Lesson 22: Optimization Problems (slides)

..

Sec on 4.5Op miza on Problems

V63.0121.011: Calculus IProfessor Ma hew Leingang

New York University

April 18, 2011

Announcements

I Quiz 5 on Sec ons4.1–4.4 April 28/29

I Final Exam Thursday May12, 2:00–3:50pm

I I am teaching Calc II MW2:00pm and Calc III TR2:00pm both Fall ’11 andSpring ’12

Objectives

I Given a problemrequiring op miza on,iden fy the objec vefunc ons, variables, andconstraints.

I Solve op miza onproblems with calculus.

Leading by ExampleExample

What is the rectangle of fixed perimeter with maximum area?

Solu on

I Draw a rectangle.

.

.ℓ

.

w



Solu on

I Draw a rectangle.

.

.ℓ

.

w



Solu on

I Draw a rectangle.

.

.ℓ

.

w



Solu on

I Draw a rectangle.

..ℓ

.

w



Solu on

I Draw a rectangle.

..ℓ

.

w

SolutionSolu on (Con nued)

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


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

I This is a func on of two variables, not one. But the perimeter isfixed.


I This is a func on of two variables, not one. But the perimeter isfixed.

I Since p = 2ℓ+ 2w, we have ℓ =p− 2w

2, so

A = ℓw =p− 2w

2· w =

12(p− 2w)(w) =

12pw− w2



2, so

A = ℓw =p− 2w

2· w =

12(p− 2w)(w) =

12pw− w2

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



2, so

A = ℓw =p− 2w

2· w =

12(p− 2w)(w) =

12pw− w2

I Now we have A as a func on of w alone (p is constant).I The natural domain of this func on is [0, p/2] (we want tomake sure A(w) ≥ 0).


I We use the Closed Interval Method for A(w) =12pw− w2 on

[0, p/2].



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



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

I To find the cri cal points, we finddAdw

=12p− 2w.



=12p− 2w.

I The cri cal points are when12p− 2w = 0, or w =

p4.



=12p− 2w.


p4.

I Since this is the only cri cal point, it must be the maximum. Inthis case ℓ =

p4as well.



=12p− 2w.


p4.

I Since this is the only cri cal point, it must be the maximum. Inthis case ℓ =

p4as well.

I 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 problem2. Devise a plan3. Carry out the plan4. Review and extend

György Pólya(Hungarian, 1887–1985)

The Text in the Box1. Understand the Problem. What is known? What is unknown?

What are the condi ons?

2. Draw a diagram.3. Introduce Nota on.4. Express the “objec ve func on” Q in terms of the other

symbols5. 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.


What are the condi ons?2. Draw a diagram.

3. Introduce Nota on.4. Express the “objec ve func on” Q in terms of the other





What are the condi ons?2. Draw a diagram.3. Introduce Nota on.

4. Express the “objec ve func on” Q in terms of the othersymbols

5. If Q is a func on of more than one “decision variable”, use thegiven informa on to eliminate all but one of them.

6. Find the absolute maximum (or minimum, depending on theproblem) of the func on on its domain.


What are the condi ons?2. Draw a diagram.3. Introduce Nota on.4. Express the “objec ve func on” Q in terms of the other

symbols

5. If Q is a func on of more than one “decision variable”, use thegiven informa on to eliminate all but one of them.





given informa on to eliminate all but one of them.







Polya’s Method in KindergartenName [_

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

The Closed Interval MethodSee Section 4.1

The Closed Interval MethodTo find the extreme values of a func on f on [a, b], we need to:

I Evaluate f at the endpoints a and bI Evaluate f at the cri cal points x where either f′(x) = 0 or f isnot differen able at x.

I The points with the largest func on value are the globalmaximum points

I The points with the smallest/most nega ve func on value arethe global minimum points.

The First Derivative TestSee 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).I If f′ changes from nega ve to posi ve at c, then c is a local

minimum.I If f′ changes from posi ve to nega ve at c, then c is a local

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

The First Derivative TestSee Section 4.3

Corollary

I If f′ < 0 for all x < c and f′(x) > 0 for all x > c, then c is theglobal minimum of f on (a, b).

I If f′ < 0 for all x > c and f′(x) > 0 for all x < c, then c is theglobal maximum of f on (a, b).

Recall: The Second Derivative TestSee 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) withf′(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.


Theorem (The Second Deriva ve Test)

Let f, f′, and f′′ be con nuous on [a, b]. Let c be in (a, b) withf′(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 deriva ve test is inconclusive (this does notmean c is neither; we just don’t know yet).


Corollary

I If f′(c) = 0 and f′′(x) > 0 for all x, then c is the global minimumof f

I If f′(c) = 0 and f′′(x) < 0 for all x, then c is the global maximumof f

Which to use when?CIM 1DT 2DT

Pro

– no need forinequali es– gets global extremaautoma cally

– works onnon-closed,non-boundedintervals– only one deriva ve

– works onnon-closed,non-boundedintervals– no need forinequali es

Con

– only for closedbounded intervals

– Uses inequali es– More work atboundary than CIM

– More deriva ves– less conclusive than1DT– more work atboundary than CIM


Pro– no need forinequali es

– gets global extremaautoma cally



Con





Pro– no need forinequali es– gets global extremaautoma cally



Con








Con– only for closedbounded intervals





– works onnon-closed,non-boundedintervals

– only one deriva ve

















– Uses inequali es

– More work atboundary than CIM












– works onnon-closed,non-boundedintervals

– no need forinequali es

















– More deriva ves

– less conclusive than1DT– more work atboundary than CIM







– More deriva ves– less conclusive than1DT

– more work atboundary than CIM

Which to use when?

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

Outline

The Text in the Box

More Examples

Another ExampleExample (The Best Fencing Plan)

A rectangular plot of farmland will be bounded on one side by ariver 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 canenclose, and what are its dimensions?

I Known: amount of fence usedI Unknown: area enclosedI Objec ve: maximize areaI Constraint: fixed fence length

SolutionSolu on

1. Everybody understand?






I Known: amount of fence usedI Unknown: area enclosed

I Objec ve: maximize areaI Constraint: fixed fence length




SolutionSolu on

1. Everybody understand?

SolutionSolu on

1. Everybody understand?2. Draw a diagram.

DiagramA rectangular plot of farmland will be bounded on one side by ariver 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 canenclose, and what are its dimensions?

.

.

..

.

w

.

ℓ

SolutionSolu on

1. Everybody understand?2. Draw a diagram.

SolutionSolu on

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


.

.

..

.

w

.

ℓ


.

.

...

w

.

ℓ

SolutionSolu on

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

SolutionSolu on

4. Q = area = ℓw.

SolutionSolu 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]

SolutionSolu 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]

6.dQdw

= p− 4w, which is zero when w =p4.

SolutionSolu on

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

Q(p4

)= p · p

4− 2 · p

2

16=

p2

8= 80, 000m2

so the cri cal point is the absolute maximum.

SolutionSolu on

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

Q(p4

)= p · p

4− 2 · p

2

16=

p2

8= 80, 000m2

so the cri cal point is the absolute maximum.8. The dimensions of the op mal rectangle are

w =p4= 200m2 and ℓ =

p2= 400m2

Your turnExample (The shortest fence)

A 216m2 rectangular pea patch is to be enclosed by a fence anddivided into two equal parts by another fence parallel to one of itssides. What dimensions for the outer rectangle will require thesmallest total length of fence? How much fence will be needed?

Solu onLet the length and width of the pea patch be ℓ and w. The amount offence needed is f = 2ℓ+ 3w. Since ℓw = A, a constant, we havef(w) = 2

Aw

+ 3w. The domain is all posi ve numbers.

Your turnExample (The shortest fence)

A 216m2 rectangular pea patch is to be enclosed by a fence anddivided into two equal parts by another fence parallel to one of itssides. What dimensions for the outer rectangle will require thesmallest total length of fence? How much fence will be needed?

Solu onLet the length and width of the pea patch be ℓ and w. The amount offence needed is f = 2ℓ+ 3w. Since ℓw = A, a constant, we havef(w) = 2

Aw

+ 3w. The domain is all posi ve numbers.

Diagram

....ℓ

.

w

f = 2ℓ+ 3w A = ℓw ≡ 216

Solution (Continued)We need to find the minimum value of f(w) =

2Aw

+3w on (0,∞).


2Aw

+3w on (0,∞).

I We havedfdw

= −2Aw2 + 3 which is zero when w =

√2A3.


2Aw

+3w on (0,∞).

I We havedfdw

= −2Aw2 + 3 which is zero when w =

√2A3.

I Since f′′(w) = 4Aw−3, which is posi ve for all posi ve w, thecri cal point is a minimum, in fact the global minimum.


2Aw

+3w on (0,∞).

I So the area is minimized when w =

√2A3

= 12 and

ℓ =Aw

=

√3A2

= 18.


2Aw

+3w on (0,∞).

I So the area is minimized when w =

√2A3

= 12 and

ℓ =Aw

=

√3A2

= 18.

I The amount of fence needed is

f

(√2A3

)= 2 ·

√3A2

+ 3√

2A3

= 2√6A = 2

√6 · 216 = 72m

Try this oneExample

An adver sement consists of a rectangular printed region plus 1 inmargins on the sides and 1.5 in margins on the top and bo om. Ifthe total area of the adver sement is to be 120 in2, whatdimensions should the adver sement be to maximize the area ofthe printed region?

AnswerThe op mal paper dimensions are 4

√5 in by 6

√5 in.

Try this oneExample

An adver sement consists of a rectangular printed region plus 1 inmargins on the sides and 1.5 in margins on the top and bo om. Ifthe total area of the adver sement is to be 120 in2, whatdimensions should the adver sement be to maximize the area ofthe printed region?

AnswerThe op mal paper dimensions are 4

√5 in by 6

√5 in.

SolutionLet the dimensions of the printed region be x and y,P the printed area, and A the paper area. We wishto maximize P = xy subject to the constraint that

A = (x+ 2)(y+ 3) ≡ 120

Isola ng y in A ≡ 120 gives y =120x+ 2

− 3 whichyields

P = x(

120x+ 2

− 3)

=120xx+ 2

− 3x

The domain of P is (0,∞).

..

Lorem ipsumdolor sit amet,consecteturadipiscing elit.Nam dapibusvehicula mollis.Proin nec tris quemi. Pellentesquequis placeratdolor. Praesent.1.5 cm.

1.5 cm

.

1cm

.

1cm

.x

.

y

Solution (Concluded)I We want to find the absolute maximum value of P.

dPdx

=(x+ 2)(120)− (120x)(1)

(x+ 2)2− 3 =

240− 3(x+ 2)2

(x+ 2)2I There is a single (posi ve) cri cal point when(x+ 2)2 = 80 =⇒ x = 4

√5− 2.

I The second deriva ve isd2Pdx2

=−480(x+ 2)3

, which is nega ve all

along the domain of P.I Hence the unique cri cal point x =

(4√5− 2

)cm is the

absolute maximum of P.

Solution (Concluded)I Hence the unique cri cal point x =

(4√5− 2

)cm is the

absolute maximum of P.I This means the paper width is 4

√5 cm.

I the paper length is1204√5= 6

√5 cm.

Summary

I Remember the checklistI Ask yourself: what is theobjec ve?

I Remember yourgeometry:

I similar trianglesI right trianglesI trigonometric func ons

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

Lesson 22: Optimization Problems (slides)

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