Vertices: (0,4), (21/11,30/11), (3,0), (0,0). Found by graphing constraints & identifying intersection points. Option C is the right choice.
Graph the constraints.
We can start by graphing each of the constraints as inequalities.
The first constraint, 2x+3y≥12, can be graphed as a blue line by converting it to slope-intercept form: y ≥ − 2/3x+4. We'll mark the points where the line intersects the axes: (0,4) and (6,0).
The second constraint, 5x+2y≥15, can be graphed as a green line:
y ≥ − 5 / 2 x+ 15 / 2. We'll mark the points where it intersects the axes: (0, 15 / 2) and (3,0).
The third constraint, x ≥ 0, is simply the non-negative side of the x-axis. We'll mark the point where it intersects the y-axis: (0,0).
The fourth constraint, y ≥ 0, is simply the non-negative side of the y-axis. We've already marked this point as (0,0).
Identify the feasible region.
The feasible region is the area that satisfies all of the constraints. In other words, it's the area where all of the inequalities are true. We can shade this area in to see it better.
Find the vertices.
The vertices are the corner points of the feasible region. They occur where the lines intersect. In this case, the lines intersect at four points:
(0,4): This is the point where the blue and green lines intersect.
(21 / 11, 30 / 11 ) : This is the point where the blue line and the x-axis intersect.
(3,0): This is the point where the green line and the x-axis intersect.
(0,0): This is the point where all four lines intersect.
These four points, (0,4), ( 21 / 11, 30 / 11 ), (3,0), and (0,0), are the vertices of the feasible region. Option C is the right choice.
Question:-
The constraints of a problem are listed below. What are the vertices of the feasible region?
2x + 3y > 12
5x + 2y >= 15
x >= 0
y >= 0
A) (0,0) --- (0,4) --- (21/11, 30/11) --- (3,0)
B) (0,0) --- (0,15/2) --- (21/11, 30/11) --- (6,0)
C) (0,4) --- (21/11, 30/11) --- (3,0)
D) (0,15/2) --- (21/11, 30/11) --- (6,0)