tion. If we use isoprofit lines, we only need to examine one corner
9. Shadow price or dual: the value of one additional unit of a
resource, such as one more hour of a scarce labor resource or one
more dollar to invest.
10. The isocost line is moved down in a minimization problem
until it no longer intersects with any constraint equation. The last
point in the feasible region that the line touches is the optimal
11. The corner point method examines the profit at every corner
point, whereas the isoprofit line method draws a series of parallel
profit lines until one line finally touches the last tip (corner point)
of the feasible region. That last point touched is the optimal solu-
tion, so other corner points need not be tested.
4. What happens if we can reduce the electronics time for Blue-
berrys to 2.5 hours?
The profit rises by $70.
278 BUSINESS ANALYTICS MODULE B LI N E A R PR O G R A M M I N G
END–OF-MODULE PROBLEMS
B.3 We solve this problem by the isocost line method:
X
Y
Maximize = 4X + 6Y
B.1
BUSINESS ANALYTICS MODULE B LIN EA R PR O G R A M M I N G 279
B.5
X
Y
0
20
11
0
3.86
4.54
B.6 (a) Let x1 = number of liver-flavored biscuits in a package
x2 = number of chicken-flavored biscuits in a package
Minimize x1 + 2x2
Subject to x1 + x2 40
2x1 + 4x2 60
x1 15
x1, x2 0
(b) Corner points are (0,40) and (15,25). Optimal solution is
(15,25) with cost of 65 cents.
(c) Minimum cost = (1)(.15) + (2)(.25) = 65 cents.
B.7
x2 = number of heaters to be produced
Maximize 25x1 + 15x2
2x1 + 1x2 140 (drilling)
x1, x2 0 (non-negativity)
Profit:
@a: (x1 = 0, x2 = 0) Obj = $0
@b: (x1 = 0, x2 = 120) Obj = 25 0 + 15 120 = $1,800
60 heaters each period. Profit will be $1,900.
280 BUSINESS ANALYTICS MODULE B LI N E A R PR O G R A M M I N G
B.8
x2 = number of Model B gates produced
Maximize 90x1 + 70x2
Subject to 125x1 + 100x2 25,000 (steel)
20x1 + 30x2 6,000 (zinc)
x1, x2 0 (nonnegativity)
Profit:
@a: (x1 = 0, x2 = 200) Obj = 90 0 + 70 200
= $14,000.00
@b: (x1 = 85.71, x2 = 142.86) Obj = 90 85.71 + 70 142.86
= $18,000.00*
* The optimal solution is to produce 200 Model A gates, and 0
Model B gates. Profit will be $18,000.
B.9 (a) Let T = number of trucks to produce per day
C = number of cars to produce per day
(c)
(d) Produce 20 trucks and 30 cars daily for a profit of
$12,600 per day.
B.10
x2 = number of Beta-2 robots
Maximize 1,200x1 + 1,800x2
Subject to 20x1 + 25x2 = 800 (total hours)
x1 10 (Alpha–1s)
x2 15 (Beta-2s)
X1 30 (pounds compost per bag)
X2 40 (pounds sewage per bag)
Corner point a:
(X1 = 30, X2 = 40) cost = 5(30) + (4)(40) = $3.10
Corner point b (which is optimal):
Point
Coordinates
Profit
O
(0, 0)
0
A
(0, 50)
11,000
C
(40 ,0)
12,000
B
Solve 2 equations in 2 unknowns
derived from constraints (1) and (2)
to obtain (20, 30)
12,600
BUSINESS ANALYTICS MODULE B LIN EA R PR O G R A M M I N G 281
Copyright ©2014 Pearson Education, Inc.
B.13 The last constraint is not linear because it contains the
square root of x and the objective function and first constraint are
not because of the x1x2 term.
B.14 (a) Using POM for Windows software, we find that the
optimal solution is:
x1 = 7.95
x2 = 5.95
x3 = 12.60
Profit = $143.76 (rounded)
(b) There is no unused time available on any of the three
machines.
EB EC EE
282 BUSINESS ANALYTICS MODULE B LI N E A R PR O G R A M M I N G
700 (number students in sector )
500 (number students in sector )
100 (number students in sector )
800 (number students in sector )
AB AC AE
BB BC BE
CB CC CE
DB DC DE
X X X
A
X X X
B
X X X
C
X X X
D
++
=
++
=
++
=
++
=
(b) Solution: XAB = 400
XAE = 300
XBB = 500
XCC = 100
XDC = 800
X1 = $ invested in Treasury notes
X2 = $ invested in bonds
Maximize ROI = 0.08X1 + 0.09X2
X1 $125,000
X2 $100,000
X1 + X2 = $250,000
X1, X2 0
Point a (X1 = 150,000, X2 = 100,000),
ROI = $21,000 (optimal solution)
Point b (X1 = 250,000, X2 = 0), ROI = $20,000
x1 + x2 + x3 + x4 + x5 + x6 (Minimize staff size)
Subject to:
12
23
34
45
12
16
9
11
xx
xx
xx
xx
+
+
+
+
400(number students in sector )
900 (school capacity)
900 (school capacity)
900 (school capacity)
EB EC EE
AB BB CB DB EB
AC BC CC DC EC
AE BE CE DE EE
X X X
E
X X X X X
B
X X X X X
C
X X X X X
E
++
=
+ + + +
+ + + +
+ + + +
X2 = bluebird houses
Maximize profit = 6X1 + 15X2
4X1 + 2X2 60
4X1 + 12X2 120
Corner Points
X1
X2
Profit
0
0
0
15
0
90
0
10
150
12
6
162 (Optimal)
The maximum value of the objective is $162, obtained by
producing 12 wren houses and 6 bluebird houses.
x1 + 2x2 12 (lengths, redwood)
where: x1 = number of coffee tables/week
x2 = number of bookcases/week
Optimal: x1 = 8, x2 = 2, Profit = $96
B.21 (a, b) Let S = number of standard bags to produce per week
D = number of deluxe bags to produce per week
Maximize profit = 10S + 8D
B.22
The original equations are:
Objective: 4x1 + 5x2 (minimize)
Subject to: x1 + 2x2 80
3x1 + x2 75
The optimal solution is found at the intersection of the two
constraints:
x1 + 2x2 80
3x1 + x2 75
To solve these equations simultaneously, begin by writing
them in the form shown below:
x1 + 2x2 = 80
3x1 + x2 = 75
Multiply the second equation by –2 and add it to the first:
1 2 1 2
1 2 1 2
1
2 80 2 80
2(3 75) 6 2 150
5 70
x x x x
x x x x
x
+ = → + =
− + = → − − = −
− = −
Thus, x1 = 70/5 = 14. Given: x1 + 2x2 = 80,
2x2 = 80 – x1 = 80 – 14, or x2 = 66/2 = 33.
The cost is given by:
C = 4x1 + 5x2 = (4 14) + (5 33) = 221
Extreme (Corner)
Points
Point
Profit
(0, 0)
$0
(0, 300)
$2,400
(360, 0)
$3,600
(240, 180)
$3,840
Optimal solution and answer
(b)
1+ 300 ( )
2
2360 ( )
3
, 0
S D A
S D B
SD
+
284 BUSINESS ANALYTICS MODULE B LI N E A R PR O G R A M M I N G
B.23 Let x1 = number of class A containers to be used
x2 = number of class K containers to be used
x3 = number of class T containers to be used
The appropriate equations are:
Maximize: 9x1 + 7x2 + 15x3
2x1 + 6x2 + 4x3 = 240 (time)
x1, x2, x3 0 (non-negativity)
Using software we find that the optimal solution is:
x1 = 0, x2 = 14.29, x3 = 38.57
and
Profit = $678.57
B.24
X2 = number of newspaper ads
Maximize exposures = 35,000X1 + 20,000X2
2
1
Subject to: 3,000 1,250 $100,000
XX
+
==
==
==
12
12
12
Point ( 5, 10) exposure = 375,000
Point ( 5, 68) exposure = 175,000 + 1,360,000
= 1,535,000 (this is optimal)
Point ( 25, 20) exposu
a X X
b X X
c X X =+
=
= = =
=
12
re 875,000 400,000
1,275,000
Point ( 25, 10) exposure 875,000 + 200,000
1,075,000
d X X
1 1 1 2 2 2 3 3 3
a b c a b c a b c
1 2 3
1 2 3
1 2 3
Subject to: 7
12
5
a a a
b b b
c c c
x x x
x x x
x x x
+ + =
+ + =
+ + =
C4: W2’s supply
+
=
BUSINESS ANALYTICS MODULE B LIN EA R PR O G R A M M I N G 285
(e) The range of optimality on Variable VA3? $6.25 to
$7.25. (To get this, look at Allowable Increase (.75)
and Decrease (.25) for VA3. Add/subtract from C3
(6.5): 6.25 = 6.5 – .25 & 7.25 = 6.5 + .75.
(f) If we could ship 10 tons less to Customer A, we might
be able to save: $65 (shadow price = $6.5/ton). Cust B
Shadow price is higher for filling Customer B’s demand
$7.25 > $6.5 and we are trying to minimize costs.
B.28 Let x1 = number of medical patients
x2 = number of surgical patients
The appropriate equations are:
Maximize 2280x1 + 1515x2
12
12
12
2
1
Subject to: 8 5 32,850 (patient days available)
3.1 2.6 15,000 (lab tests)
1 2 7,000 (X-rays)
2,800 (operations)
xx
xx
xx
x
x
+
+
+
2
, 0 (non-negativity)x
Optimal: x1 = 2790, x2 = 2104, Revenue = $9,551,659
Subject to
Finish
Finish
Finish
1
1
0
4
3
1
2
10
43
62
71
2 All , 0
A
B
C
D
E
F
G
G G D A A
G E E F F
G D D F C C
E B B i j
Y
Y
Y
Y
Y
Y
Y
X
X X Y X X Y
X X Y X X Y
X X Y X X Y
X X Y X Y
− + − + 
− +  − + 
− + − + 
− + 
278 BUSINESS ANALYTICS MODULE B LI N E A R PR O G R A M M I N G
END–OF-MODULE PROBLEMS
B.3 We solve this problem by the isocost line method:
X
Y
Maximize = 4X + 6Y
B.1
BUSINESS ANALYTICS MODULE B LIN EA R PR O G R A M M I N G 279
B.5
X
Y
0
20
11
0
3.86
4.54
B.6 (a) Let x1 = number of liver-flavored biscuits in a package
x2 = number of chicken-flavored biscuits in a package
Minimize x1 + 2x2
Subject to x1 + x2 40
2x1 + 4x2 60
x1 15
x1, x2 0
(b) Corner points are (0,40) and (15,25). Optimal solution is
(15,25) with cost of 65 cents.
(c) Minimum cost = (1)(.15) + (2)(.25) = 65 cents.
B.7
x2 = number of heaters to be produced
Maximize 25x1 + 15x2
2x1 + 1x2 140 (drilling)
x1, x2 0 (non-negativity)
Profit:
@a: (x1 = 0, x2 = 0) Obj = $0
@b: (x1 = 0, x2 = 120) Obj = 25 0 + 15 120 = $1,800
60 heaters each period. Profit will be $1,900.
280 BUSINESS ANALYTICS MODULE B LI N E A R PR O G R A M M I N G
B.8
x2 = number of Model B gates produced
Maximize 90x1 + 70x2
Subject to 125x1 + 100x2 25,000 (steel)
20x1 + 30x2 6,000 (zinc)
x1, x2 0 (nonnegativity)
Profit:
@a: (x1 = 0, x2 = 200) Obj = 90 0 + 70 200
= $14,000.00
@b: (x1 = 85.71, x2 = 142.86) Obj = 90 85.71 + 70 142.86
= $18,000.00*
* The optimal solution is to produce 200 Model A gates, and 0
Model B gates. Profit will be $18,000.
B.9 (a) Let T = number of trucks to produce per day
C = number of cars to produce per day
(c)
(d) Produce 20 trucks and 30 cars daily for a profit of
$12,600 per day.
B.10
x2 = number of Beta-2 robots
Maximize 1,200x1 + 1,800x2
Subject to 20x1 + 25x2 = 800 (total hours)
x1 10 (Alpha–1s)
x2 15 (Beta-2s)
X1 30 (pounds compost per bag)
X2 40 (pounds sewage per bag)
Corner point a:
(X1 = 30, X2 = 40) cost = 5(30) + (4)(40) = $3.10
Corner point b (which is optimal):
Point
Coordinates
Profit
O
(0, 0)
0
A
(0, 50)
11,000
C
(40 ,0)
12,000
B
Solve 2 equations in 2 unknowns
derived from constraints (1) and (2)
to obtain (20, 30)
12,600
BUSINESS ANALYTICS MODULE B LIN EA R PR O G R A M M I N G 281
Copyright ©2014 Pearson Education, Inc.
B.13 The last constraint is not linear because it contains the
square root of x and the objective function and first constraint are
not because of the x1x2 term.
B.14 (a) Using POM for Windows software, we find that the
optimal solution is:
x1 = 7.95
x2 = 5.95
x3 = 12.60
Profit = $143.76 (rounded)
(b) There is no unused time available on any of the three
machines.
EB EC EE
282 BUSINESS ANALYTICS MODULE B LI N E A R PR O G R A M M I N G
700 (number students in sector )
500 (number students in sector )
100 (number students in sector )
800 (number students in sector )
AB AC AE
BB BC BE
CB CC CE
DB DC DE
X X X
A
X X X
B
X X X
C
X X X
D
++
=
++
=
++
=
++
=
(b) Solution: XAB = 400
XAE = 300
XBB = 500
XCC = 100
XDC = 800
X1 = $ invested in Treasury notes
X2 = $ invested in bonds
Maximize ROI = 0.08X1 + 0.09X2
X1 $125,000
X2 $100,000
X1 + X2 = $250,000
X1, X2 0
Point a (X1 = 150,000, X2 = 100,000),
ROI = $21,000 (optimal solution)
Point b (X1 = 250,000, X2 = 0), ROI = $20,000
x1 + x2 + x3 + x4 + x5 + x6 (Minimize staff size)
Subject to:
12
23
34
45
12
16
9
11
xx
xx
xx
xx
+
+
+
+
400(number students in sector )
900 (school capacity)
900 (school capacity)
900 (school capacity)
EB EC EE
AB BB CB DB EB
AC BC CC DC EC
AE BE CE DE EE
X X X
E
X X X X X
B
X X X X X
C
X X X X X
E
++
=
+ + + +
+ + + +
+ + + +
X2 = bluebird houses
Maximize profit = 6X1 + 15X2
4X1 + 2X2 60
4X1 + 12X2 120
Corner Points
X1
X2
Profit
0
0
0
15
0
90
0
10
150
12
6
162 (Optimal)
The maximum value of the objective is $162, obtained by
producing 12 wren houses and 6 bluebird houses.
x1 + 2x2 12 (lengths, redwood)
where: x1 = number of coffee tables/week
x2 = number of bookcases/week
Optimal: x1 = 8, x2 = 2, Profit = $96
B.21 (a, b) Let S = number of standard bags to produce per week
D = number of deluxe bags to produce per week
Maximize profit = 10S + 8D
B.22
The original equations are:
Objective: 4x1 + 5x2 (minimize)
Subject to: x1 + 2x2 80
3x1 + x2 75
The optimal solution is found at the intersection of the two
constraints:
x1 + 2x2 80
3x1 + x2 75
To solve these equations simultaneously, begin by writing
them in the form shown below:
x1 + 2x2 = 80
3x1 + x2 = 75
Multiply the second equation by –2 and add it to the first:
1 2 1 2
1 2 1 2
1
2 80 2 80
2(3 75) 6 2 150
5 70
x x x x
x x x x
x
+ = → + =
− + = → − − = −
− = −
Thus, x1 = 70/5 = 14. Given: x1 + 2x2 = 80,
2x2 = 80 – x1 = 80 – 14, or x2 = 66/2 = 33.
The cost is given by:
C = 4x1 + 5x2 = (4 14) + (5 33) = 221
Extreme (Corner)
Points
Point
Profit
(0, 0)
$0
(0, 300)
$2,400
(360, 0)
$3,600
(240, 180)
$3,840
Optimal solution and answer
(b)
1+ 300 ( )
2
2360 ( )
3
, 0
S D A
S D B
SD
+
284 BUSINESS ANALYTICS MODULE B LI N E A R PR O G R A M M I N G
B.23 Let x1 = number of class A containers to be used
x2 = number of class K containers to be used
x3 = number of class T containers to be used
The appropriate equations are:
Maximize: 9x1 + 7x2 + 15x3
2x1 + 6x2 + 4x3 = 240 (time)
x1, x2, x3 0 (non-negativity)
Using software we find that the optimal solution is:
x1 = 0, x2 = 14.29, x3 = 38.57
and
Profit = $678.57
B.24
X2 = number of newspaper ads
Maximize exposures = 35,000X1 + 20,000X2
2
1
Subject to: 3,000 1,250 $100,000
XX
+
==
==
==
12
12
12
Point ( 5, 10) exposure = 375,000
Point ( 5, 68) exposure = 175,000 + 1,360,000
= 1,535,000 (this is optimal)
Point ( 25, 20) exposu
a X X
b X X
c X X =+
=
= = =
=
12
re 875,000 400,000
1,275,000
Point ( 25, 10) exposure 875,000 + 200,000
1,075,000
d X X
1 1 1 2 2 2 3 3 3
a b c a b c a b c
1 2 3
1 2 3
1 2 3
Subject to: 7
12
5
a a a
b b b
c c c
x x x
x x x
x x x
+ + =
+ + =
+ + =
C4: W2’s supply
+
=
BUSINESS ANALYTICS MODULE B LIN EA R PR O G R A M M I N G 285
(e) The range of optimality on Variable VA3? $6.25 to
$7.25. (To get this, look at Allowable Increase (.75)
and Decrease (.25) for VA3. Add/subtract from C3
(6.5): 6.25 = 6.5 – .25 & 7.25 = 6.5 + .75.
(f) If we could ship 10 tons less to Customer A, we might
be able to save: $65 (shadow price = $6.5/ton). Cust B
Shadow price is higher for filling Customer B’s demand
$7.25 > $6.5 and we are trying to minimize costs.
B.28 Let x1 = number of medical patients
x2 = number of surgical patients
The appropriate equations are:
Maximize 2280x1 + 1515x2
12
12
12
2
1
Subject to: 8 5 32,850 (patient days available)
3.1 2.6 15,000 (lab tests)
1 2 7,000 (X-rays)
2,800 (operations)
xx
xx
xx
x
x
+
+
+
2
, 0 (non-negativity)x
Optimal: x1 = 2790, x2 = 2104, Revenue = $9,551,659
Subject to
Finish
Finish
Finish
1
1
0
4
3
1
2
10
43
62
71
2 All , 0
A
B
C
D
E
F
G
G G D A A
G E E F F
G D D F C C
E B B i j
Y
Y
Y
Y
Y
Y
Y
X
X X Y X X Y
X X Y X X Y
X X Y X X Y
X X Y X Y
− + − + 
− +  − + 
− + − + 
− + 