This document contains lecture notes from a Calculus I class discussing optimization problems. It begins with announcements about upcoming exams and courses the professor is teaching. It then presents an example problem about finding the rectangle of a fixed perimeter with the maximum area. The solution uses calculus techniques like taking the derivative to find the critical points and determine that the optimal rectangle is a square. The notes discuss strategies for solving optimization problems and summarize the key steps to take.

1. Sec on 4.5
Op miza on Problems
V63.0121.011: Calculus I
Professor Ma hew Leingang
New York University
April 18, 2011
.

2. Announcements
Quiz 5 on Sec ons
4.1–4.4 April 28/29
Final Exam Thursday May
12, 2:00–3:50pm
I am teaching Calc II MW
2:00pm and Calc III TR
2:00pm both Fall ’11 and
Spring ’12

3. Objectives
Given a problem
requiring op miza on,
iden fy the objec ve
func ons, variables, and
constraints.
Solve op miza on
problems with calculus.

4. Leading by Example
Example
What is the rectangle of fixed perimeter with maximum area?

5. Leading by Example
Example
What is the rectangle of fixed perimeter with maximum area?
Solu on

6. Leading by Example
Example
What is the rectangle of fixed perimeter with maximum area?
Solu on
Draw a rectangle.
.

7. Leading by Example
Example
What is the rectangle of fixed perimeter with maximum area?
Solu on
Draw a rectangle.
.
ℓ

8. Leading by Example
Example
What is the rectangle of fixed perimeter with maximum area?
Solu on
Draw a rectangle.
w
.
ℓ

9. Solution
Solu on (Con nued)
Let its length be ℓ and its width be w. The objec ve func on is
area A = ℓw.

10. Solution
Solu on (Con nued)
Let its length be ℓ and its width be w. The objec ve func on is
area A = ℓw.
This is a func on of two variables, not one. But the perimeter is
fixed.

11. Solution
Solu on (Con nued)
This is a func on of two variables, not one. But the perimeter is
fixed.
p − 2w
Since p = 2ℓ + 2w, we have ℓ = , so
2
p − 2w 1 1
A = ℓw = · w = (p − 2w)(w) = pw − w2
2 2 2

12. Solution
Solu on (Con nued)
p − 2w
Since p = 2ℓ + 2w, we have ℓ = , so
2
p − 2w 1 1
A = ℓw = · w = (p − 2w)(w) = pw − w2
2 2 2
Now we have A as a func on of w alone (p is constant).

13. Solution
Solu on (Con nued)
p − 2w
Since p = 2ℓ + 2w, we have ℓ = , so
2
p − 2w 1 1
A = ℓw = · w = (p − 2w)(w) = pw − w2
2 2 2
Now we have A as a func on of w alone (p is constant).
The natural domain of this func on is [0, p/2] (we want to
make sure A(w) ≥ 0).

14. Solution
Solu on (Con nued)
1
We use the Closed Interval Method for A(w) = pw − w2 on
2
[0, p/2].

15. Solution
Solu on (Con nued)
1
We use the Closed Interval Method for A(w) = pw − w2 on
2
[0, p/2].
At the endpoints, A(0) = A(p/2) = 0.

16. Solution
Solu on (Con nued)
1
We use the Closed Interval Method for A(w) = pw − w2 on
2
[0, p/2].
At the endpoints, A(0) = A(p/2) = 0.
dA 1
To find the cri cal points, we find = p − 2w.
dw 2

17. Solution
Solu on (Con nued)
dA 1
To find the cri cal points, we find = p − 2w.
dw 2
1 p
The cri cal points are when p − 2w = 0, or w = .
2 4

18. Solution
Solu on (Con nued)
dA 1
To find the cri cal points, we find = p − 2w.
dw 2
1 p
The cri cal points are when p − 2w = 0, or w = .
2 4
Since this is the only cri cal point, it must be the maximum. In
p
this case ℓ = as well.
4

19. Solution
Solu on (Con nued)
dA 1
To find the cri cal points, we find = p − 2w.
dw 2
1 p
The cri cal points are when p − 2w = 0, or w = .
2 4
Since this is the only cri cal point, it must be the maximum. In
p
this case ℓ = as well.
4
We have a square! The maximal area is A(p/4) = p2 /16.

20. Outline
The Text in the Box
More Examples

21. Strategies for Problem Solving
1. Understand the problem
2. Devise a plan
3. Carry out the plan
4. Review and extend
György Pólya
(Hungarian, 1887–1985)

22. The Text in the Box
1. Understand the Problem. What is known? What is unknown?
What are the condi ons?

23. The Text in the Box
1. Understand the Problem. What is known? What is unknown?
What are the condi ons?
2. Draw a diagram.

24. The Text in the Box
1. Understand the Problem. What is known? What is unknown?
What are the condi ons?
2. Draw a diagram.
3. Introduce Nota on.

25. The Text in the Box
1. Understand the Problem. What is known? What is unknown?
What are the condi ons?
2. Draw a diagram.
3. Introduce Nota on.
4. Express the “objec ve func on” Q in terms of the other
symbols

26. The Text in the Box
1. Understand the Problem. What is known? What is unknown?
What are the condi ons?
2. Draw a diagram.
3. Introduce Nota on.
4. Express the “objec ve func on” Q in terms of the other
symbols
5. If Q is a func on of more than one “decision variable”, use the
given informa on to eliminate all but one of them.

27. The Text in the Box
1. Understand the Problem. What is known? What is unknown?
What are the condi ons?
2. Draw a diagram.
3. Introduce Nota on.
4. Express the “objec ve func on” Q in terms of the other
symbols
5. If Q is a func on of more than one “decision variable”, use the
given informa on to eliminate all but one of them.
6. Find the absolute maximum (or minimum, depending on the
problem) of the func on on its domain.

28. Polya’s Method in Kindergarten
Name [_
Problem Solving Strategy
Draw a Picture
Kathy had a box of 8 crayons.
She gave some crayons away.
She has 5 left.
How many crayons did Kathy give away?
UNDERSTAND
•
What do you want to find out?
Draw a line under the question.
You can draw a picture
to solve the problem.
What number do I
add to 5 to get 8?
8 - = 5
crayons 5 + 3 = 8
CHECK
Does your answer make sense?
Explain.
What number
Draw a picture to solve the problem. do I add to 3
Write how many were given away. to make 10?
I. I had 10 pencils. ft ft ft A
I gave some away. 13 ill
i :i
I
'•' I I
I have 3 left. How many i? «
11 I

29. The Closed Interval Method
See Section 4.1
The Closed Interval Method
To find the extreme values of a func on f on [a, b], we need to:
Evaluate f at the endpoints a and b
Evaluate f at the cri cal points x where either f′ (x) = 0 or f is
not differen able at x.
The points with the largest func on value are the global
maximum points
The points with the smallest/most nega ve func on value are
the global minimum points.

30. The First Derivative Test
See Section 4.3
Theorem (The First Deriva ve Test)
Let f be con nuous on (a, b) and c a cri cal point of f in (a, b).
If f′ changes from nega ve to posi ve at c, then c is a local
minimum.
If f′ changes from posi ve to nega ve at c, then c is a local
maximum.
If f′ does not change sign at c, then c is not a local extremum.

31. The First Derivative Test
See Section 4.3
Corollary
If f′ < 0 for all x < c and f′ (x) > 0 for all x > c, then c is the
global minimum of f on (a, b).
If f′ < 0 for all x > c and f′ (x) > 0 for all x < c, then c is the
global maximum of f on (a, b).

32. Recall: The Second Derivative Test
See Section 4.3
Theorem (The Second Deriva ve Test)
Let f, f′ , and f′′ be con nuous on [a, b]. Let c be in (a, b) with
f′ (c) = 0.
If f′′ (c) < 0, then f(c) is a local maximum.
If f′′ (c) > 0, then f(c) is a local minimum.

33. Recall: The Second Derivative Test
See Section 4.3
Theorem (The Second Deriva ve Test)
Let f, f′ , and f′′ be con nuous on [a, b]. Let c be in (a, b) with
f′ (c) = 0.
If f′′ (c) < 0, then f(c) is a local maximum.
If f′′ (c) > 0, then f(c) is a local minimum.
Warning
If f′′ (c) = 0, the second deriva ve test is inconclusive (this does not
mean c is neither; we just don’t know yet).

34. Recall: The Second Derivative Test
See Section 4.3
Corollary
If f′ (c) = 0 and f′′ (x) > 0 for all x, then c is the global minimum
of f
If f′ (c) = 0 and f′′ (x) < 0 for all x, then c is the global maximum
of f

35. Which to use when?
CIM 1DT 2DT
Pro
Con

36. Which to use when?
CIM 1DT 2DT
Pro
– no need for
inequali es
Con

37. Which to use when?
CIM 1DT 2DT
Pro
– no need for
inequali es
– gets global extrema
automa cally
Con

38. Which to use when?
CIM 1DT 2DT
Pro
– no need for
inequali es
– gets global extrema
automa cally
Con
– only for closed
bounded intervals

39. Which to use when?
CIM 1DT 2DT
Pro
– no need for – works on
inequali es non-closed,
– gets global extrema non-bounded
automa cally intervals
Con
– only for closed
bounded intervals

40. Which to use when?
CIM 1DT 2DT
Pro
– no need for – works on
inequali es non-closed,
– gets global extrema non-bounded
automa cally intervals
– only one deriva ve
Con
– only for closed
bounded intervals

41. Which to use when?
CIM 1DT 2DT
Pro
– no need for – works on
inequali es non-closed,
– gets global extrema non-bounded
automa cally intervals
– only one deriva ve
Con
– only for closed – Uses inequali es
bounded intervals

42. Which to use when?
CIM 1DT 2DT
Pro
– no need for – works on
inequali es non-closed,
– gets global extrema non-bounded
automa cally intervals
– only one deriva ve
Con
– only for closed – Uses inequali es
bounded intervals – More work at
boundary than CIM

43. Which to use when?
CIM 1DT 2DT
Pro
– no need for – works on – works on
inequali es non-closed, non-closed,
– gets global extrema non-bounded non-bounded
automa cally intervals intervals
– only one deriva ve
Con
– only for closed – Uses inequali es
bounded intervals – More work at
boundary than CIM

44. Which to use when?
CIM 1DT 2DT
Pro
– no need for – works on – works on
inequali es non-closed, non-closed,
– gets global extrema non-bounded non-bounded
automa cally intervals intervals
– only one deriva ve – no need for
inequali es
Con
– only for closed – Uses inequali es
bounded intervals – More work at
boundary than CIM

45. Which to use when?
CIM 1DT 2DT
Pro
– no need for – works on – works on
inequali es non-closed, non-closed,
– gets global extrema non-bounded non-bounded
automa cally intervals intervals
– only one deriva ve – no need for
inequali es
Con
– only for closed – Uses inequali es – More deriva ves
bounded intervals – More work at
boundary than CIM

46. Which to use when?
CIM 1DT 2DT
Pro
– no need for – works on – works on
inequali es non-closed, non-closed,
– gets global extrema non-bounded non-bounded
automa cally intervals intervals
– only one deriva ve – no need for
inequali es
Con
– only for closed – Uses inequali es – More deriva ves
bounded intervals – More work at – less conclusive than
boundary than CIM 1DT

47. Which to use when?
If domain is closed and bounded, use CIM.
If domain is not closed or not bounded, use 2DT if you like to
take deriva ves, or 1DT if you like to compare signs.

48. Outline
The Text in the Box
More Examples

49. Another Example
Example (The Best Fencing Plan)
A rectangular plot of farmland will be bounded on one side by a
river and on the other three sides by a single-strand electric fence.
With 800m of wire at your disposal, what is the largest area you can
enclose, and what are its dimensions?

50. Solution
Solu on
1. Everybody understand?

51. Another Example
Example (The Best Fencing Plan)
A rectangular plot of farmland will be bounded on one side by a
river and on the other three sides by a single-strand electric fence.
With 800m of wire at your disposal, what is the largest area you can
enclose, and what are its dimensions?

52. Another Example
Example (The Best Fencing Plan)
A rectangular plot of farmland will be bounded on one side by a
river and on the other three sides by a single-strand electric fence.
With 800m of wire at your disposal, what is the largest area you can
enclose, and what are its dimensions?
Known: amount of fence used
Unknown: area enclosed

53. Another Example
Example (The Best Fencing Plan)
A rectangular plot of farmland will be bounded on one side by a
river and on the other three sides by a single-strand electric fence.
With 800m of wire at your disposal, what is the largest area you can
enclose, and what are its dimensions?
Known: amount of fence used
Unknown: area enclosed
Objec ve: maximize area
Constraint: fixed fence length

54. Solution
Solu on
1. Everybody understand?

55. Solution
Solu on
1. Everybody understand?
2. Draw a diagram.

56. Diagram
A rectangular plot of farmland will be bounded on one side by a
river and on the other three sides by a single-strand electric fence.
With 800 m of wire at your disposal, what is the largest area you can
enclose, and what are its dimensions?
.
.

57. Solution
Solu on
1. Everybody understand?
2. Draw a diagram.

58. Solution
Solu on
1. Everybody understand?
2. Draw a diagram.
3. Length and width are ℓ and w. Length of wire used is p.

59. Diagram
A rectangular plot of farmland will be bounded on one side by a
river and on the other three sides by a single-strand electric fence.
With 800 m of wire at your disposal, what is the largest area you can
enclose, and what are its dimensions?
.
.

60. Diagram
A rectangular plot of farmland will be bounded on one side by a
river and on the other three sides by a single-strand electric fence.
With 800 m of wire at your disposal, what is the largest area you can
enclose, and what are its dimensions?
ℓ
w
.
.

61. Solution
Solu on
1. Everybody understand?
2. Draw a diagram.
3. Length and width are ℓ and w. Length of wire used is p.

62. Solution
Solu on
4. Q = area = ℓw.

63. Solution
Solu on
4. Q = area = ℓw.
5. Since p = ℓ + 2w, we have ℓ = p − 2w and so
Q(w) = (p − 2w)(w) = pw − 2w2
The domain of Q is [0, p/2]

64. Solution
Solu on
4. Q = area = ℓw.
5. Since p = ℓ + 2w, we have ℓ = p − 2w and so
Q(w) = (p − 2w)(w) = pw − 2w2
The domain of Q is [0, p/2]
dQ p
6. = p − 4w, which is zero when w = .
dw 4

65. Solution
Solu on
7. Q(0) = Q(p/2) = 0, but
(p) p p2 p2
Q =p· −2· = = 80, 000m2
4 4 16 8
so the cri cal point is the absolute maximum.

66. Solution
Solu on
7. Q(0) = Q(p/2) = 0, but
(p) p p2 p2
Q =p· −2· = = 80, 000m2
4 4 16 8
so the cri cal point is the absolute maximum.
8. The dimensions of the op mal rectangle are
p p
w= = 200 m2 and ℓ = = 400 m2
4 2

67. Your turn
Example (The shortest fence)
A 216m2 rectangular pea patch is to be enclosed by a fence and
divided into two equal parts by another fence parallel to one of its
sides. What dimensions for the outer rectangle will require the
smallest total length of fence? How much fence will be needed?

68. Your turn
Example (The shortest fence)
A 216m2 rectangular pea patch is to be enclosed by a fence and
divided into two equal parts by another fence parallel to one of its
sides. What dimensions for the outer rectangle will require the
smallest total length of fence? How much fence will be needed?
Solu on
Let the length and width of the pea patch be ℓ and w. The amount of
fence needed is f = 2ℓ + 3w. Since ℓw = A, a constant, we have
A
f(w) = 2 + 3w. The domain is all posi ve numbers.
w

69. Diagram
w
.
ℓ
f = 2ℓ + 3w A = ℓw ≡ 216

70. Solution (Continued)
2A
We need to find the minimum value of f(w) = + 3w on (0, ∞).
w

71. Solution (Continued)
2A
We need to find the minimum value of f(w) = + 3w on (0, ∞).
w
√
df 2A 2A
We have = − 2 + 3 which is zero when w = .
dw w 3

72. Solution (Continued)
2A
We need to find the minimum value of f(w) = + 3w on (0, ∞).
w
√
df 2A 2A
We have = − 2 + 3 which is zero when w = .
dw w 3
Since f′′ (w) = 4Aw−3 , which is posi ve for all posi ve w, the
cri cal point is a minimum, in fact the global minimum.

73. Solution (Continued)
2A
We need to find the minimum value of f(w) = + 3w on (0, ∞).
w
√
2A
So the area is minimized when w = = 12 and
√ 3
A 3A
ℓ= = = 18.
w 2

74. Solution (Continued)
2A
We need to find the minimum value of f(w) = + 3w on (0, ∞).
w
√
2A
So the area is minimized when w = = 12 and
√ 3
A 3A
ℓ= = = 18.
w 2
The amount of fence needed is
(√ ) √ √
2A 3A 2A √ √
f =2· +3 = 2 6A = 2 6 · 216 = 72m
3 2 3

75. Try this one
Example
An adver sement consists of a rectangular printed region plus 1 in
margins on the sides and 1.5 in margins on the top and bo om. If
the total area of the adver sement is to be 120 in2 , what
dimensions should the adver sement be to maximize the area of
the printed region?

76. Try this one
Example
An adver sement consists of a rectangular printed region plus 1 in
margins on the sides and 1.5 in margins on the top and bo om. If
the total area of the adver sement is to be 120 in2 , what
dimensions should the adver sement be to maximize the area of
the printed region?
Answer
√ √
The op mal paper dimensions are 4 5 in by 6 5 in.

77. Solution of the printed region be x and y,
Let the dimensions
P the printed area, and A the paper area. We wish
to maximize P = xy subject to the constraint that
1.5 cm
A = (x + 2)(y + 3) ≡ 120 Lorem ipsum
dolor sit amet,
consectetur
adipiscing elit.
1 cm
1 cm
120 y Nam dapibus
Isola ng y in A ≡ 120 gives y = − 3 which vehicula mollis.
Proin nec tris que
x+2 mi. Pellentesque
quis placerat
yields . dolor. Praesent
( ) 1.5 cm
120 120x x
P=x −3 = − 3x
x+2 x+2
The domain of P is (0, ∞).

78. Solution (Concluded)
We want to find the absolute maximum value of P.
dP (x + 2)(120) − (120x)(1) 240 − 3(x + 2)2
= −3=
dx (x + 2)2 (x + 2)2
There is a single (posi ve) √ cal point when
cri
(x + 2) = 80 =⇒ x = 4 5 − 2.
2
d2 P −480
The second deriva ve is 2 = , which is nega ve all
dx (x + 2)3
along the domain of P. ( √ )
Hence the unique cri cal point x = 4 5 − 2 cm is the
absolute maximum of P.

79. Solution (Concluded)
( √ )
Hence the unique cri cal point x = 4 5 − 2 cm is the
absolute maximum of P.
√
This means the paper width is 4 5 cm.
120 √
the paper length is √ = 6 5 cm.
4 5

80. Summary
Name [_
Problem Solving Strategy
Draw a Picture
Kathy had a box of 8 crayons.
She gave some crayons away.
Remember the checklist She has 5 left.
How many crayons did Kathy give away?
UNDERSTAND
Ask yourself: what is the
•
What do you want to find out?
Draw a line under the question.
objec ve? You can draw a picture
to solve the problem.
Remember your
geometry: crayons
What number do I
add to 5 to get 8?
8 -
5 + 3 = 8
= 5
similar triangles CHECK
Does your answer make sense?
right triangles Explain.
What number
do I add to 3
Draw a picture to solve the problem.
trigonometric func ons Write how many were given away.
I. I had 10 pencils.
to make 10?
ft ft ft A
I gave some away. 13 ill
i :i
I
'•' I I
I have 3 left. How many i? «
11 I
pencils did I give away? I
H 11
M i l
~7 U U U U> U U