Lesson 20: Optimization (handout)

Section 4.5Optimization Problems

V63.0121.006/016, Calculus I

New York University

April 7, 2010

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I Quiz 4 April 16 on §§4.1–4.4

Announcements

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I Quiz 4 April 16 on §§4.1–4.4

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

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Notes

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Section 4.5 : Optimization ProblemsV63.0121, Calculus I April 7, 2010

Evaluations: The bad

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

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Evaluations: The ugly

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A slide on slides

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I ConI “I wish he would actually use the chalkboard occasionally”I “Sometimes the slides skip steps”I “too fast”

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Notes

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2


My handwriting


A slide on slides

I ProI “Excellent slides and examples”I “clear and well-rehearsed”I “Slides are easy to follow and posted”

I ConI “I wish he would actually use the chalkboard occasionally”I “Sometimes the slides skip steps”I “too fast”

I Why I like themI Board handwriting not an issueI Easy to put online; notetaking is more than transcription

I What we can doI if you have suggestions for details to put in, I’m listeningI Feel free to ask me to fill in something on the board


Leading by Example

Example

What is the rectangle of fixed perimeter with maximum area?

Solution

I Draw a rectangle.

`

w


Notes

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3


Solution Continued

I Let its length be ` and its width be w . The objective function is areaA = `w .

I This is a function of two variables, not one. But the perimeter is fixed.

I Since p = 2` + 2w , we have ` =p − 2w

2, so

A = `w =p − 2w

2· w =

1

2(p − 2w)(w) =

1

2pw − w2

I Now we have A as a function of w alone (p is constant).

I The natural domain of this function is [0, p/2] (we want to make sureA(w) ≥ 0).


Solution Concluded

We use the Closed Interval Method for A(w) =1

2pw − w2 on [0, p/2].

I At the endpoints, A(0) = A(p/2) = 0.

I To find the critical points, we finddA

dw=

1

2p − 2w .

I The critical points are when

0 =1

2p − 2w =⇒ w =

p

4

I Since this is the only critical point, it must be the maximum. In this

case ` =p

4as well.

I We have a square! The maximal area is A(p/4) = p2/16.


Strategies for Problem Solving

1. Understand the problem

2. Devise a plan

3. Carry out the plan

4. Review and extend

Gyorgy Polya(Hungarian, 1887–1985)


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4


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 giveninformation 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 StrategyDraw 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 pictureto solve the problem.

crayons

What number do Iadd to 5 to get 8?

8 - = 55 + 3 = 8

CHECK

Does your answer make sense?Explain.

Draw a picture to solve the problem.Write how many were given away.I. I had 10 pencils.

I gave some away.I have 3 left. How manypencils did I give away?

~7

What numberdo I add to 3to make 10?

13iftill:ii ?11

ftI'•'«II

ft A

H 11M i l

U U U U> U U

I I


Recall: The Closed Interval MethodSee Section 4.1

To find the extreme values of a function f on [a, b], we need to:

I Evaluate f at the endpoints a and b

I Evaluate f at the critical points x where either f ′(x) = 0 or f is notdifferentiable at x .

I The points with the largest function value are the global maximumpoints

I The points with the smallest or most negative function value are theglobal minimum points.


Notes

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5


Recall: The First Derivative TestSee 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).

I If f ′ changes from negative to positive at c, then c is a localminimum.

I If f ′ changes from positive to negative at c, then c is a localmaximum.

I If f ′ does not change sign at c, then c is not a local extremum.


Recall: The Second Derivative TestSee 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.

I If f ′′(c) < 0, then f (c) is a local maximum.

I 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 meanc is neither; we just don’t know yet).


Which to use when?

CIM 1DT 2DT

Pro – no need forinequalities– gets global extremaautomatically

– works onnon-closed,non-boundedintervals– only one derivative

– works onnon-closed,non-boundedintervals– no need forinequalities

Con – only for closedbounded intervals

– Uses inequalities– More work atboundary than CIM

– More derivatives– less conclusive than1DT– more work atboundary than CIM

I Use CIM if it applies: the domain is a closed, bounded interval

I If domain is not closed or not bounded, use 2DT if you like to takederivatives, or 1DT if you like to compare signs.


Notes

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6


Another Example

Example (The Best Fencing Plan)

A rectangular plot of farmland will be bounded on one side by a river andon the other three sides by a single-strand electric fence. With 800m ofwire at your disposal, what is the largest area you can enclose, and whatare its dimensions?

I Known: amount of fence used

I Unknown: area enclosed

I Objective: maximize area

I Constraint: fixed fence length


Solution

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 − 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 · p

2

16=

p2

8= 80, 000m2

so the critical point is the absolute maximum.


Another Example

Example (The Best Fencing Plan)

A rectangular plot of farmland will be bounded on one side by a river andon the other three sides by a single-strand electric fence. With 800m ofwire at your disposal, what is the largest area you can enclose, and whatare its dimensions?

I Known: amount of fence used

I Unknown: area enclosed

I Objective: maximize area

I Constraint: fixed fence length


Notes

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7


Solution


2. Draw a diagram.


4. Q = area = `w .


Q(w) = (p − 2w)(w) = pw − 2w2


6.dQ


p

4. Q(0) = Q(p/2) = 0, but

Q(p

4

)= p · p

4− 2 · p

2

16=

p2

8= 80, 000m2



Diagram

A rectangular plot of farmland will be bounded on one side by a river andon the other three sides by a single-strand electric fence. With 800 m ofwire at your disposal, what is the largest area you can enclose, and whatare its dimensions?

w

`


Solution


2. Draw a diagram.


4. Q = area = `w .


Q(w) = (p − 2w)(w) = pw − 2w2


6.dQ


p

4. Q(0) = Q(p/2) = 0, but

Q(p

4

)= p · p

4− 2 · p

2

16=

p2

8= 80, 000m2



Notes

Notes

Notes

8


Diagram

A rectangular plot of farmland will be bounded on one side by a river andon the other three sides by a single-strand electric fence. With 800 m ofwire at your disposal, what is the largest area you can enclose, and whatare its dimensions?

w

`


Solution


2. Draw a diagram.


4. Q = area = `w .


Q(w) = (p − 2w)(w) = pw − 2w2


6.dQ


p

4. Q(0) = Q(p/2) = 0, but

Q(p

4

)= p · p

4− 2 · p

2

16=

p2

8= 80, 000m2



Your turn

Example (The shortest fence)

A 216m2 rectangular pea patch is to be enclosed by a fence and dividedinto two equal parts by another fence parallel to one of its sides. Whatdimensions for the outer rectangle will require the smallest total length offence? How much fence will be needed?

Solution

Let the length and width of the pea patch be ` and w. The amount offence needed is f = 2` + 3w. Since `w = A, a constant, we have

f (w) = 2A

w+ 3w .

The domain is all positive numbers.


Notes

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9


Diagram

`

w

f = 2` + 3w A = `w ≡ 216


Solution (Continued)

We need to find the minimum value of f (w) =2A

w+ 3w on (0,∞).

I We havedf

dw= −2A

w2+ 3

which is zero when w =

√2A

3.

I Since f ′′(w) = 4Aw−3, which is positive for all positive w , the criticalpoint is a minimum, in fact the global minimum.

I So the area is minimized when w =

√2A

3= 12 and

` =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 marginson the sides and 1.5 in margins on the top and bottom. If the total area ofthe advertisement is to be 120 in2, what dimensions should theadvertisement be to maximize the area of the printed region?

Answer

The optimal paper dimensions are 4√

5 in by 6√

5 in.


Notes

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10


Solution

Let the dimensions of theprinted region be x and y , Pthe printed area, and A thepaper area. We wish tomaximize P = xy subject tothe 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,∞)

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1.5 cm

1.5 cm

1cm

1cm

x

y


Solution (Concluded)

We want to find the absolute maximum value of P. Taking derivatives,

dP

dx=

(x + 2)(120)− (120x)(1)

(x + 2)2− 3 =

240− 3(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

d2P

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 is120

4√

5= 6√

5 cm.


Summary

I Remember the checklist

I Ask yourself: what is the objective?I Remember your geometry:

I similar trianglesI right trianglesI trigonometric functions


Notes

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Lesson 20: Optimization (handout)