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PROBLEM 7-14

PROBLEM 7-14. Let: X 1 = number of air conditioners to be produced X 2 = number of fans to be produced. Maximize profit = 25 X 1 + 15 X 2 subject to 3 X 1 + 2 X 2  240 (wiring)  2 X 1 + 1 X 2  140 (drilling) X 1 , X 2  0. Profit at point a ( X 1 = 0, X 2 = 0) = $0

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PROBLEM 7-14

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  1. PROBLEM 7-14

  2. Let: X1 = number of air conditioners to be produced X2 = number of fans to be produced Maximize profit = 25X1 + 15X2 subject to 3X1 + 2X2 240 (wiring)  2X1 + 1X2 140 (drilling) X1, X2 0 Profit at point a (X1 = 0, X2 = 0) = $0 Profit at point b (X1 = 0, X2 = 120) = 25(0) + (15)(120) = $1,800 Profit at point c (X1 = 40, X2 = 60) = 25(40) + (15)(60) = $1,900 Profit at point d (X1 = 70, X2 = 0) = 25(70) + (15)(0) = $1,750 The optimal solution is to produce 40 air conditioners and 60 fans during each production period. Profit will be $1,900.

  3. PROBLEM 7-15

  4. Maximize profit = 25X1 + 15X2 subject to   3X1 + 2X2 240 2X1 + 1X2 140 X1 20 X2 80 X1, X2 0 The feasible region for this problem is the combination of all of the shaded areas. Profit at point a (X1 = 20, X2 = 0) = 25(20) + (15)(0) = $500 Profit at point b (X1 = 20, X2 = 80) = 25(20) + (15)(80) = $1,700 Profit at point c (X1 = 40, X2 = 60) = 25(40) + (15)(60) = $1,900 – Optimal solution. Profit at point d (X1 = 70, X2 = 0) = 25(70) + (15)(0) = $1,750 Profit at point e (X1 = 26.67, X2 = 80) = 25(26.67) + (15)(80) = $1,867

  5. Hence, even though the shape of the feasible region changed from Problem 7-14, the optimal solution remains the same. The calculations for the slack available at each of the four constraints at the optimal solution (40, 60) are shown below. The first two have zero slack and hence are binding constraints. The third constraint of X1 ≥ 20 has a surplus (this is called a surplus instead of slack because the constraint is “>“) of 20 while the fourth constraint of X2 ≤ 80 has a slack of 20. 3X1 + 2X2 + S1 = 240 so S1 = 240 - 3X1 - 2X2 = 240 – 3(40) - 2(60) = 0 2X1 + 1X2 + S2 = 140 so S2 = 140 - 2X1 - 1X2 = 140 - 2(40) – 1(60) = 0 X1 – S3 = 20 soS3 = -20 + X1 = -20 + 40 = 20 X2 + S4 = 80 so S4 = 80 - X2 = 80 - 60 = 20

  6. 7-15. b. Maximize profit = 25X1 + 15X2 subject to   3X1 + 2X2 240 2X1 + 1X2 140 X1 30 X2 50 X1, X2 0 The feasible region for this problem is only the darker shaded area in Figure 7.15 above, It is significantly smaller and only has four corners denoted by a’, b’, c’ and d. Profit at point a’ (X1 = 30, X2 = 0) = 25(30) + (15)(0) = $750 Profit at point b’ (X1 = 30, X2 = 50) = 25(30) + (15)(50) = $1,500 Profit at point c’ (X1 = 45, X2 = 50) = 25(45) + (15)(50) = $1,875 – Optimal solution. Profit at point d (X1 = 70, X2 = 0) = 25(70) + (15)(0) = $1,750

  7. Here, the shape and size of the feasible region changed and the optimal solution changed. The calculations for the slack available at each of the four constraints at the optimal solution (45, 50) are shown below. The second and the fourth constraints have zero slack and hence are are binding constraints. The third constraint of X1 ≥ 30 has a surplus of 15 while the first constraint of 3X1 + 2X2 240 has a slack of 5. 3X1 + 2X2 + S1 = 240 so S1 = 240 - 3X1 - 2X2 = 240 – 3(45) - 2(50) = 5 2X1 + 1X2 + S2 = 140 so S2 = 140 - 2X1 - 1X2 = 140 - 2(45) – 1(50) = 0 X1 – S3 = 30 soS3 = -30 + X1 = -30 + 45 = 15 X2 + S4 = 50 so S4 = 50 - X2 = 50 - 50 = 0

  8. 7-16. Let R = number of radio ads; T = number of TV ads. Maximizeexposure = 3,000R + 7,000T Subject to: 200R + 500T 40,000 (budget) R 10 T 10 RT R, T 0 Optimal corner point R = 175, T = 10, Audience = 3,000(175) + 7,000(10) = 595,000 people

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