The document contains announcements regarding homework deadlines, a quiz schedule, and student feedback indicating the need for more examples in the Calculus I course. It includes an example problem of finding the rectangle of maximum area with a fixed perimeter, outlining the steps to solve optimization problems using objective functions and graphical representations. Additionally, it discusses methods for determining maximum or minimum values through closed interval and derivative tests.

1. Section 4.5
Optimization Problems
V63.0121.027, Calculus I
November 24, 2009
Announcements
Written HW (inc. WebAssign from 4.4) due Wednesday,
Nov. 25, 11:00am in my mailbox
Quiz 5 next week on §§4.1–4.4, 4.7
. . . . . .

2. Thank you for the evaluations
Comments and requests
Too fast, not enough examples
. . . . . .

3. Thank you for the evaluations
Comments and requests
Too fast, not enough examples
Not enough time to do everything
Class is not the only learning time (recitation and
independent study)
I try to balance
. . . . . .

4. Thank you for the evaluations
Comments and requests
Too fast, not enough examples
Not enough time to do everything
Class is not the only learning time (recitation and
independent study)
I try to balance
Too many proofs
. . . . . .

5. Thank you for the evaluations
Comments and requests
Too fast, not enough examples
Not enough time to do everything
Class is not the only learning time (recitation and
independent study)
I try to balance
Too many proofs
In this course we care about concepts
There will be conceptual problems on the exam
Concepts are the keys to overcoming templated problems
. . . . . .

6. A slide on slides
Pro
“Good use of powerpoint”
“Use of slides removes uncertainty that could arise from the
chalkboard”
“helpful when they were posted online and I could refer back
to them”
. . . . . .

7. A slide on slides
Pro
“Good use of powerpoint”
“Use of slides removes uncertainty that could arise from the
chalkboard”
“helpful when they were posted online and I could refer back
to them”
Con
“I wish he would use the chalkboard more”
“Powerpoint slides are not a good way to teach math”
“I hate powerpoint”
. . . . . .

8. A slide on slides
Pro
“Good use of powerpoint”
“Use of slides removes uncertainty that could arise from the
chalkboard”
“helpful when they were posted online and I could refer back
to them”
Con
“I wish he would use the chalkboard more”
“Powerpoint slides are not a good way to teach math”
“I hate powerpoint”
Why I like them
Board handwriting not an issue
Easy to put online; notetaking is more than transcription
Improvable—if you have suggestions I’m listening
. . . . . .

9. A slide on WebAssign
If you find a mistake, please let me know.
We are dropping 5 lowest assignments (roughly two weeks’
worth)
. . . . . .

10. Also requested, more:
Tree stretches
Music
Dancing
I’ll see what I can do!
. . . . . .

11. Outline
Leading by Example
The Text in the Box
More Examples
. . . . . .

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

13. Leading by Example
Example
What is the rectangle of fixed perimeter with maximum area?
Solution
Draw a rectangle.
.
.
. . . . . .

14. Leading by Example
Example
What is the rectangle of fixed perimeter with maximum area?
Solution
Draw a rectangle.
.
.
.
ℓ
. . . . . .

15. Leading by Example
Example
What is the rectangle of fixed perimeter with maximum area?
Solution
Draw a rectangle.
. w
.
.
.
ℓ
. . . . . .

16. Solution (Continued)
Let its length be ℓ and its width be w. The objective function
is area A = ℓw.
. . . . . .

17. Solution (Continued)
Let its length be ℓ and its width be w. The objective function
is area A = ℓw.
This is a function of two variables, not one. But the perimeter
is fixed.
. . . . . .

18. Solution (Continued)
Let its length be ℓ and its width be w. The objective function
is area A = ℓw.
This is a function of two variables, not one. But the perimeter
is fixed.
p − 2w
Since p = 2ℓ + 2w, we have ℓ = ,
2
. . . . . .

19. Solution (Continued)
Let its length be ℓ and its width be w. The objective function
is area A = ℓw.
This is a function 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
. . . . . .

20. Solution (Continued)
Let its length be ℓ and its width be w. The objective function
is area A = ℓw.
This is a function 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
Now we have A as a function of w alone (p is constant).
. . . . . .

21. Solution (Continued)
Let its length be ℓ and its width be w. The objective function
is area A = ℓw.
This is a function 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
Now we have A as a function of w alone (p is constant).
The natural domain of this function is [0, p/2] (we want to
make sure A(w) ≥ 0).
. . . . . .

22. Solution (Concluded)
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.
. . . . . .

23. Solution (Concluded)
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 critical points, we find = p − 2w.
dw 2
. . . . . .

24. Solution (Concluded)
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 critical points, we find = p − 2w.
dw 2
The critical points are when
1 p
0= p − 2w =⇒ w =
2 4
. . . . . .

25. Solution (Concluded)
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 critical points, we find = p − 2w.
dw 2
The critical points are when
1 p
0= p − 2w =⇒ w =
2 4
Since this is the only critical point, it must be the maximum.
p
In this case ℓ = as well.
4
. . . . . .

26. Solution (Concluded)
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 critical points, we find = p − 2w.
dw 2
The critical points are when
1 p
0= p − 2w =⇒ w =
2 4
Since this is the only critical point, it must be the maximum.
p
In this case ℓ = as well.
4
We have a square! The maximal area is A(p/4) = p2 /16.
. . . . . .

27. Outline
Leading by Example
The Text in the Box
More Examples
. . . . . .

28. The Text in the Box
1. Understand the Problem. What is known? What is
unknown? What are the conditions?
. . . . . .

29. The Text in the Box
1. Understand the Problem. What is known? What is
unknown? What are the conditions?
2. Draw a diagram.
. . . . . .

30. The Text in the Box
1. Understand the Problem. What is known? What is
unknown? What are the conditions?
2. Draw a diagram.
3. Introduce Notation.
. . . . . .

31. The Text in the Box
1. Understand the Problem. What is known? What is
unknown? What are the conditions?
2. Draw a diagram.
3. Introduce Notation.
4. Express the “objective function” Q in terms of the other
symbols
. . . . . .

32. The Text in the Box
1. Understand the Problem. What is known? What is
unknown? What are the conditions?
2. Draw a diagram.
3. Introduce Notation.
4. Express the “objective function” Q in terms of the other
symbols
5. If Q is a function of more than one “decision variable”, use
the given information to eliminate all but one of them.
. . . . . .

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

34. Recall: The Closed Interval Method
See Section 4.1
To find the extreme values of a function f on [a, b], we need to:
Evaluate f at the endpoints a and b
Evaluate f at the critical points x where either f′ (x) = 0 or f is
not differentiable at x.
The points with the largest function value are the global
maximum points
The points with the smallest or most negative function value
are the global minimum points.
. . . . . .

35. Recall: The First Derivative Test
See Section 4.3
Theorem (The First Derivative Test)
Let f be continuous on [a, b] and c a critical point of f in (a, b).
If f′ (x) > 0 on (a, c) and f′ (x) < 0 on (c, b), then c is a local
maximum.
If f′ (x) < 0 on (a, c) and f′ (x) > 0 on (c, b), then c is a local
minimum.
If f′ (x) has the same sign on (a, c) and (c, b), then c is not a
local extremum.
. . . . . .

36. Recall: The Second Derivative Test
See Section 4.3
Theorem (The Second Derivative Test)
Let f, f′ , and f′′ be continuous on [a, b]. Let c be be a point 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.
If f′′ (c) = 0, the second derivative test is inconclusive (this does
not mean c is neither; we just don’t know yet).
. . . . . .

37. Which to use when?
CIM 1DT 2DT
Pro – no need for – works on – works on
inequalities non-closed, non-closed,
– gets global non-bounded non-bounded
extrema intervals intervals
automatically – only one – no need for
derivative inequalities
Con – only for closed – Uses – More derivatives
bounded intervals inequalities – less conclusive
– More work at than 1DT
boundary than – more work at
CIM boundary than
CIM
. . . . . .

38. Which to use when? The bottom line
Use CIM if it applies: the domain is a closed, bounded
interval
If domain is not closed or not bounded, use 2DT if you like
to take derivatives, or 1DT if you like to compare signs.
. . . . . .

39. Outline
Leading by Example
The Text in the Box
More Examples
. . . . . .

40. 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 800 m of wire at your disposal, what is the largest
area you can enclose, and what are its dimensions?
. . . . . .

41. Solution
1. Everybody understand?
. . . . . .

42. 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 800 m of wire at your disposal, what is the largest
area you can enclose, and what are its dimensions?
. . . . . .

43. 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 800 m 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
. . . . . .

44. 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 800 m 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
Objective: maximize area
Constraint: fixed fence length
. . . . . .

45. Solution
1. Everybody understand?
. . . . . .

46. Solution
1. Everybody understand?
2. Draw a diagram.
. . . . . .

47. 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?
. .
.
.
. . . . . .

48. Solution
1. Everybody understand?
2. Draw a diagram.
. . . . . .

49. Solution
1. Everybody understand?
2. Draw a diagram.
3. Introduce notation: Length and width are ℓ and w. Length of
wire used is p.
. . . . . .

50. 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?
. .
.
.
. . . . . .

51. 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
.
. .
.
.
. . . . . .

52. Solution
1. Everybody understand?
2. Draw a diagram.
3. Introduce notation: Length and width are ℓ and w. Length of
wire used is p.
. . . . . .

53. Solution
1. Everybody understand?
2. Draw a diagram.
3. Introduce notation: Length and width are ℓ and w. Length of
wire used is p.
4. Q = area = ℓw.
. . . . . .

54. Solution
1. Everybody understand?
2. Draw a diagram.
3. Introduce notation: Length and width are ℓ and w. Length of
wire used is p.
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]
. . . . . .

55. Solution
1. Everybody understand?
2. Draw a diagram.
3. Introduce notation: Length and width are ℓ and w. Length of
wire used is p.
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
. . . . . .

56. Solution
1. Everybody understand?
2. Draw a diagram.
3. Introduce notation: Length and width are ℓ and w. Length of
wire used is p.
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
. . . . . .

57. Solution
1. Everybody understand?
2. Draw a diagram.
3. Introduce notation: Length and width are ℓ and w. Length of
wire used is p.
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
Q(0) = Q(p/2) = 0, but
(p) p p2 p2
Q =p· −2· = = 80, 000m2
4 4 16 8
so the critical point is the absolute maximum.
. . . . . .

58. 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?
. . . . . .

59. 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?
Solution
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.
w
The domain is all positive numbers.
. . . . . .

60. . .
w
.
.
.
ℓ
f = 2ℓ + 3w A = ℓw ≡ 216
. . . . . .

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

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

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

64. Solution (Continued)
2A
We need to find the minimum value of f(w) = + 3w on
w
(0, ∞).
We have
df 2A
=− 2 +3
dw w
√
2A
which is zero when w = .
3
Since f′′ (w) = 4Aw−3 , which is positive for all positive w, the
critical point is a minimum, in fact the global minimum.
√
2A
So the area is minimized when w = = 12 and
√ 3
A 3A
ℓ= = = 18. The amount of fence needed is
w 2
(√ ) √ √
2A 3A 2A √ √
f = 2· +3 = 2 6A = 2 6 · 216 = 72m
3 2 3
. . . . . .

65. Example
A Norman window has the outline of a semicircle on top of a
rectangle. Suppose there is 8 + 2π feet of wood trim available.
Find the dimensions of the rectangle and semicircle that will
maximize the area of the window.
.
. . . . . .

66. Example
A Norman window has the outline of a semicircle on top of a
rectangle. Suppose there is 8 + 2π feet of wood trim available.
Find the dimensions of the rectangle and semicircle that will
maximize the area of the window.
.
Answer
The dimensions are 4ft by 2ft.
. . . . . .

67. Solution
Let h and w be the height and width of the window. We have
π π ( w )2
L = 2h + w + w A = wh +
2 2 2
If L is fixed to be 8 + 2π , we have
16 + 4π − 2w − π w
h= ,
4
so
( )
w π 1 π
A = (16 + 4π − 2w − π w) + w2 = (π + 4)w − + w2 .
4 8 2 8
( π )
So A′ = (π + 4)w − 1 + , which is zero when
4
π+4
w= = 4 ft. The dimensions are 4ft by 2ft.
1+ π2
. . . . . .

68. Summary
Remember the checklist
Ask yourself: what is the objective?
Remember your geometry:
similar triangles
right triangles
trigonometric functions
. . . . . .