Lesson 22: Optimization Problems (slides)

Sec on 4.5
Op miza on Problems
V63.0121.011: Calculus I
Professor Ma hew Leingang
New York University

April 18, 2011

.

Announcements
Quiz 5 on Sec ons
4.1–4.4 April 28/29
Final Exam Thursday May
12, 2:00–3:50pm
I am teaching Calc II MW
2:00pm and Calc III TR
2:00pm both Fall ’11 and
Spring ’12

Objectives

Given a problem
requiring op miza on,
iden fy the objec ve
func ons, variables, and
constraints.
Solve op miza on
problems with calculus.

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

Leading by Example
Example

Solu on

Leading by Example
Example

Solu on
Draw a rectangle.

.

Leading by Example
Example

Solu on
Draw a rectangle.

.
ℓ

Leading by Example
Example

Solu on
Draw a rectangle.

w
.
ℓ

Solution
Solu on (Con nued)

Let its length be ℓ and its width be w. The objec ve func on is
area A = ℓw.

Solution
Solu on (Con nued)

Let its length be ℓ and its width be w. The objec ve func on is
area A = ℓw.
This is a func on of two variables, not one. But the perimeter is
ﬁxed.

Solution
Solu on (Con nued)

This is a func on 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

Solution
Solu on (Con nued)

p − 2w
2
p − 2w 1 1
A = ℓw = · w = (p − 2w)(w) = pw − w2
2 2 2
Now we have A as a func on of w alone (p is constant).

Solution
Solu on (Con nued)

p − 2w
2
p − 2w 1 1
A = ℓw = · w = (p − 2w)(w) = pw − w2
2 2 2
Now we have A as a func on of w alone (p is constant).
The natural domain of this func on is [0, p/2] (we want to
make sure A(w) ≥ 0).

Solution
Solu on (Con nued)

1
We use the Closed Interval Method for A(w) = pw − w2 on
2
[0, p/2].

Solution
Solu on (Con nued)

1
2
[0, p/2].
At the endpoints, A(0) = A(p/2) = 0.

Solution
Solu on (Con nued)

1
2
[0, p/2].
At the endpoints, A(0) = A(p/2) = 0.
dA 1
To ﬁnd the cri cal points, we ﬁnd = p − 2w.
dw 2

Solution
Solu on (Con nued)

dA 1
dw 2
1 p
The cri cal points are when p − 2w = 0, or w = .
2 4

Solution
Solu on (Con nued)

dA 1
dw 2
1 p
2 4
Since this is the only cri cal point, it must be the maximum. In
p
this case ℓ = as well.
4

Solution
Solu on (Con nued)

dA 1
dw 2
1 p
2 4
Since this is the only cri cal point, it must be the maximum. In
p
this case ℓ = as well.
4
We have a square! The maximal area is A(p/4) = p2 /16.

Outline

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 condi ons?

The Text in the Box
2. Draw a diagram.

The Text in the Box
2. Draw a diagram.
3. Introduce Nota on.

The Text in the Box
2. Draw a diagram.
4. Express the “objec ve func on” Q in terms of the other
symbols

The Text in the Box
2. Draw a diagram.
symbols
5. If Q is a func on of more than one “decision variable”, use the
given informa on to eliminate all but one of them.

The Text in the Box
2. Draw a diagram.
symbols
5. If Q is a func on of more than one “decision variable”, use the
given informa on to eliminate all but one of them.
6. Find the absolute maximum (or minimum, depending on the
problem) of the func on 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.

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

The Closed Interval Method
See Section 4.1
The Closed Interval Method
To ﬁnd the extreme values of a func on f on [a, b], we need to:
Evaluate f at the endpoints a and b
Evaluate f at the cri cal points x where either f′ (x) = 0 or f is
not diﬀeren able at x.
The points with the largest func on value are the global
maximum points
The points with the smallest/most nega ve func on value are
the global minimum points.

The First Derivative Test
See Section 4.3
Theorem (The First Deriva ve Test)
Let f be con nuous on (a, b) and c a cri cal point of f in (a, b).
If f′ changes from nega ve to posi ve at c, then c is a local
minimum.
If f′ changes from posi ve to nega ve at c, then c is a local
maximum.
If f′ does not change sign at c, then c is not a local extremum.

The First Derivative Test
See Section 4.3

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 Deriva ve Test)
Let f, f′ , and f′′ be con nuous 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
Theorem (The Second Deriva ve Test)
Let f, f′ , and f′′ be con nuous 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 deriva ve test is inconclusive (this does not
mean c is neither; we just don’t know yet).

See Section 4.3

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

Con

Which to use when?
CIM 1DT 2DT
Pro
– no need for
inequali es

Con

Which to use when?
CIM 1DT 2DT
Pro
– no need for
inequali es
– gets global extrema
automa cally

Con

Which to use when?
CIM 1DT 2DT
Pro
– no need for
inequali es
– gets global extrema
automa cally

Con
– only for closed
bounded intervals

Which to use when?
CIM 1DT 2DT
Pro
– no need for – works on
inequali es non-closed,
– gets global extrema non-bounded
automa cally intervals

Con
– only for closed
bounded intervals

Which to use when?
CIM 1DT 2DT
Pro
– only one deriva ve

Con
– only for closed
bounded intervals

Which to use when?
CIM 1DT 2DT
Pro

Con
– only for closed – Uses inequali es
bounded intervals

Which to use when?
CIM 1DT 2DT
Pro

Con
bounded intervals – More work at
boundary than CIM

Which to use when?
CIM 1DT 2DT
Pro
– no need for – works on – works on
inequali es non-closed, non-closed,
– gets global extrema non-bounded non-bounded
automa cally intervals intervals

Con
boundary than CIM

Which to use when?
CIM 1DT 2DT
Pro
– only one deriva ve – no need for
inequali es
Con
boundary than CIM

Which to use when?
CIM 1DT 2DT
Pro
inequali es
Con
– only for closed – Uses inequali es – More deriva ves
boundary than CIM

Which to use when?
CIM 1DT 2DT
Pro
inequali es
Con
– only for closed – Uses inequali es – More deriva ves
bounded intervals – More work at – less conclusive than
boundary than CIM 1DT

Which to use when?

If domain is closed and bounded, use CIM.
If domain is not closed or not bounded, use 2DT if you like to
take deriva ves, or 1DT if you like to compare signs.

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
Solu on
1. Everybody understand?

Another Example

Known: amount of fence used
Unknown: area enclosed

Another Example

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

Solution
Solu on
2. Draw a diagram.

Diagram
With 800 m of wire at your disposal, what is the largest area you can

.
.

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

Diagram
With 800 m of wire at your disposal, what is the largest area you can
ℓ

w
.
.

Solution
Solu on

4. Q = area = ℓw.

Solution
Solu on

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
Solu on

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

Solution
Solu on

7. Q(0) = Q(p/2) = 0, but
(p) p p2 p2
Q =p· −2· = = 80, 000m2
4 4 16 8
so the cri cal point is the absolute maximum.

Solution
Solu on

7. Q(0) = Q(p/2) = 0, but
(p) p p2 p2
Q =p· −2· = = 80, 000m2
4 4 16 8
so the cri cal point is the absolute maximum.
8. The dimensions of the op mal rectangle are
p p
w= = 200 m2 and ℓ = = 400 m2
4 2

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?

Solu on
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. The domain is all posi ve numbers.
w

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

2A
w
√
df 2A 2A
We have = − 2 + 3 which is zero when w = .
dw w 3
Since f′′ (w) = 4Aw−3 , which is posi ve for all posi ve w, the
cri cal point is a minimum, in fact the global minimum.

2A
w
√
2A
So the area is minimized when w = = 12 and
√ 3
A 3A
ℓ= = = 18.
w 2

2A
w
√
2A
So the area is minimized when w = = 12 and
√ 3
A 3A
ℓ= = = 18.
w 2
The amount of fence needed is
(√ ) √ √
2A 3A 2A √ √
f =2· +3 = 2 6A = 2 6 · 216 = 72m
3 2 3

Try this one
Example
An adver sement consists of a rectangular printed region plus 1 in
margins on the sides and 1.5 in margins on the top and bo om. If
the total area of the adver sement is to be 120 in2 , what
dimensions should the adver sement be to maximize the area of
the printed region?

Try this one
Example
An adver sement consists of a rectangular printed region plus 1 in
margins on the sides and 1.5 in margins on the top and bo om. If
the total area of the adver sement is to be 120 in2 , what
dimensions should the adver sement be to maximize the area of
the printed region?

Answer
√ √
The op mal paper dimensions are 4 5 in by 6 5 in.

Solution of the printed region be x and y,
Let the dimensions
P the printed area, and A the paper area. We wish
to maximize P = xy subject to the constraint that
1.5 cm
A = (x + 2)(y + 3) ≡ 120 Lorem ipsum
dolor sit amet,
consectetur
adipiscing elit.

1 cm

1 cm
120 y Nam dapibus

Isola ng y in A ≡ 120 gives y = − 3 which vehicula mollis.
Proin nec tris que
x+2 mi. Pellentesque
quis placerat
yields . dolor. Praesent

( ) 1.5 cm
120 120x x
P=x −3 = − 3x
x+2 x+2

The domain of P is (0, ∞).

Solution (Concluded)
We want to ﬁnd the absolute maximum value of P.
dP (x + 2)(120) − (120x)(1) 240 − 3(x + 2)2
= −3=
dx (x + 2)2 (x + 2)2
There is a single (posi ve) √ cal point when
cri
(x + 2) = 80 =⇒ x = 4 5 − 2.
2

d2 P −480
The second deriva ve is 2 = , which is nega ve all
dx (x + 2)3
along the domain of P. ( √ )
Hence the unique cri cal point x = 4 5 − 2 cm is the
absolute maximum of P.

Solution (Concluded)
( √ )
Hence the unique cri cal point x = 4 5 − 2 cm is the
absolute maximum of P.
√
This means the paper width is 4 5 cm.
120 √
the paper length is √ = 6 5 cm.
4 5

Summary
Name [_
Problem Solving Strategy
Draw a Picture
Kathy had a box of 8 crayons.
She gave some crayons away.

Remember the checklist She has 5 left.
How many crayons did Kathy give away?

UNDERSTAND

Ask yourself: what is the
•
What do you want to find out?
Draw a line under the question.

objec ve? You can draw a picture
to solve the problem.

Remember your
geometry: crayons
What number do I
add to 5 to get 8?
8 -
5 + 3 = 8
= 5

similar triangles CHECK
Does your answer make sense?

right triangles Explain.
What number
do I add to 3
Draw a picture to solve the problem.
trigonometric func ons Write how many were given away.
I. I had 10 pencils.
to make 10?

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

Lesson 22: Optimization Problems (slides)

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