \ge 0\)Identify corner points by finding the intersections of the boundary lines.Point A (y-axis intersection): Set x = 0. Constraints: \(y \ge 3\) and \(2y \ge 6 \implies y \ge 3\). Intersection: (0, 3).Point B (x-axis intersection): Set y = 0. Constraints: \(2x \ge 3 \implies x \ge 1.5\) and \(x \ge 6\). Intersection: (6, 0).Point C (Intersection of \(2x + y = 3\) and \(x + 2y = 6\)): From \(2x + y = 3\), \(y = 3 - 2x\). Substitute into the second equation: \(x + 2(3 - 2x) = 6 \implies x + 6 - 4x = 6 \implies -3x = 0 \implies x = 0\). If \(x = 0\), then \(y = 3 - 2(0) = 3\). Intersection: (0, 3), which is Point A.The corner points of the unbounded feasible region are (0, 3) and (6, 0).Evaluate Z = x + 2y at these points:At (0, 3): \(Z = 0 + 2(3) = 6\).At (6, 0): \(Z = 6 + 2(0) = 6\).The minimum Z value at the corner points is 6, occurring at both (0, 3) and (6, 0).To confirm for the unbounded region, check if Z can be less than 6 by examining the region \(x + 2y<6\). This region is disjoint from the feasible region due to the constraint \(x + 2y \ge 6\). Therefore, the minimum value is indeed 6.Step 4: Conclusion:
The minimum value of Z is 6. This minimum is achieved at both corner points (0, 3) and (6, 0). Consequently, the optimal feasible solution is located at both (6, 0) and (0, 3).