Uncountably many problems in life and nature can be expressed in terms of an optimization principle. We look at the process and find a few good examples.
* Find zeros of polynomial functions
* Use the Fundamental Theorem of Algebra to find a function that satisfies given conditions
* Find all zeros of a polynomial function
* Find zeros of polynomial functions
* Use the Fundamental Theorem of Algebra to find a function that satisfies given conditions
* Find all zeros of a polynomial function
MATH Lesson Plan sample for demo teaching preyaleandrina
This is my first made lesson plan ...
i thought before that its hard to make lesson plan but being just resourceful and with the help of different methods and strategies in teaching we can have our guide for highly and better teaching instruction:)..
Detailed Lesson Plan (ENGLISH, MATH, SCIENCE, FILIPINO)Junnie Salud
Thanks everybody! The lesson plans presented were actually outdated and can still be improved. I was also a college student when I did these. There were minor errors but the important thing is, the structure and flow of activities (for an hour-long class) are included here. I appreciate all of your comments! Please like my fan page on facebook search for JUNNIE SALUD.
*The detailed LP for English is from Ms. Juliana Patricia Tenzasas. I just revised it a little.
For questions about education-related matters, you can directly email me at mr_junniesalud@yahoo.com
Uncountably many problems in life and nature can be expressed in terms of an optimization principle. We look at the process and find a few good examples.
Uncountably many problems in life and nature can be expressed in terms of an optimization principle. We look at the process and find a few good examples.
Uncountably many problems in life and nature can be expressed in terms of an optimization principle. We look at the process and find a few good examples.
Uncountably many problems in life and nature can be expressed in terms of an optimization principle. We look at the process and find a few good examples.
Uncountably many problems in life and nature can be expressed in terms of an optimization principle. We look at the process and find a few more good examples.
Lesson 22: Optimization II (Section 041 handout)Matthew Leingang
Uncountably many problems in life and nature can be expressed in terms of an optimization principle. We look at the process and find a few more good examples.
Lesson 19: The Mean Value Theorem (Section 021 handout)Matthew Leingang
The Mean Value Theorem is the most important theorem in calculus. It is the first theorem which allows us to infer information about a function from information about its derivative. From the MVT we can derive tests for the monotonicity (increase or decrease) and concavity of a function.
Uncountably many problems in life and nature can be expressed in terms of an optimization principle. We look at the process and find a few more good examples.
Lesson 18: Maximum and Minimum Values (Section 021 handout)Matthew Leingang
There are various reasons why we would want to find the extreme (maximum and minimum values) of a function. Fermat's Theorem tells us we can find local extreme points by looking at critical points. This process is known as the Closed Interval Method.
Lesson 19: The Mean Value Theorem (Section 041 handout)Matthew Leingang
The Mean Value Theorem is the most important theorem in calculus. It is the first theorem which allows us to infer information about a function from information about its derivative. From the MVT we can derive tests for the monotonicity (increase or decrease) and concavity of a function.
Lesson 24: Areas, Distances, the Integral (Section 041 slides)Matthew Leingang
We can define the area of a curved region by a process similar to that by which we determined the slope of a curve: approximation by what we know and a limit.
Lesson 18: Maximum and Minimum Values (Section 021 slides)Matthew Leingang
There are various reasons why we would want to find the extreme (maximum and minimum values) of a function. Fermat's Theorem tells us we can find local extreme points by looking at critical points. This process is known as the Closed Interval Method.
Lesson 18: Maximum and Minimum Values (Section 041 slides)Matthew Leingang
There are various reasons why we would want to find the extreme (maximum and minimum values) of a function. Fermat's Theorem tells us we can find local extreme points by looking at critical points. This process is known as the Closed Interval Method.
Lesson 19: The Mean Value Theorem (Section 021 slides)Matthew Leingang
The Mean Value Theorem is the most important theorem in calculus. It is the first theorem which allows us to infer information about a function from information about its derivative. From the MVT we can derive tests for the monotonicity (increase or decrease) and concavity of a function.
Similar to Lesson 22: Optimization (Section 041 slides) (20)
Streamlining assessment, feedback, and archival with auto-multiple-choiceMatthew Leingang
Auto-multiple-choice (AMC) is an open-source optical mark recognition software package built with Perl, LaTeX, XML, and sqlite. I use it for all my in-class quizzes and exams. Unique papers are created for each student, fixed-response items are scored automatically, and free-response problems, after manual scoring, have marks recorded in the same process. In the first part of the talk I will discuss AMC’s many features and why I feel it’s ideal for a mathematics course. My contributions to the AMC workflow include some scripts designed to automate the process of returning scored papers
back to students electronically. AMC provides an email gateway, but I have written programs to return graded papers via the DAV protocol to student’s dropboxes on our (Sakai) learning management systems. I will also show how graded papers can be archived, with appropriate metadata tags, into an Evernote notebook.
Integration by substitution is the chain rule in reverse.
NOTE: the final location is section specific. Section 1 (morning) is in SILV 703, Section 11 (afternoon) is in CANT 200
Lesson 24: Areas and Distances, The Definite Integral (handout)Matthew Leingang
We can define the area of a curved region by a process similar to that by which we determined the slope of a curve: approximation by what we know and a limit.
Lesson 24: Areas and Distances, The Definite Integral (slides)Matthew Leingang
We can define the area of a curved region by a process similar to that by which we determined the slope of a curve: approximation by what we know and a limit.
At times it is useful to consider a function whose derivative is a given function. We look at the general idea of reversing the differentiation process and its applications to rectilinear motion.
At times it is useful to consider a function whose derivative is a given function. We look at the general idea of reversing the differentiation process and its applications to rectilinear motion.
Uncountably many problems in life and nature can be expressed in terms of an optimization principle. We look at the process and find a few good examples.
Uncountably many problems in life and nature can be expressed in terms of an optimization principle. We look at the process and find a few good examples.
Lesson 20: Derivatives and the Shapes of Curves (handout)
Lesson 22: Optimization (Section 041 slides)
1. Section 4.4
Optimization Problems
V63.0121.041, Calculus I
New York University
November 22, 2010
Announcements
There is class on Wednesday, November 24
Turn in HW anytime between now and November 24, 2pm
No recitation this week
Quiz 4 on §§4.1–4.4 next week in recitation
. . . . . .
2. . . . . . .
Announcements
There is class on
Wednesday, November 24
Turn in HW anytime
between now and
November 24, 2pm
No recitation this week
Quiz 4 on §§4.1–4.4 next
week in recitation
V63.0121.041, Calculus I (NYU) Section 4.4 Optimization Problems November 22, 2010 2 / 31
3. . . . . . .
Objectives
Given a problem requiring
optimization, identify the
objective functions,
variables, and constraints.
Solve optimization
problems with calculus.
V63.0121.041, Calculus I (NYU) Section 4.4 Optimization Problems November 22, 2010 3 / 31
4. . . . . . .
Outline
Leading by Example
The Text in the Box
More Examples
V63.0121.041, Calculus I (NYU) Section 4.4 Optimization Problems November 22, 2010 4 / 31
5. . . . . . .
Leading by Example
Example
What is the rectangle of fixed perimeter with maximum area?
V63.0121.041, Calculus I (NYU) Section 4.4 Optimization Problems November 22, 2010 5 / 31
6. . . . . . .
Leading by Example
Example
What is the rectangle of fixed perimeter with maximum area?
Solution
Draw a rectangle.
.
V63.0121.041, Calculus I (NYU) Section 4.4 Optimization Problems November 22, 2010 5 / 31
7. . . . . . .
Leading by Example
Example
What is the rectangle of fixed perimeter with maximum area?
Solution
Draw a rectangle.
..
ℓ
V63.0121.041, Calculus I (NYU) Section 4.4 Optimization Problems November 22, 2010 5 / 31
8. . . . . . .
Leading by Example
Example
What is the rectangle of fixed perimeter with maximum area?
Solution
Draw a rectangle.
..
ℓ
.
w
V63.0121.041, Calculus I (NYU) Section 4.4 Optimization Problems November 22, 2010 5 / 31
9. . . . . . .
Solution Continued
Let its length be ℓ and its width be w. The objective function is
area A = ℓw.
V63.0121.041, Calculus I (NYU) Section 4.4 Optimization Problems November 22, 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.
V63.0121.041, Calculus I (NYU) Section 4.4 Optimization Problems November 22, 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.
Since p = 2ℓ + 2w, we have ℓ =
p − 2w
2
,
V63.0121.041, Calculus I (NYU) Section 4.4 Optimization Problems November 22, 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.
Since p = 2ℓ + 2w, we have ℓ =
p − 2w
2
, so
A = ℓw =
p − 2w
2
· w =
1
2
(p − 2w)(w) =
1
2
pw − w2
V63.0121.041, Calculus I (NYU) Section 4.4 Optimization Problems November 22, 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.
Since p = 2ℓ + 2w, we have ℓ =
p − 2w
2
, so
A = ℓw =
p − 2w
2
· w =
1
2
(p − 2w)(w) =
1
2
pw − w2
Now we have A as a function of w alone (p is constant).
V63.0121.041, Calculus I (NYU) Section 4.4 Optimization Problems November 22, 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.
Since p = 2ℓ + 2w, we have ℓ =
p − 2w
2
, so
A = ℓw =
p − 2w
2
· w =
1
2
(p − 2w)(w) =
1
2
pw − w2
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).
V63.0121.041, Calculus I (NYU) Section 4.4 Optimization Problems November 22, 2010 6 / 31
15. . . . . . .
Solution Concluded
We use the Closed Interval Method for A(w) =
1
2
pw − w2
on [0, p/2].
At the endpoints, A(0) = A(p/2) = 0.
V63.0121.041, Calculus I (NYU) Section 4.4 Optimization Problems November 22, 2010 7 / 31
16. . . . . . .
Solution Concluded
We use the Closed Interval Method for A(w) =
1
2
pw − w2
on [0, p/2].
At the endpoints, A(0) = A(p/2) = 0.
To find the critical points, we find
dA
dw
=
1
2
p − 2w.
V63.0121.041, Calculus I (NYU) Section 4.4 Optimization Problems November 22, 2010 7 / 31
17. . . . . . .
Solution Concluded
We use the Closed Interval Method for A(w) =
1
2
pw − w2
on [0, p/2].
At the endpoints, A(0) = A(p/2) = 0.
To find the critical points, we find
dA
dw
=
1
2
p − 2w.
The critical points are when
0 =
1
2
p − 2w =⇒ w =
p
4
V63.0121.041, Calculus I (NYU) Section 4.4 Optimization Problems November 22, 2010 7 / 31
18. . . . . . .
Solution Concluded
We use the Closed Interval Method for A(w) =
1
2
pw − w2
on [0, p/2].
At the endpoints, A(0) = A(p/2) = 0.
To find the critical points, we find
dA
dw
=
1
2
p − 2w.
The critical points are when
0 =
1
2
p − 2w =⇒ w =
p
4
Since this is the only critical point, it must be the maximum. In this
case ℓ =
p
4
as well.
V63.0121.041, Calculus I (NYU) Section 4.4 Optimization Problems November 22, 2010 7 / 31
19. . . . . . .
Solution Concluded
We use the Closed Interval Method for A(w) =
1
2
pw − w2
on [0, p/2].
At the endpoints, A(0) = A(p/2) = 0.
To find the critical points, we find
dA
dw
=
1
2
p − 2w.
The critical points are when
0 =
1
2
p − 2w =⇒ w =
p
4
Since this is the only critical point, it must be the maximum. In this
case ℓ =
p
4
as well.
We have a square! The maximal area is A(p/4) = p2
/16.
V63.0121.041, Calculus I (NYU) Section 4.4 Optimization Problems November 22, 2010 7 / 31
20. . . . . . .
Outline
Leading by Example
The Text in the Box
More Examples
V63.0121.041, Calculus I (NYU) Section 4.4 Optimization Problems November 22, 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)
V63.0121.041, Calculus I (NYU) Section 4.4 Optimization Problems November 22, 2010 9 / 31
22. . . . . . .
The Text in the Box
1. Understand the Problem. What is known? What is unknown?
What are the conditions?
V63.0121.041, Calculus I (NYU) Section 4.4 Optimization Problems November 22, 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.
V63.0121.041, Calculus I (NYU) Section 4.4 Optimization Problems November 22, 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.
V63.0121.041, Calculus I (NYU) Section 4.4 Optimization Problems November 22, 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
V63.0121.041, Calculus I (NYU) Section 4.4 Optimization Problems November 22, 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.
V63.0121.041, Calculus I (NYU) Section 4.4 Optimization Problems November 22, 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.
V63.0121.041, Calculus I (NYU) Section 4.4 Optimization Problems November 22, 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.
crayons
What number do I
add to 5 to get 8?
8 - = 5
5 + 3 = 8
CHECK
Does your answer make sense?
Explain.
Draw a picture to solve the problem.
Write how many were given away.
I. I had10 pencils.
I gave some away.
I have 3left. How many
pencils did I give away?
~7
What number
do I add to 3
to make 10?
13
i
ft
ill
:i
i ?
11
ft
I
'•'
«
I
I
ft A
H 11
M i l
U U U U> U U
I I
V63.0121.041, Calculus I (NYU) Section 4.4 Optimization Problems November 22, 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.
V63.0121.041, Calculus I (NYU) Section 4.4 Optimization Problems November 22, 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.
V63.0121.041, Calculus I (NYU) Section 4.4 Optimization Problems November 22, 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).
V63.0121.041, Calculus I (NYU) Section 4.4 Optimization Problems November 22, 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.
V63.0121.041, Calculus I (NYU) Section 4.4 Optimization Problems November 22, 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).
V63.0121.041, Calculus I (NYU) Section 4.4 Optimization Problems November 22, 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
V63.0121.041, Calculus I (NYU) Section 4.4 Optimization Problems November 22, 2010 14 / 31
35. . . . . . .
Which to use when?
CIM 1DT 2DT
Pro – no need for
inequalities
– gets global
extrema
automatically
– works on
non-closed,
non-bounded
intervals
– only one derivative
– works on
non-closed,
non-bounded
intervals
– no need for
inequalities
Con – only for closed
bounded intervals
– Uses inequalities
– More work at
boundary than CIM
– More derivatives
– less conclusive
than 1DT
– more work at
boundary than CIM
V63.0121.041, Calculus I (NYU) Section 4.4 Optimization Problems November 22, 2010 15 / 31
36. . . . . . .
Which to use when?
CIM 1DT 2DT
Pro – no need for
inequalities
– gets global
extrema
automatically
– works on
non-closed,
non-bounded
intervals
– only one derivative
– works on
non-closed,
non-bounded
intervals
– no need for
inequalities
Con – only for closed
bounded intervals
– Uses inequalities
– More work at
boundary than CIM
– More derivatives
– less conclusive
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.
V63.0121.041, Calculus I (NYU) Section 4.4 Optimization Problems November 22, 2010 15 / 31
37. . . . . . .
Outline
Leading by Example
The Text in the Box
More Examples
V63.0121.041, Calculus I (NYU) Section 4.4 Optimization Problems November 22, 2010 16 / 31
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?
V63.0121.041, Calculus I (NYU) Section 4.4 Optimization Problems November 22, 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?
V63.0121.041, Calculus I (NYU) Section 4.4 Optimization Problems November 22, 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
V63.0121.041, Calculus I (NYU) Section 4.4 Optimization Problems November 22, 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
V63.0121.041, Calculus I (NYU) Section 4.4 Optimization Problems November 22, 2010 19 / 31
44. . . . . . .
Solution
1. Everybody understand?
2. Draw a diagram.
V63.0121.041, Calculus I (NYU) Section 4.4 Optimization Problems November 22, 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?
.
.
..
V63.0121.041, Calculus I (NYU) Section 4.4 Optimization Problems November 22, 2010 21 / 31
46. . . . . . .
Solution
1. Everybody understand?
2. Draw a diagram.
V63.0121.041, Calculus I (NYU) Section 4.4 Optimization Problems November 22, 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.
V63.0121.041, Calculus I (NYU) Section 4.4 Optimization Problems November 22, 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?
.
.
..
V63.0121.041, Calculus I (NYU) Section 4.4 Optimization Problems November 22, 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?
.
.
...
w
.
ℓ
V63.0121.041, Calculus I (NYU) Section 4.4 Optimization Problems November 22, 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.
V63.0121.041, Calculus I (NYU) Section 4.4 Optimization Problems November 22, 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.
V63.0121.041, Calculus I (NYU) Section 4.4 Optimization Problems November 22, 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]
V63.0121.041, Calculus I (NYU) Section 4.4 Optimization Problems November 22, 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]
6.
dQ
dw
= p − 4w, which is zero when w =
p
4
.
V63.0121.041, Calculus I (NYU) Section 4.4 Optimization Problems November 22, 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]
6.
dQ
dw
= p − 4w, which is zero when w =
p
4
. Q(0) = Q(p/2) = 0, but
Q
(p
4
)
= p ·
p
4
− 2 ·
p2
16
=
p2
8
= 80, 000m2
so the critical point is the absolute maximum.
V63.0121.041, Calculus I (NYU) Section 4.4 Optimization Problems November 22, 2010 24 / 31
55. . . . . . .
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?
V63.0121.041, Calculus I (NYU) Section 4.4 Optimization Problems November 22, 2010 25 / 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?
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
f(w) = 2
A
w
+ 3w.
The domain is all positive numbers.
V63.0121.041, Calculus I (NYU) Section 4.4 Optimization Problems November 22, 2010 25 / 31
57. . . . . . .
Diagram
....
ℓ
.
w
f = 2ℓ + 3w A = ℓw ≡ 216
V63.0121.041, Calculus I (NYU) Section 4.4 Optimization Problems November 22, 2010 26 / 31
58. . . . . . .
Solution (Continued)
We need to find the minimum value of f(w) =
2A
w
+ 3w on (0, ∞).
V63.0121.041, Calculus I (NYU) Section 4.4 Optimization Problems November 22, 2010 27 / 31
59. . . . . . .
Solution (Continued)
We need to find the minimum value of f(w) =
2A
w
+ 3w on (0, ∞).
We have
df
dw
= −
2A
w2
+ 3
which is zero when w =
√
2A
3
.
V63.0121.041, Calculus I (NYU) Section 4.4 Optimization Problems November 22, 2010 27 / 31
60. . . . . . .
Solution (Continued)
We need to find the minimum value of f(w) =
2A
w
+ 3w on (0, ∞).
We have
df
dw
= −
2A
w2
+ 3
which is zero when w =
√
2A
3
.
Since f′′
(w) = 4Aw−3
, which is positive for all positive w, the
critical point is a minimum, in fact the global minimum.
V63.0121.041, Calculus I (NYU) Section 4.4 Optimization Problems November 22, 2010 27 / 31
61. . . . . . .
Solution (Continued)
We need to find the minimum value of f(w) =
2A
w
+ 3w on (0, ∞).
We have
df
dw
= −
2A
w2
+ 3
which is zero when w =
√
2A
3
.
Since f′′
(w) = 4Aw−3
, which is positive for all positive w, the
critical point is a minimum, in fact the global minimum.
So the area is minimized when w =
√
2A
3
= 12 and
ℓ =
A
w
=
√
3A
2
= 18. The amount of fence needed is
f
(√
2A
3
)
= 2 ·
√
3A
2
+ 3
√
2A
3
= 2
√
6A = 2
√
6 · 216 = 72m
V63.0121.041, Calculus I (NYU) Section 4.4 Optimization Problems November 22, 2010 27 / 31
62. . . . . . .
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?
V63.0121.041, Calculus I (NYU) Section 4.4 Optimization Problems November 22, 2010 28 / 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?
Answer
The optimal paper dimensions are 4
√
5 in by 6
√
5 in.
V63.0121.041, Calculus I (NYU) Section 4.4 Optimization Problems November 22, 2010 28 / 31
64. . . . . . .
Solution
Let the dimensions of the
printed region be x and y, P
the printed area, and A the
paper area. We wish to
maximize P = xy subject to
the constraint that
A = (x + 2)(y + 3) ≡ 120
Isolating y in A ≡ 120 gives
y =
120
x + 2
− 3 which yields
P = x
(
120
x + 2
− 3
)
=
120x
x + 2
−3x
The domain of P is (0, ∞).
..
Lorem ipsum dolor sit amet,
consectetur adipiscing elit. Nam
dapibus vehicula mollis. Proin nec
tristique mi. Pellentesque quis
placerat dolor. Praesent a nisl diam.
Phasellus ut elit eu ligula accumsan
euismod. Nunc condimentum
lacinia risus a sodales. Morbi nunc
risus, tincidunt in tristique sit amet,
ultrices eu eros. Proin pellentesque
aliquam nibh ut lobortis. Ut et
sollicitudin ipsum. Proin gravida
ligula eget odio molestie rhoncus
sed nec massa. In ante lorem,
imperdiet eget tincidunt at, pharetra
sit amet felis. Nunc nisi velit,
tempus ac suscipit quis, blandit
vitae mauris. Vestibulum ante ipsum
primis in faucibus orci luctus et
ultrices posuere cubilia Curae;
.
1.5 cm
.
1.5 cm
.1cm.
1cm
.
x
.
y
V63.0121.041, Calculus I (NYU) Section 4.4 Optimization Problems November 22, 2010 29 / 31
65. . . . . . .
Solution (Concluded)
We want to find the absolute maximum value of P.
dP
dx
=
(x + 2)(120) − (120x)(1)
(x + 2)2
− 3 =
240 − 3(x + 2)2
(x + 2)2
There is a single (positive) critical point when
(x + 2)2
= 80 =⇒ x = 4
√
5 − 2.
The second derivative is
d2
P
dx2
=
−480
(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 the paper width is 4
√
5 cm, and the paper length is
120
4
√
5
= 6
√
5 cm.
V63.0121.041, Calculus I (NYU) Section 4.4 Optimization Problems November 22, 2010 30 / 31
66. . . . . . .
Summary
Remember the checklist
Ask yourself: what is the
objective?
Remember your geometry:
similar triangles
right triangles
trigonometric functions
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.
crayons
What number do I
add to 5 to get 8?
8 - = 5
5 + 3 = 8
CHECK
Does your answer make sense?
Explain.
Draw a picture to solve the problem.
Write how many were given away.
I. I had10 pencils.
I gave some away.
I have 3left. How many
pencils did I give away?
~7
What number
do I add to 3
to make 10?
13
i
ft
ill
:i
i ?
11
ft
I
'•'
«
I
I
ft A
H 11
M i l
U U U U> U U
I I
V63.0121.041, Calculus I (NYU) Section 4.4 Optimization Problems November 22, 2010 31 / 31