PROBLEM 7-14

Let: X1 = number of air conditioners to be producedX2 = 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) = $0Profit at point b (X1 = 0, X2 = 120)

= 25(0) + (15)(120) = $1,800Profit at point c (X1 = 40, X2 = 60)

= 25(40) + (15)(60) = $1,900Profit 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.

PROBLEM 7-15

Maximize profit = 25X1 + 15X2

subject to 3X1 + 2X2 2402X1 + 1X2 140

X1 20X2 80X1, 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

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) = 02X1 + 1X2 + S2 = 140 so S2 = 140 - 2X1 - 1X2 = 140 - 2(40) – 1(60) = 0 X1 – S3 = 20 so S3 = -20 + X1 = -20 + 40 = 20 X2 + S4 = 80 so S4 = 80 - X2 = 80 - 60 = 20

7-15. b.Maximize profit = 25X1 + 15X2

subject to 3X1 + 2X2 2402X1 + 1X2 140

X1 30X2 50X1, 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) = $750Profit at point b’ (X1 = 30, X2 = 50)

= 25(30) + (15)(50) = $1,500Profit 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

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) = 52X1 + 1X2 + S2 = 140 so S2 = 140 - 2X1 - 1X2 = 140 - 2(45) – 1(50) = 0 X1 – S3 = 30 so S3 = -30 + X1 = -30 + 45 = 15 X2 + S4 = 50 so S4 = 50 - X2 = 50 - 50 = 0

7-16. Let R = number of radio ads; T = number of TV ads.Maximize exposure = 3,000R + 7,000TSubject to: 200R + 500T 40,000 (budget)

R 10T 10R TR, T 0

Optimal corner point R = 175, T = 10,Audience = 3,000(175) + 7,000(10) = 595,000 people