Final answer:
The minimum value of f(x) = 3x + y for the feasible region defined by the inequalities y ≥ 0, x ≥ 0, and y ≤ −2x + 4 is 0. This occurs at the vertex (0,0) of the feasible region on a coordinate plane.
Step-by-step explanation:
We need to find the minimum value of f(x) = 3x + y for the feasible region defined by the system of inequalities: y ≥ 0, x ≥ 0, and y ≤ −2x + 4. We will also determine which value of f(x) corresponds to this minimum.
Firstly, let's visualize the feasible region on a coordinate plane. As x and y are both greater than or equal to 0, this region will be in the first quadrant. The inequality y ≤ −2x + 4 represents a line that will intersect the y-axis at y=4 and the x-axis at x=2. The feasible region is below this line and above both the x and y-axes.
The objective function f(x) will achieve its minimum value at one of the vertices of this feasible region, which are (0,0), (2,0), and (0,4). Plugging these points into the function gives us:
- f(0,0) = 3(0) + (0) = 0
- f(2,0) = 3(2) + (0) = 6
- f(0,4) = 3(0) + (4) = 4
The minimum value of f(x) is therefore 0, which occurs at the vertex (0,0).