Final answer:
The maximum value of the function f(x) = 3x + y for the given feasible region is 6, found by evaluating the function at the vertices of the feasible region, which is bounded by the system of inequalities y ≥ 0, x ≥ 0, and y ≤ –2x + 4.
Step-by-step explanation:
The goal is to find the maximum value of the function f(x) = 3x + y for the feasible region defined by the system of inequalities y ≥ 0, x ≥ 0, and y ≤ –2x + 4. To find this maximum value, we need to determine the coordinates of the vertices of the feasible region, which is the area that satisfies all of these inequalities on the xy-plane.
Firstly, since x ≥ 0 and y ≥ 0, the feasible region is in the first quadrant. The inequality y ≤ –2x + 4 forms a boundary line, and the feasible region will be below this line. The vertices of the feasible region (where the lines intersect) are likely to give the maximum value of the function.
To find the vertices, we intersect the lines defined by the equalities corresponding to our inequalities: y = 0 (the x-axis), x = 0 (the y-axis), and y = –2x + 4. We quickly find that the vertices are (0,0), (0,4), and (2,0).
Next, we evaluate f(x) at these vertices:
- At (0,0): f(x) = 3(0) + 0 = 0
- At (0,4): f(x) = 3(0) + 4 = 4
- At (2,0): f(x) = 3(2) + 0 = 6
The maximum f(x) value in the feasible region is 6, which corresponds to option D.