Maximize Z = 3x + 5y Subject to: x – y ≥ 2 x + y ≥ 4 x, y ≥ 0 What can be concluded about the feasible region?

To determine the nature of the feasible region and the existence of a maximum for the objective function Z=3x+5y subject to the given constraints, let's analyze the inequalities graphically. The constraints are: x−y≥2 x+y≥4 x≥0 y≥0 Consider the boundary lines: x−y=2 x+y=4 x=0 (y-axis) y=0 (x-axis) Let's find the intersection points of these lines: Intersection of x−y=2 and x+y=4: Adding the two equations: (x−y)+(x+y)=2+4⟹2x=6⟹x=3. Substituting x=3 into x+y=4: 3+y=4⟹y=1. Intersection point: (3,1). Now, let's consider the regions defined by the inequalities: x−y≥2: Test point (3,0)⟹3−0≥2 (True). Region includes points below the line x−y=2. x+y≥4: Test point (4,0)⟹4+0≥4 (True). Region includes points above the line x+y=4. x≥0: Right half-plane. y≥0: Upper half-plane. The feasible region is the intersection of all these regions. The intersection of x−y≥2 and x+y≥4 occurs at (3,1). Let's visualize the region. The line x−y=2 passes through (2,0) and (0,−2). The region x−y≥2 is to the right of this line. The line x+y=4 passes through (4,0) and (0,4). The region x+y≥4 is above this line. The feasible region is the area in the first quadrant (x≥0,y≥0) that satisfies both x−y≥2 and x+y≥4. This region starts at the intersection point (3,1) and extends outwards. Consider the objective function Z=3x+5y. To maximize Z, we look for points in the feasible region where 3x+5y is as large as possible. Since the feasible region extends infinitely (it is not enclosed by any finite boundaries), it is unbounded. Now, let's check if a maximum value for Z exists. As we move to larger values of x and y within the feasible region, the value of Z=3x+5y will also increase without bound. For instance, consider points far out in the feasible region; both x and y can be arbitrarily large while satisfying the inequalities. Therefore, the feasible region is unbounded, and no maximum value for Z exists.