Graph the OPPOSITE of the four constraint relations on the same set of axes. The unshaded portion in the center of your graph will be the feasibility polygon. The ordered pair denoting at least one of the vertices of the feasibility polygon (in your case on of the four vertices of your feasibility quadrilateral) will represent the optimum values. Test the coordinates in each of the vertices in the objective function and chose the coordinate values of the vertex that produces the largest value.
The reason that you want to graph the opposites (e.g. graph x + 2y > 1000, instead of x + 2y <= 1000) is that it produces a result where the feasibility area is an easy to discern unshaded area, rather than trying to figure out the area where all four constraint relations overlap.
John
My calculator said it, I believe it, that settles it
From
I > Ø
