We know the procedure for optimization. When applied to more real-world problems there is a wealth of questions to answer.

1. Section 4.5
Optimization Problems
V63.0121, Calculus I
April 5, 2009
Announcements
Quiz 5 is next week, covering Sections 4.1–4.4
I am moving to WWH 624 sometime next week (April 13th)
Happy Opening Day!
. .
Image credit: wallyg
. . . . . .

2. Office Hours and other help
In addition to recitation
Day Time Who/What Where in WWH
M 1:00–2:00 Leingang OH 718/624
3:30–4:30 Katarina OH 707
5:00–7:00 Curto PS 517
T 1:00–2:00 Leingang OH 718/624
4:00–5:50 Curto PS 317
W 1:00–2:00 Katarina OH 707
2:00–3:00 Leingang OH 718/624
R 9:00–10:00am Leingang OH 718/624
5:00–7:00pm Maria OH 807
F 2:00–4:00 Curto OH 1310
I am moving to WWH 624 sometime next week (April 13th)
. . . . . .

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

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?
Solution
Draw a rectangle.
.
.
. . . . . .

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

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

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

9. 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.
. . . . . .

10. 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
. . . . . .

11. 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
· w = (p − 2w)(w) = pw − w2
A = ℓw =
2 2 2
. . . . . .

12. 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
· w = (p − 2w)(w) = pw − w2
A = ℓw =
2 2 2
Now we have A as a function of w alone (p is constant).
. . . . . .

13. 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
· w = (p − 2w)(w) = pw − w2
A = ℓw =
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).
. . . . . .

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

15. Solution (Concluded)
1
pw − w2 on
We use the Closed Interval Method for A(w) =
2
[0, p/2].
At the endpoints, A(0) = A(p/2) = 0.
dA 1
= p − 2w.
To find the critical points, we find
dr 2
. . . . . .

16. Solution (Concluded)
1
pw − w2 on
We use the Closed Interval Method for A(w) =
2
[0, p/2].
At the endpoints, A(0) = A(p/2) = 0.
dA 1
= p − 2w.
To find the critical points, we find
dr 2
The critical points are when
p
1
p − 2w =⇒ w =
0=
2 4
. . . . . .

17. Solution (Concluded)
1
pw − w2 on
We use the Closed Interval Method for A(w) =
2
[0, p/2].
At the endpoints, A(0) = A(p/2) = 0.
dA 1
= p − 2w.
To find the critical points, we find
dr 2
The critical points are when
p
1
p − 2w =⇒ w =
0=
2 4
Since this is the only critical point, it must be the maximum.
p
In this case ℓ = as well.
4
. . . . . .

18. Solution (Concluded)
1
pw − w2 on
We use the Closed Interval Method for A(w) =
2
[0, p/2].
At the endpoints, A(0) = A(p/2) = 0.
dA 1
= p − 2w.
To find the critical points, we find
dr 2
The critical points are when
p
1
p − 2w =⇒ w =
0=
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.
. . . . . .

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

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

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

22. 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.
. . . . . .

23. 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
. . . . . .

24. 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.
. . . . . .

25. 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.
. . . . . .

26. The Closed Interval Method
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.
. . . . . .

27. The First Derivative Test
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.
. . . . . .

28. 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).
. . . . . .

29. Which to use when?
CIM 1DT 2DT
Pro no need for in- w o r k s on w o r k s on
equalities non-closed, non-closed,
gets global ex- non-bounded non-bounded
trema automati- intervals intervals
cally only one deriva- no need for in-
tive equalities
Con only for closed Uses inequalities More derivatives
bounded inter- More work at less conclusive
vals boundary than than 1DT
CIM more work at
boundary than
CIM
. . . . . .

30. 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.
. . . . . .

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

32. 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?
. . . . . .

33. Solution
1. Everybody understand?
. . . . . .

34. 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?
. . . . . .

35. 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
. . . . . .

36. 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
. . . . . .

37. Solution
1. Everybody understand?
. . . . . .

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

39. Diagram
.
.
.
.
. . . . . .

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

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

42. Diagram
.
.
.
.
. . . . . .

43. Diagram
.
ℓ
w
.
.
.
.
.
. . . . . .

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

45. 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.
. . . . . .

46. 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
. . . . . .

47. 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
dQ p
= p − 4w, which is zero when w = .
6.
dw 4
(p) p2 p2
p
= 80000m2
=p· −2·
Q =
4 4 16 8
Since Q(0) = Q(p/2) = 0, this critical point is a maximum.
. . . . . .

48. 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.
. . . . . .

49. . .
w
.
.
.
ℓ
A = ℓw ≡ 216
f = 2ℓ + 3w
. . . . . .

50. Solution (Continued)
So
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
= 18. The amount of fence needed is
ℓ= =
w 2
(√ ) √ √
√ √
2A 2A 2A
=2· = 2 6A = 2 6 · 216 = 72m
f +3
3 2 3
. . . . . .

51. 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.
.
. . . . . .

52. 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.
. . . . . .

53. Solution
Let h and w be the height and width of the window. We have
π ( w )2
π
L = 2h + w + w A = wh +
2 22
If L is fixed to be 8 + 2π, we have
16 + 4π − 2w − πw
h= ,
4
so
( )
w 1π
π
A = (16 + 4π − 2w − πw) + w2 = (π + 4)w − w2 .
+
4 8 28
( )
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
. . . . . .

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