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SOLUTIONS + PowerPoint Slides
, Solution Manual – Optimization Modelling
CONTENT
Page#
Chapter 1 5
Chapter 2 8
Chapter 3 10
Chapter 4 19
Chapter 5 32
Chapter 6 41
Chapter 7 45
Chapter 10 49
Chapter 11 58
Chapter 12 62
,Solution Manual – Optimization Modelling
, Solution Manual – Optimization Modelling
Chapter 1
Solution to Exercises
1.1 Jenny will run an ice cream stand in the coming week-long
multicultural event. She believes the fixed cost per day of running
the stand is $60. Her best guess is that she can sell up to 250 ice
creams per day at $1.50 per ice cream. The cost of each ice cream is
$0.85. Find an expression for the daily profit, and hence find the
breakeven point (no profit–no loss point).
Solution:
Suppose x the number of ice creams Jenny can sell in a
day. The cost of x ice creams ($) = 0.85x
Jenny’s cost per day ($) = 60 + 0.85x
Daily revenue from ice cream sale ($) = 1.50x
Expression for daily profit ($) P = 1.50x – (60 + 0.85x) = 0.65x –
60 At breakeven point, 0.65x – 60 = 0
So, x = 60/0.65 = 92.31 ice creams
1.2 The total cost of producing x items per day is 45x + 27 dollars, and
the price per item at which each may be sold is 60 – 0.5x dollars.
Find an expression for the daily profit, and hence find the maximum
possible profit.
Solution:
Daily revenue = x(60 – 0.5x) = 60x – 0.5x2
The expression for daily profit, P = 60x – 0.5x2 – (45x + 27)
= 15x – 0.5x2 – 27
Differentiating the profit function, we get:
dP
15 x 0, that means x = 15. So, the optimal profit is $85.5.
dx
The profit function looks like as follows:
95
85
75
65
55
45
35
25
4 9 14 19 24
Val ue of X
1.3 A stone is thrown upwards so that at any time x seconds after
throwing, the height of the stone is y = 100 + 10x – 5x2 meters. Find
, Solution Manual – Optimization Modelling
the maximum height reached.
, Solution Manual – Optimization Modelling
Solution:
Differentiating the expression of height with respect to time, we get:
dy
10 10x that means x = 1.
0,
dx
So the corresponding /optimal height is (100 + 10 – 5) = 105
meters. You can draw the function y in Excel to check the
result.
1.4 A manufacturer finds that the cost C(x) = 2x2 – 8x + 15, where x is the
number of machines operating. Find how many machines he should
operate in order to minimize the total cost of production. What is the
optimal cost of production?
Solution:
Differentiating the cost function with respect to x, we get:
dC(x)
4x 8 0, that means x = 2.
dx
So the optimal cost = 8 – 16 + 15 = $7
1.5 A string 72 cm long is to be cut into two pieces. One piece is used to
form a circle and the other a square. What should be the perimeter
of the square in order to minimize the sum of two areas?
Solution:
Let us assume that each side of the square is x cm
long. The perimeter of the square is 4x.
The circumference of the circle will be 72 – 4x = 2 r, where r is the
radius of the circle. So, r = (72 – 4x)/(2 .
The area of the square = x2
The area of the circle = r2
The sum of two areas, A(x) = x2 + r2 = x2 + (72 – 4x)/(2 ]2
=x +2 (36 – 2x)/( ]2
= x2 + (1/ )(36 – 2x) 2
= x2 + (4/ )(18 – x) 2
= (4/ )(182) –(4/ )36x + ((4/ )x2
Differentiating A(x) with respect to x, we get:
dA(x)
( )36 ( 1)2x 0,
4 4
dx
4 4
or ( 1)2
x ( )36,
or (4 )x 72,
72
or x
(4 )
,, Solution Manual – Optimization Modelling
or x 10.08
So the perimeter is 40.32 cm.
, Solution Manual – Optimization Modelling
The areas of the square and the circle and combined square and
circle with the length (integer value) of square are shown below.
A(Square)
A(Circle)
A(Square) + A(Circle)
300
Area (sq. cm)
250
200
150
100
1.6 Find the maximum or minimum values of the following quadratic
functions, and the values of x for which they occur:
(a) f(x) = x2 – 4x + 7
(b) f(x) = 3 + 8x – x2
Solution:
df (x)
(a) 2x 4 0, or x = 2
dx
d 2 f (x)
2, so x = 2 is the minimum value.
dx2
df (x)
(b) 8 2x 0, or x = 4
dx
d 2 f (x)
2, so x = 4 is the maximum value.
dx2
, Solution Manual – Optimization Modelling
Chapter 2
Solution to Exercises
2.1 The lift users of a multistoried building have complained about the
delay in getting an elevator. Being the property manager, how do
you define the problem in order to solve it? In other words, what is
your problem precisely which you intend to solve?
The problem definition may vary from person to person for such a
situation. If you cannot define the problem appropriately it is
unlikely that it will be solved. For example, the problem may be
thought as:
1. Minimizing the waiting time by using better and efficient
elevator which would require an expensive re-engineering
of the elevator system.
2. Minimizing the people movement by studying the reasons
for frequent elevator usage and reduce them taking
appropriate action. For example, having laundry in each
floor instead of a common laundry at the basement.
3. Minimizing or eliminating the complaints using simple but
innovative means such as putting mirrors on the walls around
the lobby of the building. This would not change the waiting
time of the elevators and people movement, but will change
the perception, because people became occupied with another
activity. So the complaints will be disappeared.
4. You may add another option!
Most people would choose the first one as the problem definition
(minimizing waiting time) and suggest an expensive re-engineering
as the solution. Which one you would choose and why?
Solution: The minimization or elimination of the complaints, as in
option (3), could be an appropriate problem definition for some
property managers. The corresponding solution would be least
expensive.
2.2 A manufacturing company produces several products in its shopfloor
and sells them, directly to their customers, through its retail section.
Although the production capacity is fixed and known, the demand of
each product varies from period to period. As a result, few products
are experiencing shortages in some periods whereas some other
products are having excess stocks. The retail manager knows that the
overall performance of the company can be improved by applying
optimization techniques. The company is currently performing very
well financially. The top management is neither familiar with
optimization techniques nor intended to make any changes in its
current production schedule. As the retail manager, how would you
convenience the top management to study the current system using
, Solution Manual – Optimization Modelling
optimization techniques?
Solution: There is no fixed answer for this question. You may try as
follows: perform a pilot study on your own and show the financial
benefits, due to application of optimization techniques, from this pilot
study. The top management will need to be convinced that the
proposed solution performs
SOLUTIONS + PowerPoint Slides
, Solution Manual – Optimization Modelling
CONTENT
Page#
Chapter 1 5
Chapter 2 8
Chapter 3 10
Chapter 4 19
Chapter 5 32
Chapter 6 41
Chapter 7 45
Chapter 10 49
Chapter 11 58
Chapter 12 62
,Solution Manual – Optimization Modelling
, Solution Manual – Optimization Modelling
Chapter 1
Solution to Exercises
1.1 Jenny will run an ice cream stand in the coming week-long
multicultural event. She believes the fixed cost per day of running
the stand is $60. Her best guess is that she can sell up to 250 ice
creams per day at $1.50 per ice cream. The cost of each ice cream is
$0.85. Find an expression for the daily profit, and hence find the
breakeven point (no profit–no loss point).
Solution:
Suppose x the number of ice creams Jenny can sell in a
day. The cost of x ice creams ($) = 0.85x
Jenny’s cost per day ($) = 60 + 0.85x
Daily revenue from ice cream sale ($) = 1.50x
Expression for daily profit ($) P = 1.50x – (60 + 0.85x) = 0.65x –
60 At breakeven point, 0.65x – 60 = 0
So, x = 60/0.65 = 92.31 ice creams
1.2 The total cost of producing x items per day is 45x + 27 dollars, and
the price per item at which each may be sold is 60 – 0.5x dollars.
Find an expression for the daily profit, and hence find the maximum
possible profit.
Solution:
Daily revenue = x(60 – 0.5x) = 60x – 0.5x2
The expression for daily profit, P = 60x – 0.5x2 – (45x + 27)
= 15x – 0.5x2 – 27
Differentiating the profit function, we get:
dP
15 x 0, that means x = 15. So, the optimal profit is $85.5.
dx
The profit function looks like as follows:
95
85
75
65
55
45
35
25
4 9 14 19 24
Val ue of X
1.3 A stone is thrown upwards so that at any time x seconds after
throwing, the height of the stone is y = 100 + 10x – 5x2 meters. Find
, Solution Manual – Optimization Modelling
the maximum height reached.
, Solution Manual – Optimization Modelling
Solution:
Differentiating the expression of height with respect to time, we get:
dy
10 10x that means x = 1.
0,
dx
So the corresponding /optimal height is (100 + 10 – 5) = 105
meters. You can draw the function y in Excel to check the
result.
1.4 A manufacturer finds that the cost C(x) = 2x2 – 8x + 15, where x is the
number of machines operating. Find how many machines he should
operate in order to minimize the total cost of production. What is the
optimal cost of production?
Solution:
Differentiating the cost function with respect to x, we get:
dC(x)
4x 8 0, that means x = 2.
dx
So the optimal cost = 8 – 16 + 15 = $7
1.5 A string 72 cm long is to be cut into two pieces. One piece is used to
form a circle and the other a square. What should be the perimeter
of the square in order to minimize the sum of two areas?
Solution:
Let us assume that each side of the square is x cm
long. The perimeter of the square is 4x.
The circumference of the circle will be 72 – 4x = 2 r, where r is the
radius of the circle. So, r = (72 – 4x)/(2 .
The area of the square = x2
The area of the circle = r2
The sum of two areas, A(x) = x2 + r2 = x2 + (72 – 4x)/(2 ]2
=x +2 (36 – 2x)/( ]2
= x2 + (1/ )(36 – 2x) 2
= x2 + (4/ )(18 – x) 2
= (4/ )(182) –(4/ )36x + ((4/ )x2
Differentiating A(x) with respect to x, we get:
dA(x)
( )36 ( 1)2x 0,
4 4
dx
4 4
or ( 1)2
x ( )36,
or (4 )x 72,
72
or x
(4 )
,, Solution Manual – Optimization Modelling
or x 10.08
So the perimeter is 40.32 cm.
, Solution Manual – Optimization Modelling
The areas of the square and the circle and combined square and
circle with the length (integer value) of square are shown below.
A(Square)
A(Circle)
A(Square) + A(Circle)
300
Area (sq. cm)
250
200
150
100
1.6 Find the maximum or minimum values of the following quadratic
functions, and the values of x for which they occur:
(a) f(x) = x2 – 4x + 7
(b) f(x) = 3 + 8x – x2
Solution:
df (x)
(a) 2x 4 0, or x = 2
dx
d 2 f (x)
2, so x = 2 is the minimum value.
dx2
df (x)
(b) 8 2x 0, or x = 4
dx
d 2 f (x)
2, so x = 4 is the maximum value.
dx2
, Solution Manual – Optimization Modelling
Chapter 2
Solution to Exercises
2.1 The lift users of a multistoried building have complained about the
delay in getting an elevator. Being the property manager, how do
you define the problem in order to solve it? In other words, what is
your problem precisely which you intend to solve?
The problem definition may vary from person to person for such a
situation. If you cannot define the problem appropriately it is
unlikely that it will be solved. For example, the problem may be
thought as:
1. Minimizing the waiting time by using better and efficient
elevator which would require an expensive re-engineering
of the elevator system.
2. Minimizing the people movement by studying the reasons
for frequent elevator usage and reduce them taking
appropriate action. For example, having laundry in each
floor instead of a common laundry at the basement.
3. Minimizing or eliminating the complaints using simple but
innovative means such as putting mirrors on the walls around
the lobby of the building. This would not change the waiting
time of the elevators and people movement, but will change
the perception, because people became occupied with another
activity. So the complaints will be disappeared.
4. You may add another option!
Most people would choose the first one as the problem definition
(minimizing waiting time) and suggest an expensive re-engineering
as the solution. Which one you would choose and why?
Solution: The minimization or elimination of the complaints, as in
option (3), could be an appropriate problem definition for some
property managers. The corresponding solution would be least
expensive.
2.2 A manufacturing company produces several products in its shopfloor
and sells them, directly to their customers, through its retail section.
Although the production capacity is fixed and known, the demand of
each product varies from period to period. As a result, few products
are experiencing shortages in some periods whereas some other
products are having excess stocks. The retail manager knows that the
overall performance of the company can be improved by applying
optimization techniques. The company is currently performing very
well financially. The top management is neither familiar with
optimization techniques nor intended to make any changes in its
current production schedule. As the retail manager, how would you
convenience the top management to study the current system using
, Solution Manual – Optimization Modelling
optimization techniques?
Solution: There is no fixed answer for this question. You may try as
follows: perform a pilot study on your own and show the financial
benefits, due to application of optimization techniques, from this pilot
study. The top management will need to be convinced that the
proposed solution performs
