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Homework I: Consider the graph at the right containing three lines. Which of the following equations matches the lines A, B, and C? 1. 2. 3. 4. 5. 6. y = 4 + (1/3) x y=6 y = -1/3 + 4 x y=x+6 y = 4 - (1/3) x y = (3/2)x + 2 Line A is equivalent to #6 (y = (3/2) x + 2) Line B is equivalent to #2 (y = 6) Line C is equivalent to #5 (y = 4 - (1/3) x) Homework II: Consider the graph containing two lines, which of the following equations matches the lines A and B. 1. 2. 3. 4. 5. 6. y=4-2x y = (2/3)x + 14 y = 1/2 x + 4 y=4+2x y = 14 - (3/2) x y = 14 - (2/3) x Line A is equivalent to #4 (y = 4 + 2 x) Line B is equivalent to #6 (y = 14 – (2/3) x) Answer: It is a bounded feasible region. There are four corner points indicated by the letters ‘A’, ‘B’, ‘C’, and ‘D’. The coordinates (X1, X2) for these points were given in the problem above. What about this problem? x1>=4 4x1+2x2<=8 x2 >=6 All variables are non-negative Blue line = x1=4 Red line = 4x1+2x2=8 Green line = x2=6 Answer: This graph shows an infeasible region. There are no feasible points satisfying all the constraints. Therefore, there are no corner points Answer: It is an unbounded feasible region. The feasible region is above the line X2 = 2 and to the right side of X1 = 4. This region extends “forever”. It has one vertex at point A with coordinates (X1= 4, X2 = 2). Corner points C X1 4 X2 2 Max X1 + 2X2 Subject to: 2X1 + X2 ≤10 X1 + X ≤ 6 -X1 + X2 ≤ 2 -2X1 + X2 ≤ 1 X1 ≥ 0, X2 ≥ 0 By trying all feasible vertices we will find that, the optimal solution occurs at vertex X1 = 2, X2 = 4, with optima value of X1 + 2X2 = 10, Answer: Since the feasible region is bounded, you can find all of the corner points. By finding all 6 feasible vertices, you can evaluate the objective function at each point by substituting its coordinate X1 and X2 into the objective function (X1 + 2X2). By doing so, you will find that, the optimal solution occurs at vertex X1 = 2, X2 = 4, with optima value of X1 + 2X2 = 10. Problem 2.7 from the book