ANSWERS TO HOMEWORK 5 (with additional problems for study potential)
7-14.
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.
7-15. a.
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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 in Figure
7.15 above,
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.
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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
so S3 = -20 + X1 = -20 + 40 = 20
X2 S4 80
so S4 80 - X2 
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)
= $1,875 – Optimal solution.
=25(45) + (15)(50)
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) = 5
2X1 + 1X2 + S2 = 140
so S2 = 140 - 2X1 - 1X2 = 140 - 2(45) – 1(50) = 0
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X1 – S3 = 30
so S3 = -30 + X1 = -30 + 45 = 15
X2 S4 50
so S4 50 - X2 
7-16. Let R = number of radio ads; T = number of TV ads.
Maximize exposure = 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
7-17. X1 = number of benches produced
X2 = number of tables produced
Maximize profit  $9X1  $20X2
subject to
4X1 + 6X2  1,200 hours
10X1 + 35X2  3,500 feet
X1, X2  0
Profit at point a (X1 = 0, X2 = 100) = $2,000
Profit at point b (X1 = 262.5, X2 = 25) = $2,862.50
Profit at point c (X1 = 300, X2 = 0) = $2,700
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7-18.
X1 = number of undergraduate courses
X2 = number of graduate courses
Minimize cost  $2,500X1  $3,000X2
subject to
X1
 30
X2
 20
X1 + X2  60
Total cost at point a (X1  40, X2  20)
= 2,500(40) + (3,000)(20)
= $160,000
Total cost at point b (X1  30, X2  30)
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= 2,500(30) + (3,000)(30)
= $165,000
Point a is optimal.
7-19.
X1 = number of Alpha 4 computers
X2 = number of Beta 5 computers
Maximize profit  $1,200X1  $1,800X2
subject to 20X1  25X2  800 hours
(total hours  5 workers  160 hours each)
X1  10
X2  15
Corner points: a(X1 = 10, X2 = 24), profit = $55,200
b(X1 = 21 1 , X2 = 15), profit = $52,500
4
Point a is optimal.
7-20. Let P = dollars invested in petrochemical; U = dollars invested in utility
Maximize return = 0.12P + 0.06U
Subject to:
P + U = 50,000 total investment is $50,000
9P + 4U  6(50,000) average risk must be  6 [or total  6(50,000)]
P, U  0
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Corner points
Return =
P
U
0.12P + 0.06U
0
50,000
3,000
20,000
30,000
4,200
The maximum return is $4,200.
The total risk is 9(20,000) + 4(30,000) = 300,000, so
average risk = 300,000/(50,000) = 6
7-21. Let P = dollars invested in petrochemical; U = dollars invested in utility
Minimize risk = 9P + 4U
Subject to:
P + U = 50,000 total investment is $50,000
0.12P + 0.06U  0.08(50,000) return must be at least 8%
P, U  0
Corner points
Risk =
P
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9P + 4U
U
7-7
50,000
0
450,000
16,666.67
33,333.33
283,333.3
The minimum risk is 283,333.33 on $50,000 so the average risk is
283,333.33/50,000 = 5.67. The return would be 0.12(16,666.67) + 0.06(33,333.33) = $4,000 (or
8% of $50,000)
7-22.
Note that this problem has one constraint with a negative sign. This may cause the beginning student some confusion in plotting the line. As for the slack of each of the constraints, the first and
third constraints are binding and, hence, have zero slack. The second constraint of X – 2Y ≤ 10
has a slack of 28.75.
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7-23. Point a lies at intersection of constraints (see figure below):
3X + 2Y = 120
X + 3Y = 90
Multiply the second equation by 3 and add it to the first (the method of simultaneous equations):
3X  2Y  120
3X  9Y  270
7Y  150  Y  21.43 and X  25.71
Cost = $1X + $2Y = $1(25.71) + ($2)(21.43)
 $68.57
7-24. X1 = $ invested in Louisiana Gas and Power
X2 = $ invested in Trimex Insulation Co.
Minimize total investment  X1  X2
subject to $0.36X1  $0.24X2  $720
$1.67X1  $1.50X2  $5,000
0.04X1  0.08X2  $200
Investment at a is $3,333.
Investment at b is $3,179.  optimal solution
Investment at c is $5,000.
Short-term growth is $926.09.
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Intermediate-term growth is $5,000.
Dividends are $200.
See graph.
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