Chapter 8 Study Guide Solutions
Selected Solutions to Problems in Chapter 8
Problem 1
A local company manufactures two lines of rocking chairs called type a and b. Each
rocking chair requires hand-assembled materials, which totals the amount of 3,000
pounds of wood and 800 pounds of nails. Type A requires 500 pounds of wood and 40
pounds of nails. Further each A chair is worth $450. Type B requires 150 pounds of
wood, 50 pounds of nails, and each B chair is worth $350. These chairs are handmade
with a high quality. The company would like to maximize their profit, help them to find
the best combination of rocking chairs.
Lbs per
Lbs per Availability
Chair A Chair B
a)
Wood
500
150
3,000
Nails
40
50
800
Profit
$450
$350
What are the decision variables?
X1 = # of rocking chairs A
X2 = # of rocking chair B
b)
What is the objective function?
Max Profit = $450X1+$350X2
c)
What are the constraints?
500X1+150X2<3,000 (Wood)
40X1+50X2<800 (Nails)
d)
What are the non-negativity assumptions?
X1, X2 >0
e)
Write the LP in modified standard form
SG8-1
Chapter 8 Study Guide Solutions
Max Profit = $450X1+$350X2
S.T.
500X1+150X2<3,000 (Wood)
40X1+50X2<800 (Nails)
X1, X2 >0
f)
Solve the LP graphically
Corner point feasible solution (a)
Set X2=0 and use the Wood equation
500X1+150X2<3000
500X1=3000
X1=6
X2=0
Corner point feasible solution (b)
Set, X1=0 and use the Nail equation
40X1+50X2<800
50X2=800
X2=16
X1=0
Corner point feasible solution (c) is obtained by solving the following
equations simultaneously
500X1+150X2=3,000 and 40X1+50X2=800
therefore, to solve, multiply the second equation by -3:
500X1+150X2 =3000
-120X1-150X2 = -2400
380X1= 600
X1= 1.58
Thus X1 =1.58 and substituting back into the Wood equation,
500X1+150X2=3000
150X2=2210
X2=14.73
Therefore, corner point feasible solution (c): X1=1.58 , X2=14.73
SG8-2
Chapter 8 Study Guide Solutions
Corner Point
Feasible Solution b
Rocking Chair LP Graph
20
Corner Point
Feasible Solution c
18
16
x2-Chair B
14
12
10
8
Feasible
Region
6
Corner Point
Feasible Solution a
4
2
0
0
2
4
10
8
6
12
14
x1-Chair A
Wood
g)
Nails
Find the optimal solution
Max profit = $450X1+$350X2
1. At a, Profit a = $450(6)+$350(0)
Profit a = $2,700
2. At b, Profit b = $450(0)+$350(16)
Profit b = $5,600
3. At c, Profit c = $450(1.58)+$350(14.73)
Profit c = $5,866.50
Hence the optimal solution is when; X1=1.58 , X2=14.73 and Profit c =
$5,866.50.
Managerial Implication:
The solution makes mathematical sense, but how can a manager build 1.58
Type A chairs and 14.73Type B chairs?
SG8-3
Chapter 8 Study Guide Solutions
Problem 2
A small manufacturer, which produces 2 types of new perfumes labeled A and B in the
table below, needs to decide how much of each perfume to produce to maximize profit.
The profit earned from the sales of perfume A is $30, and the profit earned from the sales
of perfume B is $50. The maximum Filling capacity is 180 and the maximum Boxing
capacity is 75. The perfume must go through two steps: Filling and Boxing. Perfume A
takes 30 units of Filling capacity per bottle of perfume processed, whereas Perfume B
only takes 20 units of Filling capacity per bottle processed. Perfume A takes 15 units of
Boxing capacity per bottle of perfume processed, whereas Perfume B only takes 5 units of
Boxing capacity per bottle processed. The table below summarizes the findings.
Process
Units per
Units per
Maximum available
Perfume A Perfume B
a)
Filling
30
20
180
Boxing
15
5
75
Unit Profit
$30
$50
What are the decision variables?
X1= # of perfume A
X2= # of perfume B
b)
What is the objective function?
Max Profit =$30X1+$50X2
c)
What are the constraints?
30X1+20X2<180 (Filling)
15X1+5X2<75 (Boxing)
d)
What are the non-negativity assumptions?
X1, X2 >0
e)
capacity
Write the LP in modified standard form
Max Profit =$30X1+$50X2
S.T.
SG8-4
Chapter 8 Study Guide Solutions
f)
30X1+20X2<180 (Filling)
15X1+5X2<75 (Boxing)
X1, X2 >0
Solve the LP graphically
Corner point feasible solution (a)
Set X1=0 and use the Filling equation
30X1+20X2<180
30X1=180
X2=9
X1=0
Corner point feasible solution (b)
Set X2=0 and use the Boxing equation
15X1+5X2<75
15X1=75
X1=5
X2=0
Corner point feasible solution (c) is obtained by solving the following
equations simultaneously
30X1+20X2=180 and 15X1+5X2=75
therefore, multiply the second equation by -2:
30X1+20X2 = 180
-30X1-10X2 = -150
10X2= 30
X2= 3
Thus X2 =3 and substituting back into the Filling equation,
30X1+20X2=180
30X1=120
X1=4
Therefore, corner point feasible solution (c): X1=4, X2=3
SG8-5
Chapter 8 Study Guide Solutions
Perfume Production
Corner Point
Feasible Solution a
16
14
x2-Perfume B
12
Corner Point
Feasible Solution c
10
8
Corner Point
Feasible Solution b
6
4
Feasible
Region
2
0
0
1
2
3
4
5
6
x1-Perfume A
Filling 1
Capacity
g)
Boxing 2
Capacity
Find the optimal solution
Max profit = $30X1+$50X2
1. At a, Profit a = $30(0)+$50(9)
Profit a = $450
2. At b, Profit b = $30(5)+$50(0)
Profit b = $150
3. At c, Profit c = $30(4)+$50(3)
Profit c = $270
Hence the optimal solution is when; X1=0, X2=9 and Profit =$450.
Managerial Implication:
Produce none of Perfume A and 9 of Perfume B for a profit of $450.
SG8-6
7
Chapter 8 Study Guide Solutions
Problem 3
Mark owns one-acre farmland in Waimanalo for the production of strawberries and
tomatoes. To make sure the plants grow properly and that he will make a profit, Mark has
invested in fertilizers. Help Mark find the most profitable combination of Strawberries
and Tomatoes.
Units per
Units per
Strawberries Tomatoes
a)
Labor
10
6
50
Fertilizer
5
2
20
Profit
$5
$3
What are the decision variables?
X1= Amount of strawberries
X2= Amount of tomatoes
b)
What is the objective function?
Max Profit =$5X1+$3X2
c)
What are the constraints?
10X1+6X2<50 (Labor)
5X1+2X2<20 (Fertilizer)
d)
What are the non-negativity assumptions?
X1 , X2 >0
e)
Write the LP in modified standard form.
Max Profit =$5X1+$3X2
S.T.
10X1+6X2<50 (Labor)
5X1+2X2<20 (Fertilizer)
X1 , X2 >0
f)
Availability
Solve the LP graphically
SG8-7
Chapter 8 Study Guide Solutions
Corner point feasible solution (a)
Set X1=0 and use the Labor equation
10X1+6X2=50
6X2=50
X2=8.33
X1=0
Corner point feasible solution (b)
Set X2=0 and use the Fertilizer equation
5X1+2X2=20
5X1=20
X1=4
X2=0
Corner point feasible solution (c) is obtained by solving the following
equations simultaneously
10X1+6X2=50 and 5X1+2X2=20
therefore, multiply the second equation by -2:
10X1+6X2 =50
-10X1-4X2 =-40
2X2=10
X2= 5
Thus X2 =5 and substituting back into the Labor equation,
10X1+6X2=50
X1=2
Therefore, corner point feasible solution (c): X1=2, X2=5
g)
Find the optimal solution
Max profit = $5X1+$3X2
1. At a, Profit a = $5(0)+$3(8.33)
Profit a = $24.99
2. At b, Profit b = $5(4)+$3(0)
Profit = $20
3. At c, Profit c = $5(2)+$3(5)
Profit c = $25
SG8-8
Chapter 8 Study Guide Solutions
Hence the optimal solution is when; X1=2, X2=5 or X1=0, X2=8.333
and Profit = $25
Corner Point
Feasible Solution a
Mark's Waimanalo Farm
12
10
Corner Point
Feasible Solution c
x2-Tomatoes
8
Corner Point
Feasible Solution
b
6
4
Feasible
Region
2
0
0
1
2
3
4
5
x1-Strawberries
Labor
Fertilizer
Managerial Implication:
Produce either 2 units of strawberries and 5 units of tomatoes or 0 units of
strawberries and 8.33 units of tomatoes as each will yield a Profit of $25.
NOTE:
In fact, a manager could chose to produce at any level between
those two points, i.e., anywhere on the Labor equation between
X1=2, X2=5 and X1=0, X2=8.333, to achieve the same Profit of
$25.
SG8-9
6
Chapter 8 Study Guide Solutions
Problem 4
The read-a-lot book company is facing difficulties in deciding on how to best
manufacture their books for maximum profit. They produce two type of books, hardcover
and paperback. It requires 10 pounds of corrugated paper to make one paperback and 240
pounds to produce one hard cover. The books also need a special solution for preserving
the books. Hard cover books require 30 gallon and paperbacks use 20 gallons. Each
hardcover brings in a profit of $35 dollars and each paperback brings in $19 dollars. We
currently have on-hand 12,000 pounds of corrugated paper and 1,300 gallons of solution.
We need you help to find the best production mix.
Units per
Units per
Available
Paperbacks Hardcover Quantity
Corrugated Paper
10
240
12,000
Preservation Solution
30
20
1,300
Profit
$19
$35
a)
What are the decision variables?
X1 = # of paperbacks
X2 = # of hardcover
b)
What is the objective function?
Max Profit =$19X1+35X2
c)
What are the constraints?
10X1+240X2<12,000 (Corrugated Paper)
30X1+20X2<1,300 (Preservation Solution)
d)
What are the non-negativity assumptions?
X1, X2 >0
e)
Write the LP in modified standard form.
Max Profit =$19X1+35X2
S.T.
10X1+240X2<12,000 (Corrugated Paper)
30X1+20X2<1,300 (Preservation Solution)
SG8-10
Chapter 8 Study Guide Solutions
X1, X2 >0
f)
Solve the LP graphically
Corner point feasible solution (a)
Set X1=0 and use the Corrugated Paper equation 10X1+240X2=12,000
240X2=12,000
X2=50
X1=0
Corner point feasible point (b)
Set X2=0 and use the Preservation Solution equation
30X1+20X2=1,300
30 X1=1,300
X1=43.333
X2=0
Corner point (c) is obtained by solving the following equations
simultaneously
10X1+240X2=12,000 and 30X1+20X2=1,300
Multiply the first equation by -3:
30X1+720X2 = 36,000
-30X1 - 20X2 = -1,300
700X2 = 34,700
X2 = 49.57
Thus X2 =49.57 and substituting back into the Corrugated paper
equation,
10X1+240X2=12,000
10X1 = 103.2
X1= 10.32
Therefore, corner point feasible solution (c): X1 = 10.32, X2 = 49.57
SG8-11
Chapter 8 Study Guide Solutions
Read-A-Book Company
Corner Point
Feasible Solution a
70
60
Corner Point
Feasible Solution c
x2-Hardcover
50
40
30
Feasible
Region
20
10
Corner Point
Feasible Solution b
0
0
10
20
30
40
50
60
70
x1-Paperbacks
Corrugated Paper
g)
Preservation Solution
Find the optimal solution
Max profit = $19X1+$35X2
1. At a, Profit a = $19(0)+$35(50)
Profit a = $1,750
2. At b, Profit b = $19(43.33)+$35(0)
Profit b = $823.33
3. At c, Profit c = $19(10.3)+$35(49.57)
Profit c = $1,931.03
Hence the optimal solution is when: X1=10.32, X2=49.57 and Profit c =
$1,931.03.
Managerial Implication:
The solution makes mathematical sense, but how can a manager
manufacture 10.32 paperbacks and 49.57 hardcover books?
SG8-12
80
Chapter 8 Study Guide Solutions
Problem 5
The LTD Inc is making two types of calculators, the TI-83 and HP-G49. The LTD Inc
would like to figure out which combination would be the best solution for them to
produce as they would like to see if it is possible to increase sales and profit. The
maximum capacity is 500 units per month for plant A and 200 units for operation B.
LTD
TI-83
HP-G49
A
100
60
500
B
50
20
200
$259
$130
Profit Margin
a)
Units per Units per Maximum
What are the decision variables?
X1= # of TI-83 calculators
X2= # of HP-G49 calculators
b)
What is the objective function?
Max Profit = $259X1+130X2
c)
What are the constraints?
100X1+60X2<500 (Capacity A)
50X1+20X2<200 (Capacity B)
d)
What are the non-negativity assumptions?
X1, X2 >0
e)
Write the LP in modified standard form.
Max Profit = $259X1+130X2
S.T.
100X1+60X2<500 (Capacity A)
50X1+20X2<200 (Capacity B)
X1, X2 >0
f)
Solve the LP graphically
SG8-13
Capacity
Chapter 8 Study Guide Solutions
Corner point feasible solution (a)
Set X1=0 and use the Capacity A equation
100X1+60X2=500
60X2=500
X2=8.33
X1=0
Corner point feasible solution (b)
Set X2=0 and use the Capacity B equation 50X1+20X2=200
50X1=200
X2=4.0
X2=0
Corner point feasible solution (c) is obtained by solving the following
equations simultaneously
100X1+60X2=500 and 50X1+20X2=200
therefore, multiply the second equation by -2:
100X1+60X2 =500
-100X1- 40X2 =-400
20X2=100
X2= 5.0
Thus X2=5.0 and substituting back into the Capacity A equation,
100X1+60X2=500
X1=2.0
X2=5.0
Therefore, corner point feasible solution (c): X1=2.0 , X2=5.0
SG8-14
Chapter 8 Study Guide Solutions
Corner Point
Feasible Solution a
LTD Company
12
10
Corner Point
Feasible Solution c
x2-HP-G49
8
Corner Point
Feasible Solution
b
6
4
Feasible
Region
2
0
0
1
2
3
4
5
x1-TI-83
Capacity A
g)
Capacity B
Find the optimal solution
Max profit = $259X1+$130X2
1. At a, Profit a = $259(0)+$130(8.33)
Profit a = $1083.33
2. At b, Profit b = $259(4)+$130(0)
Profit b = $1036.00
3. At c, Profit c = $259(2)+$130(5)
Profit c = $1168.00
Hence the optimal solution is when; X1=2 TI-83’s, X2=5 HP-G49’s and
Profit c = $1168.00
Managerial Implication:
Produce 2 TI-83’s and 5 HP-G49’s to yield a Profit of $1168.00.
SG8-15
6
Chapter 8 Study Guide Solutions
Problem 6
The following report has been generated in excel using the solver.
The company needs some help to have the output explained to them.
a)
What is the objective function optimal value?
5816.67
SG8-16
Chapter 8 Study Guide Solutions
b)
What are the final values of the decision variables at this optimal value?
(3.33,18.33)
c)
Explain which constraints are binding and not binding and what does that
mean?
Binding means there is no more left, whereas, not-binding means
there is availability to produce more goods. The Processing Time and
Hand Finishing constraints are labeled as binding in the Solver
output, therefore, they are all used up. However, the Gov’t Contract
constraint is labeled Not Binding, which means the constraint is not
all used up.
d)
What is the maximum increase in the objective coefficient of the first
variable that can occur without changing the optimal product mix?
There can be a 1300 unit increase in the objective coefficient of the
first variable.
e)
If you could increase the value of constraint #2 by one unit what would be
the effect on the objective function?
It would still be within the given range, as there is an allowable
increase of 120 for constraint 2. It would increase the objective
coefficient by approximately 1.11 units since the shadow price
associated the Hand Finishing constraint is 1.11.
SG8-17
Chapter 8 Study Guide Solutions
Problem 7
Big Gaming is a company that wishes to produce two new video game consoles to place
on the new market. They are a new company and wish to appeal to the public investors
with the sales of their first two gaming consoles. They wish to maximize their profit. The
two consoles both require a certain number of hours of Assembly and Packaging to
ensure that the products are not damaged before reaching consumers. Since all the
electronic parts are brought in from other companies. Big Gaming does not have to set
aside time for electronics. Since they are a relatively new company their work hours is
very limited. There is only a certain amount of hours that can be spent on each gaming
system. Also, twice as many A consoles must be produced than B consoles. From the
chart below, determine the best combination of the two products that will maximize Big
Gaming Profits.
Department
a)
Units per
Units per
Console A Console B
Available hrs
per week
Assembly
2
4
200
Packaging
2 2/3
1
200
Profit per unit
$210
$290
What are the decision variables?
X1= # of console A
X2= # of console B
b)
What is the objective function?
Max Profit = $210+$290
c)
What are the constraints?
2X1+4X2<480 (Assembly)
2 2/3X1+1X2<200 (Packaging)
X1 - 2X2>0 (At least twice as many A’s as B’s)
d)
What are the non-negativity assumptions?
SG8-18
Chapter 8 Study Guide Solutions
X1, X2 >0
e)
Write the LP in modified standard form.
Max Profit = $210+$290
S.T.
2X1+4X2<480 (Assembly)
2 2/3X1+1X2<200 (Packaging)
X1 - 2X2>0 (At least twice as many A’s as B’s)
X1, X2 >0
f)
Solve the LP graphically using the corner point method?
Corner point feasible solution (a) is obtained by solving the following
equations simultaneously
2X1+4X2=200 and X1-2X2=0
therefore, multiply the second by –2:
2X1+4X2 = 200
-2X1+4X2 = 0
8X2 = 200
X2 = 25
Thus X2 =25 and substituting back into the Assembling equation
2X1+4X2=200
2X1=100
X1=50
Therefore, corner point (a): X1=50, X2=25
Corner point feasible solution (b)
Set X2=0 and use the Packing equation
2 2/3X1+1X2=200
2 2/3X1=200
X1=75
X2=0
Corner point feasible solution (c) is obtained by solving the following
equations simultaneously
2X1+4X2=200 and 2 2/3X1+1X2=200
SG8-19
Chapter 8 Study Guide Solutions
therefore, multiply the second equation by –4:
2X1+4X2 = 200
-10 2/3X1-4X2 = -800
-8 2/3X1= -600
X1= 70
Thus X1 =70 and substituting back into the Assembling equation
2X1+4X2=200
4X2=60
X2=15
Therefore, corner point feasible solution (c): X1=70, X2=15
Big Gaming Company
100
x2-Console B
75
Corner Point
Feasible Solution c
Corner Point
Feasible Solution b
Corner Point
Feasible Solution a
50
25
Feasible
Region
0
0
10
20
30
40
50
60
70
80
x1-Console A
Assembly
g)
Packaging
Find the optimal solution
Max profit = $210X1+$290X2
1. At a, Profit a = $210(50)+$290(25)
Profit a = $17,750
2. At b, Profit b = $210(75)+$290(0)
SG8-20
Twice A than B
90
100
110
Chapter 8 Study Guide Solutions
Profit b = $15,750
3. At c, Profit c = $210(70)+$290(15)
Profit c = $19,050
Hence the optimal solution is when; X1=70, X2=15and Profit = $19,050
Managerial Implication:
Produce 70 console A’s and 15 console B’s to yield a Profit of $19,050.
SG8-21
Chapter 8 Study Guide Solutions
Problem 8
Whit’s Appliances produces two products. In order to produce, each product it requires a
certain amount of production time from Employee A, grinding, and Employee B,
polishing. The chart shows the amount of hours available per employee and the amount of
time for production.
Units per Units per
Y
Employee A-Grinding
2
4
40
Employee B-Polishing
4
3
50
$200
$305
Profit
a)
Available Hrs
X
What are the decision variables?
X1= # of product X
X2= # of product Y
b)
What is the objective function?
Max Profit = $200X1+$305X2
c)
What are the constraints?
2X1+4X2<40 (Employee A-Grinding)
4X1+3X2<50 (Employee B-Polishing)
d)
What are the non-negativity functions?
X1, X2 >0
e)
Write the LP in modified standard form.
Max Profit = $200X1+$305X2
S.T.
2X1+4X2<40 (Employee A-Grinding)
4X1+3X2<50 (Employee B-Polishing)
X1, X2 >0
f)
Solve the LP graphically using the corner point method?
SG8-22
Chapter 8 Study Guide Solutions
Corner point feasible solution (a)
Set X1=0 and use the Employee A equation
2X1+4X2<40
4X2=40
X2=10
X1=0
Corner point feasible solution (b)
Set X2=0 and use the Employee B equation
4X1+3X2<50
4X1=50
X1=12.5
Corner point (c) is obtained by solving the following equations
simultaneously
2X1+4X2=40 and 4X1+3X2=50
therefore, multiply the first equation by -2:
4X1+8X2 = 80
-4X1-3X2 = -50
5X2= 30
X2= 6
Thus X2 = 6 and substituting back into the Employee A equation
2X1+4X2=40
2X1=16
X1=8
Therefore, corner point feasible solution (c): X1=8, X2=6
g)
Find the optimal solution
Max profit = $200X1+$305X2
1. At a, Profit a = $200(0)+$305(10)
Profit a = $3,500
2. At b, Profit b = $200(12.5)+$305(0)
Profit b = $2,500
3. At c, Profit c = $200(8)+$305(6)
SG8-23
Chapter 8 Study Guide Solutions
Profit c = $3,430
Hence the optimal solution is when; X1=0, X2=10 and Profit = $3,500
Whit's Applicance
Corner Point
Feasible Solution a
18
16
14
Corner Point
Feasible Solution c
x2-Product Y
12
10
8
Corner Point
Feasible Solution b
6
Feasible
Region
4
2
0
0
5
10
15
x1-Product X
Employee A
Employee B
Managerial Implication:
Produce 0 product X and 10 product Y to yield a Profit of $3,500.
SG8-24
20
25
Chapter 8 Study Guide Solutions
Problem 9
A new company is involved in the production of two secret items (S1 and S2) both of
which are in liquid form. Each product requires hand mixing and automatic processing.
The table below gives the number of hours required a gallon of each item. The company
has 40 hours of processing time available in the next working week but only 35 hours of
hand mixing time. In addition, the company has signed a contract stating that they must
provide at least 5 gallons of S2 to the government. The company would like to maximize
its profits.
Units per Units per Available Hrs.
Gal. S1
Gal. S2
Processing time
1
2
40
Hand mixing time
5
1
35
$150
$290
Profit
a)
What are the decision variables?
X1= gallon of product S1
X2= gallon of product S2
b)
What is the objective function?
Max profit = $150+$290
c)
What are the constraints?
1X1+2X2<40 (Processing Time)
5X1+1X2<35 (Hand Finishing Time)
X2>5 (S2 Government Contract)
d)
What are the non-negativity assumptions?
X1, X2 >0
e)
Write the LP in modified standard form.
Max profit = $150+$290
S.T.
1X1+2X2<40 (Processing Time)
SG8-25
Chapter 8 Study Guide Solutions
5X1+1X2<35 (Hand Finishing Time)
X2>5 (S2 Government Contract)
X1, X2 >0
f)
Solve the LP graphically using the corner point method?
Corner point feasible solution (a)
Set X1=0 and use the Processing equation
1X1+2X2<40
2X2=40
X2=20
X1=0
Corner point feasible solution (b)
Set X2=5 (as we know we must provide at least 5 gallons of S2) and
use the Hand Mixing equation
5X1+1X2<35
5X1=30
X1=6
X2=5
Corner point (c) is obtained by solving the following equations
simultaneously
1X1+2X2=40 and 5X1+1X2=35
therefore, multiply the first equation by 5:
5X1+10X2 = 200
-5X1 - 1X2 = -35
9X2= 165
X2= 18 1/3
Thus X2 = 18 1/3 and substituting back into the Processing equation
1X1+2X2=40
1X1=3 1/3
X1=3 1/3
Therefore, the corner point feasible point (c): X1=3 1/3, X2=18 1/3
g)
Find the optimal solution
Max profit = $150X1+$290X2
1. At a, Profit a =$150 (0)+$290(20)
SG8-26
Chapter 8 Study Guide Solutions
Profit a = $5,800.00
2. At b, Profit b =$150(6)+$290(5)
Profit =$2,350.00
3. At c, Profit c =$150(3 1/3)+$290(18 1/3)
Profit c =$5,816.67
Hence the optimal solution is when; X1=3 1/3, X2=18 1/3 and Profit =
$5,816.67
Secret Items Company
Corner Point
Feasible Solution a
25
Corner Point
Feasible Solution c
x2-Secret Item S2
20
15
Feasible
Region
Corner Point Feasible
Solution b
10
5
0
0
2
4
6
8
10
12
14
x1-Secret Item S1
Processing
Hand Mixing
S2 Contract
Managerial Implication:
The solution makes mathematical sense and since it is feasible to
manufacture 3 1/3 gallons of S1 (1/3 of a gallon can be produced and sold,
whereas 1/3 of a table or chair cannot be produced and then sold). The
Secret Items Company should produce 3 1/3 gallons of S1 and 18 1/3 gallons
of S2 to yield a profit of $5,816.67.
SG8-27
Chapter 8 Study Guide Solutions
Problem 10
In producing their new golf clubs, Knight must utilize rubber for the grips and titanium
for the golf heads. The driver set requires 12 inches of rubber material and 3 pounds of
titanium. The iron set requires 10 inches of rubber and 2 pounds (lbs) of titanium. Knight
must supply a prime customer with at least 10 iron sets. Their profit is $190 for the
driver set and $155 for the iron set. What is our best way to maximize profits for the sets
of clubs?
Units per
Driver set
Iron set
Titanium
3
2
1500 (lbs)
Rubber
12
10
400 (inches)
$190
$155
Profit
a)
Units per Available Products
What are the decision variables?
X1= # of drivers
X2= # of irons
b)
What is the objective function?
Max profit =$190X1+$155X2
c)
What are the constraints?
3X1+2X2<150 (Titanium)
12X1+10X2<400 (Rubber)
X2>10 (Prime Customer Contract)
d)
What are the non-negativity assumptions?
X1, X2 >0
e)
Write the LP in modified standard form.
Max profit =$190X1+$155X2
S.T.
3X1+2X2<150 (Titanium)
12X1+10X2<400 (Rubber)
SG8-28
Chapter 8 Study Guide Solutions
X2>10 (Prime Customer Contract)
X1, X2 >0
f)
Solve the LP graphically using the corner point method?
Corner point feasible solution (a)
Set X1=0 and use the Rubber equation
12X1+10X2<400
10X2=400
X2=40
X1=0
Corner point feasible solution (b)
Set X2=10 and use the Rubber equation
12X1+10X2<400
12X1=300
X1=25
X2=10
Knight Golf Club
80
70
60
Corner Point
Feasible Solution a
x2-Irons
50
Corner Point
Feasible Solution b
40
30
Feasible
Region
20
10
0
0
10
20
30
40
x1-Drivers
Titanium
g)
Rubber
Find the optimal solution
Max profit = $190X1+$155X2
SG8-29
Prime Customer
50
60
Chapter 8 Study Guide Solutions
1. At a, Profit a = $190(0)+$155(40)
Profit a = $6,200
2. At b, Profit b = $190(25)+$155(10)
Profit b = $6,300
Hence the optimal solution is when; X1= 25, X2=10 and Profit =
$6,300.
Managerial Implication:
It appears that the Knight should provide the minimum number of Iron sets
required to the Prime Customer and then produce as many Driver sets as
possible. Therefore, Knight should produce 10 sets of Irons and 25 sets of
Drivers to yield a profit of $6,300. In addition, it appears that the company
has more than enough titanium. This can be seen in the graph as the
titanium constraint is not intersected by another constraint or by simply
plugging in the optimal decision variables into the titanium constraint to
yield the amount left-over/slack.
SG8-30
Chapter 8 Study Guide Solutions
Problem 11
A carpenter assembles tables and chair sets for a larger company. He can sell each table
for a profit of $43 and each chair set for a profit of $20. Each table requires 2 hours of
labor and each chair set requires 2 hours of labor. The carpenter also needs screws for
producing each item. The carpenter also wishes to produce 1 1/2 times as many chair sets
as tables as (he likes to make chair sets more than tables). The chart illustrates the
different variables for production. The carpenter would like to know how many of each
to assembly to maximize his profit.
Units per Table
Units per
Available
Chair Set
a)
Labor
20
8
512
Screws
16
24
1,000
Profit
$43
$20
What are the decision variables?
X1= # of tables
X2= # of chair sets
b)
What is the objective function?
Max profit = $43X1+$20X2
c)
What are the constraints?
20X1+8X2<512 (Labor)
16X1+24X2<1,000 (Screws)
1.5X1- X2 <0 (1 1/2 x Chair Sets)
d)
What are the non-negativity assumptions?
X1 , X2 >0
e)
Write the LP in modified standard form.
Max profit = $43X1+$20X2
S.T.
SG8-31
Chapter 8 Study Guide Solutions
20X1+8X2<512 (Labor)
16X1+24X2<1,000 (Screws)
1.5X1- X2 <0 (1 1/2 x Chair Sets)
X1 , X2 >0
f)
Solve the LP graphically using the corner point method?
Corner point feasible solution (a)
Set X1=0 and use the Screws equations
20X1+8X2<512
8X2=512
X2=40 2/3
X1=0
Corner point feasible solution (b) is obtained by solving the following
equations using substitution
20X1+8X2=512 and 1.5X1-X2=0
Rearranging the second equation we have
1.5X1 =X2
Using this relationship for X2 substitute into the first equation, we
have
20X1+8(1.5X1)=512
32X1=512
X1=16
X2=24
Therefore, the corner point feasible point (b): X1=16, X2=24
Corner point (c) is obtained by solving the following equations
simultaneously
20X1+8X2=512 and 16X1+24X2=1,000
therefore, multiply the first equation by 3:
60X1+24X2 = 1,536
-16X1- 24X2 = -1,000
44X1=536
X1= 12.18
SG8-32
Chapter 8 Study Guide Solutions
Thus X1 = 12.18 and substituting back into the Labor equation
20X1+8X2=512
8X2=268.4
X2=33.55
Therefore, the corner point feasible point (c): X1=12.18, X2=33.55
Corner Point
Feasible Solution a
Carpenter Assembly Inc.
55
Corner Point
Feasible Solution c
50
45
x2-Chair Sets
40
35
30
Feasible
Region
25
20
15
10
Corner Point
Feasible Solution b
5
0
0
5
10
15
x1-Tables
Labor
g)
Screws
1 1/2 x Chair Sets
Find the optimal solution
Max profit = $50X1+$20X2
1. At a, Profit a = $50(0)+$20(41 2/3)
Profit a = $833.33
2. At b, Profit b = $50(16)+$20(24)
Profit b = $1,280
3. At c, Profit c = $50(12.18)+$20(33.55)
SG8-33
20
25
Chapter 8 Study Guide Solutions
Profit c = $1,280
Hence the optimal solution is when: X1=16, X2=24 or X1=12.18,
X2=33.55 and Profit =$1,280.
Managerial Implication:
Produce either 16 Tables and 24 Chair Sets or 12.18 Tables and 33.55 Chair
Sets as each will yield a Profit of $25. However, it does not make sense for
the carpenter to make 0.18 fraction of a table or 0.55 fraction of a Chair Set.
It would make more business sense to produce in whole units of Tables and
Chair Sets, therefore, produce 16 Tables and 24 Chair Sets (see note).
NOTE:
In fact, a manager could chose to produce at any level
between those two points, i.e., anywhere on the Labor
equation between X1=16, X2=24 and X1=12.18,
X2=33.55, to achieve the same Profit of $1,280.
SG8-34
Chapter 8 Study Guide Solutions
Problem 12
Firebridge’s new radial tires come in two styles. The Fire Hawk uses 20 pounds of rubber
and 6 pounds of steel. The Tiger Claw requires 15 pounds of rubber and 1 pounds of
steel. They currently have 3,000 pounds of rubber and 300 pounds of steel. Firebridge is
obligated to provide these tires to the government in the following minimum quantities:
at least 20 Tiger Claws and at least 10 Fire Hawks. However, the government stipulates
that of all the tires provided at least twice as many must be Tiger Claws than Fire Hawks.
In addition, Firebridge must provide at least 100 tires total. At a cost of $10 for the fire
hawk and $10 for a tiger claw, what is the best solution to minimize cost?
Units per
Units per
Available
Fire Hawk Tiger Claw Products
Steel
6
Rubber 20
Cost
a)
$10
1
300
15
3,000
$10
What are the decision variables?
X1= # of Fire Hawks
X2= # of Tiger Claws
b)
What is the objective function?
Min Cost = $10X1+$10X2
c)
What are the constraints?
6X1+1X2<300 (Steel)
20X1+15X2<3,000 (Rubber)
X2 > 20 (At least 20 Tiger Claws)
X1 > 10 (At least 10 Fire Hawks)
2X1 - X2 < 0 (At least 2 x as many Tiger Claws as Fire Hawks)
X1 + X2 > 100 (At least 100 tires total)
d)
What are the non-negativity assumptions?
SG8-35
Chapter 8 Study Guide Solutions
X1, X2 >0
e)
Write the LP in modified standard form.
Min Cost = $10X1+$10X2
S.T.
6X1+1X2<300 (Steel)
20X1+15X2<3,000 (Rubber)
X2 > 20 (At least 20 Tiger Claws)
X1 > 10 (At least 10 Fire Hawks)
2X1 - X2 < 0 (At least 2 x as many Tiger Claws as Fire Hawks)
X1 + X2 > 100 (At least 100 tires total)
X1, X2 >0
f)
Solve the LP graphically using the corner point method?
Corner point (a)
Set X1=10, and use the Rubber equation
20X1+15X2=3,000
15X2=2,800
X2=186.67
X1=10
Corner point (b)
Set X1=10, and use the At Least 100 Tires equation
X1+X2=100
X2=90
X1=10
Corner point (c) is obtained by solving the following equations
simultaneously
6X1+X2=300 and 20X1+15X2=3,000
Multiply the first equation by -15:
-90X1-15X2 = -4,500
20X1+ 15X2 = 3,000
-70X1= -1,500
X1= 21.43
Thus X2 =21.43 and substituting back into the Steel equation
SG8-36
Chapter 8 Study Guide Solutions
6X1+X2=300
X2=171.43
Therefore the corner point feasible solution (c): X1=21.43, X2=171.43
Corner point (d) is obtained by solving the following equations
simultaneously
6X1+X2=300 and 2X1-X2=0
Adding the two equations together:
6X1+X2=300
2X1-X2=0
8X1= 300
X1= 37.5
Thus X1 =37.5 and substituting back into the Steel equation
6X1+X2=300
X2=75
Therefore, the corner point feasible solution (c): X1=37.5, X2=75
Corner point (e) is obtained by solving the following equations
simultaneously
2X1-X2=0 and X1+X2=100
Add the two equations together:
2X1-X2 = 0
X1+X2=100
3X1= 100
X1= 33.33
Thus X1 =33.33 and substituting back into the Total Tire equation
X1+X2=100
X2=66.67
Therefore, the corner point feasible solution (e): X1=33.33, X2=66.67
SG8-37
Chapter 8 Study Guide Solutions
Firebridge Inc.
Corner Point
Feasible Solution a
250
225
Corner Point
Feasible Solution c
200
x2-Tiger Claw
175
Corner Point
Feasible Solution d
150
Feasible
Region
125
100
75
50
Corner Point
Feasible Solution b
25
0
0
10
20
30
x1Fire Hawk-
Steel
g)
Rubber
Fire Hawk
2 x Tiger Claw
40
50
Corner Point
Feasible Solution e
At least 100 Tires
Find the optimal solution
Min Cost = $10X1+$10X2
1. At a, Cost a = $10(10)+$10(186.67)
Cost a = $1,966.70
2. At b, Cost b = $10(10)+$10(90)
Cost b = $1,000.00
3. At c, Cost c = $10(21.43)+$10(171.43)
Cost c = $1,928.60
4. At d, Cost d = $10(37.5)+$10(75)
Cost d = $1,125.00
5. At e, Cost e = $10(33.33)+$10(66.67)
Cost e = $1,000
Hence the optimal solution is when: X1=10, X2=90 or X1=33.33,
X2=66.67 and Cost = $1,000.
SG8-38
Chapter 8 Study Guide Solutions
Managerial Implication:
Produce either 10 Fire Hawks and 90 Tiger Claws or 33.33 Fire Hawks and
66.67 Tiger Claws as each will yield the lowest Cost of $1,000. However, it
does not make sense for the tire company to produce 1/3 of a Fire Hawk or
2/3 of a Tiger Claw. It would make more business sense to produce in whole
units of tires, therefore, produce 10 Fire Hawks and 90 Tiger Claws (see
note).
NOTE:
In fact, a manager could chose to produce at any level between
those two points, i.e., anywhere on the At least 100 tires
equation between X1=10, X2=90 and X1=33.33, X2=66.67, to
achieve the same Profit of $1,000. However, the only points
that make “business” sense would be integer combinations of
X1and X2 on that line between the aforementioned points (e.g.,
X1=30, X2=70).
SG8-39