From the "x > 0" and "y > 0" constraints, you know that the origin is going to be one of your corner points, so "(0, 0)" is correct. The other lines may be rewritten as (and are more easily graphed as:
. . . . .y < -(1/2)x + 5
. . . . .y < -3x + 15
Looking at a graph of x = 0, y = 0, y = -(1/2)x + 5, and y = -3x + 15, the other corner points are going to be where y = -(1/2)x + 5 and y = -3x + 15 intersect, where y = -(1/2)x + 5 crosses the y-axis (so plug in "0" for "x"), and where y = -3x + 15 crosses the x-axis (so plug in "0" for "y").
That is, you'll have four corner points.
Also, since:
. . . . .y = -(1/2)x + 5
. . . . .y = -(1/2)0 + 5
. . . . .y = 0 + 5
. . . . .y = 5
...then (0, 4) is not where the line is going to cross the y-axis. The other point does not look correct, either.
If you get stuck working this exercise, please reply showing all of your steps. Thank you.
Eliz.
