Consider the LPP min Z = x –y, subject to the conditions x+y≤3, y-x≤1, x≥0, y≥0, then minimum value of objective functions exists at the point:

Correct option is C

​$$\text{We are given the linear programming problem:} \\\text{Minimize } Z = x - y, \text{ subject to:} \\x + y \leq 3 \\y - x \leq 1 \\x \geq 0, \quad y \geq 0 \\[8pt]\text{We first identify the feasible region defined by the intersection of the constraints. The constraint } x + y = 3 \\\text{is a line from } (0, 3) \text{ to } (3, 0), \text{ and } y = x + 1 \text{ intersects it at } x + (x + 1) = 3 \Rightarrow x = 1, y = 2. \\\text{Other feasible boundary points are the intercepts: } (0, 0), (0, 1), (3, 0), \text{ all lying in the first quadrant.} \\[8pt]\text{The feasible region is thus a convex polygon bounded by the points } (0, 0), (0, 1), (1, 2), (3, 0). \\[8pt]\text{We evaluate the objective function } Z = x - y \text{ at the corner points:} \\\text{At } (0, 0), \quad Z = 0 \\\text{At } (0, 1), \quad Z = -1 \\\text{At } (1, 2), \quad Z = -1 \\\text{At } (3, 0), \quad Z = 3 \\[8pt]\text{The minimum value of } Z = -1, \text{ occurs at two points, } (0, 1) \text{ and } (1, 2). \\\text{This indicates multiple optimal solutions along the line segment joining these two points.}$$​​

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