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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
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Solution Manual – Optimization Modelling
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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
the maximum height reached.
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Acadexas Stuvia | Page 4 of 639
Covers All Chapters
SOLUTIONS + PowerPoint Slides
Acadexas Stuvia | Page 1 of 639
, STUDY NOTES & REFERENCE GUIDE
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
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Acadexas Stuvia | Page 2 of 639
, STUDY NOTES & REFERENCE GUIDE
Solution Manual – Optimization Modelling
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Acadexas Stuvia | Page 3 of 639
, STUDY NOTES & REFERENCE GUIDE
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
the maximum height reached.
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@SSeeisismmi5cicisisoolalatitoionn
Acadexas Stuvia | Page 4 of 639
