Practice finding the Range of Optimality, Dual Value, and Range of Feasibility for Problems #1 and #3. 3. a. 10 & 4 X=2. Y=3 Optimal Solution X-3, 7-2 83)+1242)48 2 X 10 b. The same extreme point, X= 3 and Y = 2, remains optimal. The value of the objective function becomes 6(3) + 12(2) = 42. c. A new extreme point, X = 2 and Y = 3, becomes optimal. The value of the objective function becomes 8(2) + 6(3) = 34. d. The objective coefficient range for variable X is 4 to 12. Since the change in part (b) is within this range, we know that the optimal solution, X = 3 and Y = 2, will not change. The objective coefficient range for variable Y is 8 to 24. Since the change in part (c) is outside this range, we have to re-solve the problem to find the new optimal solution