1. Section 4.5
Optimization Problems
V63.0121.002.2010Su, Calculus I
New York University
June 14, 2010
Announcements
The midterm is graded!
Quiz 4 Thursday on 4.1–4.4
Guest speaker on Thursday: Arjun Krishnan
. . . . . .
2. Announcements
The midterm is graded!
Quiz 4 Thursday on
4.1–4.4
Guest speaker on
Thursday: Arjun Krishnan
. . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 4.5 Optimization Problems June 14, 2010 2 / 31
3. Objectives
Given a problem requiring
optimization, identify the
objective functions,
variables, and constraints.
Solve optimization
problems with calculus.
. . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 4.5 Optimization Problems June 14, 2010 3 / 31
4. Outline
Leading by Example
The Text in the Box
More Examples
. . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 4.5 Optimization Problems June 14, 2010 4 / 31
5. Leading by Example
Example
What is the rectangle of fixed perimeter with maximum area?
. . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 4.5 Optimization Problems June 14, 2010 5 / 31
6. Leading by Example
Example
What is the rectangle of fixed perimeter with maximum area?
Solution
Draw a rectangle.
.
. . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 4.5 Optimization Problems June 14, 2010 5 / 31
7. Leading by Example
Example
What is the rectangle of fixed perimeter with maximum area?
Solution
Draw a rectangle.
.
.
ℓ
. . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 4.5 Optimization Problems June 14, 2010 5 / 31
8. Leading by Example
Example
What is the rectangle of fixed perimeter with maximum area?
Solution
Draw a rectangle.
w
.
.
.
ℓ
. . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 4.5 Optimization Problems June 14, 2010 5 / 31
9. Solution Continued
Let its length be ℓ and its width be w. The objective function is
area A = ℓw.
. . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 4.5 Optimization Problems June 14, 2010 6 / 31
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.
. . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 4.5 Optimization Problems June 14, 2010 6 / 31
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 = ,
2
. . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 4.5 Optimization Problems June 14, 2010 6 / 31
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
A = ℓw = · w = (p − 2w)(w) = pw − w2
2 2 2
. . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 4.5 Optimization Problems June 14, 2010 6 / 31
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
A = ℓw = · w = (p − 2w)(w) = pw − w2
2 2 2
Now we have A as a function of w alone (p is constant).
. . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 4.5 Optimization Problems June 14, 2010 6 / 31
14. 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).
. . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 4.5 Optimization Problems June 14, 2010 6 / 31
15. Solution Concluded
1
We use the Closed Interval Method for A(w) = pw − w2 on [0, p/2].
2
At the endpoints, A(0) = A(p/2) = 0.
. . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 4.5 Optimization Problems June 14, 2010 7 / 31
16. Solution Concluded
1
We use the Closed Interval Method for A(w) = pw − w2 on [0, p/2].
2
At the endpoints, A(0) = A(p/2) = 0.
dA 1
To find the critical points, we find = p − 2w.
dw 2
. . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 4.5 Optimization Problems June 14, 2010 7 / 31
17. Solution Concluded
1
We use the Closed Interval Method for A(w) = pw − w2 on [0, p/2].
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
. . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 4.5 Optimization Problems June 14, 2010 7 / 31
18. Solution Concluded
1
We use the Closed Interval Method for A(w) = pw − w2 on [0, p/2].
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. In this
p
case ℓ = as well.
4
. . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 4.5 Optimization Problems June 14, 2010 7 / 31
19. Solution Concluded
1
We use the Closed Interval Method for A(w) = pw − w2 on [0, p/2].
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. In this
p
case ℓ = as well.
4
We have a square! The maximal area is A(p/4) = p2 /16.
. . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 4.5 Optimization Problems June 14, 2010 7 / 31
20. Outline
Leading by Example
The Text in the Box
More Examples
. . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 4.5 Optimization Problems June 14, 2010 8 / 31
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)
. . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 4.5 Optimization Problems June 14, 2010 9 / 31
22. The Text in the Box
1. Understand the Problem. What is known? What is unknown?
What are the conditions?
. . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 4.5 Optimization Problems June 14, 2010 10 / 31
23. The Text in the Box
1. Understand the Problem. What is known? What is unknown?
What are the conditions?
2. Draw a diagram.
. . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 4.5 Optimization Problems June 14, 2010 10 / 31
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.
. . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 4.5 Optimization Problems June 14, 2010 10 / 31
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
. . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 4.5 Optimization Problems June 14, 2010 10 / 31
26. 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.
. . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 4.5 Optimization Problems June 14, 2010 10 / 31
27. 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.
. . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 4.5 Optimization Problems June 14, 2010 10 / 31
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
pencils did I give away? I
. H 11
. . . . .
M i l
V63.0121.002.2010Su, Calculus I (NYU) ~7 Section 4.5 Optimization Problems
U U U U> U U
June 14, 2010 11 / 31
29. Recall: The Closed Interval Method
See Section 4.1
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/most negative function value are the
global minimum points.
. . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 4.5 Optimization Problems June 14, 2010 12 / 31
30. 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′ changes from negative to positive at c, then c is a local
minimum.
If f′ changes from positive to negative at c, then c is a local
maximum.
If f′ does not change sign at c, then c is not a local extremum.
. . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 4.5 Optimization Problems June 14, 2010 13 / 31
31. 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′ changes from negative to positive at c, then c is a local
minimum.
If f′ changes from positive to negative at c, then c is a local
maximum.
If f′ does not change sign at c, then c is not a local extremum.
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).
. . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 4.5 Optimization Problems June 14, 2010 13 / 31
32. 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 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.
. . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 4.5 Optimization Problems June 14, 2010 14 / 31
33. 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 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 derivative test is inconclusive (this does not
mean c is neither; we just don’t know yet).
. . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 4.5 Optimization Problems June 14, 2010 14 / 31
34. 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 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 derivative test is inconclusive (this does not
mean c is neither; we just don’t know yet).
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
. . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 4.5 Optimization Problems June 14, 2010 14 / 31
35. 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 derivative – no need for
inequalities
Con – only for closed – Uses inequalities – More derivatives
bounded intervals – More work at – less conclusive
boundary than CIM than 1DT
– more work at
boundary than CIM
. . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 4.5 Optimization Problems June 14, 2010 15 / 31
36. 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 derivative – no need for
inequalities
Con – only for closed – Uses inequalities – More derivatives
bounded intervals – More work at – less conclusive
boundary than CIM than 1DT
– more work at
boundary than CIM
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.
. . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 4.5 Optimization Problems June 14, 2010 15 / 31
37. Outline
Leading by Example
The Text in the Box
More Examples
. . . . . .
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38. 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?
. . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 4.5 Optimization Problems June 14, 2010 17 / 31
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
800m of wire at your disposal, what is the largest area you can
enclose, and what are its dimensions?
. . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 4.5 Optimization Problems June 14, 2010 19 / 31
41. 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
. . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 4.5 Optimization Problems June 14, 2010 19 / 31
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
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
Objective: maximize area
Constraint: fixed fence length
. . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 4.5 Optimization Problems June 14, 2010 19 / 31
44. Solution
1. Everybody understand?
2. Draw a diagram.
. . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 4.5 Optimization Problems June 14, 2010 20 / 31
45. 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?
. .
.
.
. . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 4.5 Optimization Problems June 14, 2010 21 / 31
46. Solution
1. Everybody understand?
2. Draw a diagram.
. . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 4.5 Optimization Problems June 14, 2010 22 / 31
47. Solution
1. Everybody understand?
2. Draw a diagram.
3. Length and width are ℓ and w. Length of wire used is p.
. . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 4.5 Optimization Problems June 14, 2010 22 / 31
48. 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?
. .
.
.
. . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 4.5 Optimization Problems June 14, 2010 23 / 31
49. 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
.
. .
.
.
. . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 4.5 Optimization Problems June 14, 2010 23 / 31
50. Solution
1. Everybody understand?
2. Draw a diagram.
3. Length and width are ℓ and w. Length of wire used is p.
. . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 4.5 Optimization Problems June 14, 2010 24 / 31
51. Solution
1. Everybody understand?
2. Draw a diagram.
3. Length and width are ℓ and w. Length of wire used is p.
4. Q = area = ℓw.
. . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 4.5 Optimization Problems June 14, 2010 24 / 31
52. Solution
1. Everybody understand?
2. Draw a diagram.
3. 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]
. . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 4.5 Optimization Problems June 14, 2010 24 / 31
53. Solution
1. Everybody understand?
2. Draw a diagram.
3. 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
. . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 4.5 Optimization Problems June 14, 2010 24 / 31
54. Solution
1. Everybody understand?
2. Draw a diagram.
3. 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
. . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 4.5 Optimization Problems June 14, 2010 24 / 31
55. Solution
1. Everybody understand?
2. Draw a diagram.
3. 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 = . Q(0) = Q(p/2) = 0, but
dw 4
(p) p p2 p2
Q =p· −2· = = 80, 000m2
4 4 16 8
so the critical point is the absolute maximum.
. . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 4.5 Optimization Problems June 14, 2010 24 / 31
56. 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?
. . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 4.5 Optimization Problems June 14, 2010 25 / 31
57. 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.
. . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 4.5 Optimization Problems June 14, 2010 25 / 31
58. Diagram
. .
w
.
.
.
ℓ
f = 2ℓ + 3w A = ℓw ≡ 216
. . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 4.5 Optimization Problems June 14, 2010 26 / 31
59. Solution (Continued)
2A
We need to find the minimum value of f(w) = + 3w on (0, ∞).
w
. . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 4.5 Optimization Problems June 14, 2010 27 / 31
60. Solution (Continued)
2A
We need to find the minimum value of f(w) = + 3w on (0, ∞).
w
We have
df 2A
=− 2 +3
dw w
√
2A
which is zero when w = .
3
. . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 4.5 Optimization Problems June 14, 2010 27 / 31
61. Solution (Continued)
2A
We need to find the minimum value of f(w) = + 3w on (0, ∞).
w
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.
. . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 4.5 Optimization Problems June 14, 2010 27 / 31
62. Solution (Continued)
2A
We need to find the minimum value of f(w) = + 3w on (0, ∞).
w
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
. . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 4.5 Optimization Problems June 14, 2010 27 / 31
63. Try this one
Example
An advertisement consists of a rectangular printed region plus 1 in
margins on the sides and 1.5 in margins on the top and bottom. If the
total area of the advertisement is to be 120 in2 , what dimensions should
the advertisement be to maximize the area of the printed region?
. . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 4.5 Optimization Problems June 14, 2010 28 / 31
64. Try this one
Example
An advertisement consists of a rectangular printed region plus 1 in
margins on the sides and 1.5 in margins on the top and bottom. If the
total area of the advertisement is to be 120 in2 , what dimensions should
the advertisement be to maximize the area of the printed region?
Answer
√ √
The optimal paper dimensions are 4 5 in by 6 5 in.
. . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 4.5 Optimization Problems June 14, 2010 28 / 31
65. Solution
Let the dimensions of the
printed region be x and y, P 1
. .5 cm
the printed area, and A the .
Lorem ipsum dolor sit amet,
consectetur adipiscing elit. Nam
paper area. We wish to dapibus vehicula mollis. Proin nec
tristique mi. Pellentesque quis
maximize P = xy subject to placerat dolor. Praesent a nisl diam.
the constraint that Phasellus ut elit eu ligula accumsan
euismod. Nunc condimentum
lacinia risus a sodales. Morbi nunc
risus, tincidunt in tristique sit amet,
A = (x + 2)(y + 3) ≡ 120
. cm
. cm
y
. ultrices eu eros. Proin pellentesque
aliquam nibh ut lobortis. Ut et
1
1
sollicitudin ipsum. Proin gravida
Isolating y in A ≡ 120 gives ligula eget odio molestie rhoncus
sed nec massa. In ante lorem,
120 imperdiet eget tincidunt at, pharetra
y= − 3 which yields sit amet felis. Nunc nisi velit,
x+2 tempus ac suscipit quis, blandit
vitae mauris. Vestibulum ante ipsum
( ) primis in faucibus orci luctus et
120 120x . ultrices posuere cubilia Curae;
P=x −3 = −3x
x+2 x+2 1
. .5 cm
The domain of P is (0, ∞) x
.
. . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 4.5 Optimization Problems June 14, 2010 29 / 31
66. Solution (Concluded)
We want to find the absolute maximum value of P. Taking derivatives,
dP (x + 2)(120) − (120x)(1) 240 − 3(x + 2)2
= −3=
dx (x + 2)2 (x + 2)2
There is a single critical point when
√
(x + 2)2 = 80 =⇒ x = 4 5 − 2
(the negative critical point doesn’t count). The second derivative is
d2 P −480
2
=
dx (x + 2)3
which is negative all along the domain of P. Hence the unique critical
( √ )
point x = 4 5 − 2 cm is the absolute maximum of P. This means
√ 120 √
the paper width is 4 5 cm, and the paper length is √ = 6 5 cm.
4 5 . . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 4.5 Optimization Problems June 14, 2010 30 / 31
67. Summary
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?
Remember the checklist UNDERSTAND
What do you want to find out?
•
Draw a line under the question.
Ask yourself: what is the
objective? You can draw a picture
to solve the problem.
Remember your geometry:
What number do I
add to 5 to get 8?
8 - = 5
similar triangles crayons 5 + 3 = 8
right triangles CHECK
Does your answer make sense?
trigonometric functions 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
pencils did I give away? I
H 11
M i l
~7 U U U U> U U
. . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 4.5 Optimization Problems June 14, 2010 31 / 31
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