Lesson 20: Optimization (slides)

Section 4.5
Optimization Problems

V63.0121.006/016, Calculus I

New York University

April 6, 2010

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Quiz 4 April 16 on §§4.1–4.4
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Announcements

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Quiz 4 April 16 on §§4.1–4.4

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V63.0121, Calculus I (NYU) Section 4.5 Optimization Problems April 6, 2010 2 / 36

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In this course we care about concepts
There will be conceptual problems on the exam
Concepts are the keys to overcoming templated problems

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A slide on slides

Pro
Con
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Why I like them
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My handwriting

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A slide on slides

Pro
Con
“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

. . . . . .


Outline

Leading by Example

The Text in the Box

More Examples

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Leading by Example

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

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Leading by Example

Example

Solution

Draw a rectangle.

.

. . . . . .

Leading by Example

Example

Solution

Draw a rectangle.

.
.
ℓ

. . . . . .

Leading by Example

Example

Solution

Draw a rectangle.

w
.

.
.
ℓ

. . . . . .


Solution Continued

Let its length be ℓ and its width be w. The objective function is
area A = ℓw.

. . . . . .


Solution Continued

area A = ℓw.
This is a function of two variables, not one. But the perimeter is
fixed.

. . . . . .


Solution Continued

area A = ℓw.
fixed.
p − 2w
Since p = 2ℓ + 2w, we have ℓ = ,
2

. . . . . .


Solution Continued

area A = ℓw.
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

. . . . . .


Solution Continued

area A = ℓw.
fixed.
p − 2w
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).

. . . . . .


Solution Continued

area A = ℓw.
fixed.
p − 2w
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).

. . . . . .


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.

. . . . . .


Solution Concluded

1
2
dA 1
To find the critical points, we find = p − 2w.
dw 2

. . . . . .


Solution Concluded

1
2
dA 1
dw 2
The critical points are when

1 p
0= p − 2w =⇒ w =
2 4

. . . . . .


Solution Concluded

1
2
dA 1
dw 2

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

. . . . . .


Solution Concluded

1
2
dA 1
dw 2

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.

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Outline

Leading by Example

The Text in the Box

More Examples

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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)
. . . . . .


The Text in the Box

1. Understand the Problem. What is known? What is unknown?
What are the conditions?

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The Text in the Box

2. Draw a diagram.

. . . . . .


The Text in the Box

2. Draw a diagram.
3. Introduce Notation.

. . . . . .


The Text in the Box

2. Draw a diagram.
4. Express the “objective function” Q in terms of the other symbols

. . . . . .


The Text in the Box

2. Draw a diagram.
5. If Q is a function of more than one “decision variable”, use the
given information to eliminate all but one of them.

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The Text in the Box

2. Draw a diagram.
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
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

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.

. . . . . .


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.

. . . . . .


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.

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?

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

. . . . . .


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.
. . . . . .


Outline

Leading by Example

The Text in the Box

More Examples

. . . . . .


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?

. . . . . .


Solution

1. Everybody understand?

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Another Example


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Another Example


Known: amount of fence used
Unknown: area enclosed

. . . . . .


Another Example


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

. . . . . .


Solution


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Solution

2. Draw a diagram.

. . . . . .


Diagram

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?

. .

.
.

. . . . . .


Solution

2. Draw a diagram.

. . . . . .


Solution

2. Draw a diagram.
3. Length and width are ℓ and w. Length of wire used is p.

. . . . . .


Diagram


. .

.
.

. . . . . .


Diagram

.
ℓ

w
.

. .

.
.

. . . . . .


Solution

2. Draw a diagram.

. . . . . .


Solution

2. Draw a diagram.
4. Q = area = ℓw.

. . . . . .


Solution

2. Draw a diagram.
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]

. . . . . .


Solution

2. Draw a diagram.
4. Q = area = ℓw.

Q(w) = (p − 2w)(w) = pw − 2w2

dQ p
6. = p − 4w, which is zero when w = .
dw 4

. . . . . .


Solution

2. Draw a diagram.
4. Q = area = ℓw.

Q(w) = (p − 2w)(w) = pw − 2w2

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.
. . . . . .


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?

. . . . . .


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.

. . . . . .


Diagram

. .

w
.

.
.
ℓ

f = 2ℓ + 3w A = ℓw ≡ 216

. . . . . .


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

. . . . . .


2A
w
We have
df 2A
=− 2 +3
dw w
√
2A
which is zero when w = .
3

. . . . . .


2A
w
We have
df 2A
=− 2 +3
dw 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.

. . . . . .


2A
w
We have
df 2A
=− 2 +3
dw 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.
√
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
. . . . . .


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?

. . . . . .


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.

. . . . . .


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
.
. . . . . .


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 . . . . . .


Summary

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

. . . . . .


Lesson 20: Optimization (slides)

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