Lesson 20: Optimization (handout)

V63.0121, Calculus I Section 4.5 : Optimization Problems April 7, 2010

Notes
Section 4.5
Optimization Problems

V63.0121.006/016, Calculus I

New York University

April 7, 2010

Announcements

Thank you for the evaluations
Quiz 4 April 16 on §§4.1–4.4

Announcements
Notes

Thank you for the evaluations
Quiz 4 April 16 on §§4.1–4.4

V63.0121, Calculus I (NYU) Section 4.5 Optimization Problems April 7, 2010 2 / 36

Evaluations: The good
Notes

“Very knowledgeable”
“Knows how to teach”
“Very good at projecting voice”
“Oﬃce hours are accessible”
“Clean”
“Great syllabus”


1


Evaluations: The bad
Notes

Too fast, not enough examples
Not enough time to do everything
Lecture is not the only learning time (recitation and independent study)
I try to balance concept and procedure
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


Evaluations: The ugly
Notes

“The projector blows.”
“Sometimes condescending/rude.”
“Can’t pick his nose without checking his notes, and he still gets it
wrong the first time.”
“If I were chained to a desk and forced to see this guy teach, I would
chew my arm off in order to get free.”


A slide on slides
Notes

Pro
“Excellent slides and examples”
“clear and well-rehearsed”
“Slides are easy to follow and posted”
Con
“I wish he would actually use the chalkboard occasionally”
“Sometimes the slides skip steps”
“too fast”
Why I like them
Board handwriting not an issue
Easy to put online; notetaking is more than transcription
What we can do
if you have suggestions for details to put in, I’m listening
Feel free to ask me to fill in something on the board


2


My handwriting
Notes


A slide on slides
Notes

Pro
“Excellent slides and examples”
“clear and well-rehearsed”
“Slides are easy to follow and posted”
Con
“I wish he would actually use the chalkboard occasionally”
“Sometimes the slides skip steps”
“too fast”
Why I like them
Board handwriting not an issue
Easy to put online; notetaking is more than transcription
What we can do
if you have suggestions for details to put in, I’m listening
Feel free to ask me to ﬁll in something on the board


Leading by Example
Notes

Example
What is the rectangle of ﬁxed perimeter with maximum area?

Solution

Draw a rectangle.

w


3


Solution Continued
Notes

Let its length be and its width be w . The objective function is area
A = w.
This is a function of two variables, not one. But the perimeter is fixed.
p − 2w
Since p = 2 + 2w , we have = , so
2
p − 2w 1 1
A= w = · w = (p − 2w )(w ) = pw − w 2
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).


Solution Concluded
Notes
1
We use the Closed Interval Method for A(w ) = pw − w 2 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 = as well.
4
We have a square! The maximal area is A(p/4) = p 2 /16.


Strategies for Problem Solving
Notes

1. Understand the problem
2. Devise a plan
3. Carry out the plan
4. Review and extend

Gy¨rgy P´lya
o o
(Hungarian, 1887–1985)


4


The Text in the Box
Notes

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.


Name [_
Problem Solving Strategy
Draw a Picture Notes
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
~7 U U U U> U U


Recall: The Closed Interval Method
See Section 4.1 Notes

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.


5


Recall: 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 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.


Recall: The Second Derivative Test

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.

Warning
If f (c) = 0, the second derivative test is inconclusive (this does not mean
c is neither; we just don’t know yet).


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


6


Another Example
Notes

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


Solution
Notes
1. Everybody understand?
2. Draw a diagram.
3. Length and width are and w . Length of wire used is p.
4. Q = area = w .
5. Since p = + 2w , we have = p − 2w and so

Q(w ) = (p − 2w )(w ) = pw − 2w 2

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.

Another Example
Notes

Example (The Best Fencing Plan)
on the other three sides by a single-strand electric fence. With 800m of
are its dimensions?

Known: amount of fence used
Unknown: area enclosed
Objective: maximize area
Constraint: ﬁxed fence length


7


Solution
Notes
2. Draw a diagram.
4. Q = area = w .

Q(w ) = (p − 2w )(w ) = pw − 2w 2

dQ p
dw 4
p p p2 p2
Q =p· −2· = = 80, 000m2
4 4 16 8

Diagram
Notes
on the other three sides by a single-strand electric fence. With 800 m of
are its dimensions?

w


Solution
Notes
2. Draw a diagram.
4. Q = area = w .

Q(w ) = (p − 2w )(w ) = pw − 2w 2

dQ p
dw 4
p p p2 p2
Q =p· −2· = = 80, 000m2
4 4 16 8

8


Diagram
Notes
on the other three sides by a single-strand electric fence. With 800 m of
are its dimensions?

w


Solution
Notes
2. Draw a diagram.
4. Q = area = w .

Q(w ) = (p − 2w )(w ) = pw − 2w 2

dQ p
dw 4
p p p2 p2
Q =p· −2· = = 80, 000m2
4 4 16 8

Your turn
Notes

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 and w . The amount of
fence needed is f = 2 + 3w . Since w = A, a constant, we have
A
f (w ) = 2 + 3w .
w
The domain is all positive numbers.


9


Diagram
Notes

w

f = 2 + 3w A = w ≡ 216


Solution (Continued)
2A Notes
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
−3
Since f (w ) = 4Aw , 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
w 2

2A 3A 2A √ √
f =2· +3 = 2 6A = 2 6 · 216 = 72m
3 2 3

Try this one
Notes

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.


10


Solution
Notes
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, con-
sectetur adipiscing elit. Nam
paper area. We wish to dapibus vehicula mollis. Proin nec
tristique mi. Pellentesque quis plac-
maximize P = xy subject to erat 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, ultri-

1 cm

1 cm
A = (x + 2)(y + 3) ≡ 120 y ces eu eros. Proin pellentesque ali-
quam nibh ut lobortis. Ut et sol-
licitudin ipsum. Proin gravida ligula
Isolating y in A ≡ 120 gives eget odio molestie rhoncus sed nec
massa. In ante lorem, imperdiet eget
120 tincidunt at, pharetra sit amet felis.
y= − 3 which yields Nunc nisi velit, tempus ac suscipit
x +2 quis, blandit vitae mauris. Vestibu-
lum ante ipsum primis in faucibus
orci luctus et ultrices posuere cubilia
120 120x Curae;
P=x −3 = −3x
x +2 x +2 1.5 cm
The domain of P is (0, ∞) x

Solution (Concluded)
We want to ﬁnd the absolute maximum value of P. Taking derivatives, Notes

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

d 2P −480
=
dx 2 (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
√ 120 √
paper width is 4 5 cm, and the paper length is √ = 6 5 cm.
4 5

Summary
Notes

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


11

Lesson 20: Optimization (handout)

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