Problem

	Max	3A + 4B
	s.t.
		2A + 4B 12
		6A + 4B 24
		A,B 0

Graph

a. Find the optimal solution using the graphical solution procedure.

		The optimal solution
		A =
		B =
		Value of Objective Function =
b. If the objective function is changed to 2A + 6B, what will the optimal solution be?
		The optimal solution
		A =
		B =
		Value of Objective Function =
c. How many extreme points are there? What are the values of A and B at each extreme point?
		The number of extreme points =
		The values of A and B =>	A	B