Lesson 22: Optimization (Section 021 slides)

Section 4.5
Optimization Problems
V63.0121.021, Calculus I
New York University
November 23, 2010
Announcements
Turn in HW anytime between now and November 24, 2pm
No Thursday recitation this week
Quiz 5 on §§4.1–4.4 next week in recitation
. . . . . .

. . . . . .
Announcements
Turn in HW anytime
between now and
November 24, 2pm
No Thursday recitation this
week
Quiz 5 on §§4.1–4.4 next
week in recitation
V63.0121.021, Calculus I (NYU) Section 4.5 Optimization Problems November 23, 2010 2 / 31

. . . . . .
Objectives
Given a problem requiring
optimization, identify the
objective functions,
variables, and constraints.
Solve optimization
problems with calculus.

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

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

. . . . . .
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.
..
ℓ
.
w

. . . . . .
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.
Since p = 2ℓ + 2w, we have ℓ =
p − 2w
2
,

. . . . . .
Solution Continued
area A = ℓw.
fixed.
p − 2w
2
, so
A = ℓw =
p − 2w
2
· w =
1
2
(p − 2w)(w) =
1
2
pw − w2

. . . . . .
Solution Continued
area A = ℓw.
fixed.
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).

. . . . . .
Solution Continued
area A = ℓw.
fixed.
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).

. . . . . .
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.

. . . . . .
Solution Concluded
1
2
pw − w2
on [0, p/2].
To find the critical points, we find
dA
dw
=
1
2
p − 2w.

. . . . . .
Solution Concluded
1
2
pw − w2
on [0, p/2].
dA
dw
=
1
2
p − 2w.
The critical points are when
0 =
1
2
p − 2w =⇒ w =
p
4

. . . . . .
Solution Concluded
1
2
pw − w2
on [0, p/2].
dA
dw
=
1
2
p − 2w.
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.

. . . . . .
Solution Concluded
1
2
pw − w2
on [0, p/2].
dA
dw
=
1
2
p − 2w.
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.

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

. . . . . .
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?

. . . . . .
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.

. . . . . .
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.

. . . . . .
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

. . . . . .
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.

. . . . . .
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 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).

. . . . . .
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.

. . . . . .
See Section 4.3
Let f, f′
, and f′′
(c) = 0.
If f′′
If f′′
Warning
If f′′
(c) = 0, the second derivative test is inconclusive (this does not
mean c is neither; we just don’t know yet).

. . . . . .
See Section 4.3
Let f, f′
, and f′′
(c) = 0.
If f′′
If f′′
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

. . . . . .
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

. . . . . .
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.

. . . . . .
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?

. . . . . .
Another Example

. . . . . .
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

. . . . . .
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
.
.
...
w
.
ℓ

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

. . . . . .
Solution
2. Draw a diagram.
4. Q = area = ℓw.
Q(w) = (p − 2w)(w) = pw − 2w2
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.

. . . . . .
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
f(w) = 2
A
w
+ 3w.
The domain is all positive numbers.

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

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

. . . . . .
2A
w
+ 3w on (0, ∞).
We have
df
dw
= −
2A
w2
+ 3
which is zero when w =
√
2A
3
.

. . . . . .
2A
w
+ 3w on (0, ∞).
We have
df
dw
= −
2A
w2
+ 3
√
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
+ 3w on (0, ∞).
We have
df
dw
= −
2A
w2
+ 3
√
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

. . . . . .
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
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

. . . . . .
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.

. . . . . .
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

Lesson 22: Optimization (Section 021 slides)

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