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Gabriel Erosa OMG 423 2/16/14
Module 4 Homework
3. a.
Y
Optimal Solution
2
4
6
8
10
2 4 6 8
10
X
X = 3, Y = 2
X = 2, Y = 3
b. The same extreme point, X = 3 and Y = 2, remains optimal. The value of the objective
function becomes 6(3) + 12(2) = 42. The vertices of the feasible region are (0, 9), (2, 3),
(3,2), and (9, 0). You really should confirm these algebraically by finding where the lines
intersect. Then you need to evaluate the objective function at each of these points, and
choose the point that gives the minimum value for the objective function.

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