lesson 24: optimization

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Section4.5OptimizationProblems

V63.0121.027, CalculusI

November24, 2009

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I WrittenHW (inc. WebAssignfrom4.4)dueWednesday,Nov. 25, 11:00aminmymailbox

I Quiz5nextweekon§§4.1–4.4, 4.7

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I ToomanyproofsI InthiscoursewecareaboutconceptsI TherewillbeconceptualproblemsontheexamI Conceptsarethekeystoovercomingtemplatedproblems

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I Toofast, notenoughexamplesI NotenoughtimetodoeverythingI Classisnottheonlylearningtime(recitationandindependentstudy)

I I trytobalance


. . . . . .



I I trytobalance

I Toomanyproofs

I InthiscoursewecareaboutconceptsI TherewillbeconceptualproblemsontheexamI Conceptsarethekeystoovercomingtemplatedproblems

. . . . . .



I I trytobalance


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Outline

LeadingbyExample

TheTextintheBox

MoreExamples

. . . . . .

LeadingbyExample

ExampleWhatistherectangleoffixedperimeterwithmaximumarea?

Solution

I Drawarectangle.

.

.

.ℓ

.w

. . . . . .

Solution(Continued)

I Letitslengthbe ℓ anditswidthbe w. Theobjectivefunctionisarea A = ℓw.

I Thisisafunctionoftwovariables, notone. Buttheperimeterisfixed.

I Since p = 2ℓ+ 2w, wehave ℓ =p− 2w

2, so

A = ℓw =p− 2w

2·w =

12(p− 2w)(w) =

12pw−w2

I Nowwehave A asafunctionof w alone(p isconstant).I Thenaturaldomainofthisfunctionis [0,p/2] (wewantto

makesure A(w) ≥ 0).

. . . . . .

Solution(Continued)




2,

so

A = ℓw =p− 2w

2·w =

12(p− 2w)(w) =

12pw−w2



. . . . . .

Solution(Continued)




2, so

A = ℓw =p− 2w

2·w =

12(p− 2w)(w) =

12pw−w2



. . . . . .

Solution(Continued)




2, so

A = ℓw =p− 2w

2·w =

12(p− 2w)(w) =

12pw−w2

I Nowwehave A asafunctionof w alone(p isconstant).

I Thenaturaldomainofthisfunctionis [0,p/2] (wewanttomakesure A(w) ≥ 0).

. . . . . .

Solution(Continued)




2, so

A = ℓw =p− 2w

2·w =

12(p− 2w)(w) =

12pw−w2



. . . . . .

Solution(Concluded)WeusetheClosedIntervalMethodfor A(w) =

12pw−w2 on

[0,p/2].I Attheendpoints, A(0) = A(p/2) = 0.

I Tofindthecriticalpoints, wefinddAdw

=12p− 2w.

I Thecriticalpointsarewhen

0 =12p− 2w =⇒ w =

p4

I Sincethisistheonlycriticalpoint, itmustbethemaximum.

Inthiscase ℓ =p4aswell.

I Wehaveasquare! Themaximalareais A(p/4) = p2/16.

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Outline

LeadingbyExample

TheTextintheBox

MoreExamples

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TheTextintheBox

1. UnderstandtheProblem. Whatisknown? Whatisunknown? Whataretheconditions?

2. Drawadiagram.

3. IntroduceNotation.

4. Expressthe“objectivefunction” Q intermsoftheothersymbols

5. If Q isafunctionofmorethanone“decisionvariable”, usethegiveninformationtoeliminateallbutoneofthem.

6. Findtheabsolutemaximum(orminimum, dependingontheproblem)ofthefunctiononitsdomain.

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Recall: TheClosedIntervalMethodSeeSection4.1

Tofindtheextremevaluesofafunction f on [a,b], weneedto:I Evaluate f atthe endpoints a and bI Evaluate f atthe criticalpoints x whereeither f′(x) = 0 or f isnotdifferentiableat x.

I Thepointswiththelargestfunctionvaluearetheglobalmaximumpoints

I Thepointswiththesmallestormostnegativefunctionvaluearetheglobalminimumpoints.

. . . . . .

Recall: TheFirstDerivativeTestSeeSection4.3

Theorem(TheFirstDerivativeTest)Let f becontinuouson [a,b] and c acriticalpointof f in (a,b).

I If f′(x) > 0 on (a, c) and f′(x) < 0 on (c,b), then c isalocalmaximum.

I If f′(x) < 0 on (a, c) and f′(x) > 0 on (c,b), then c isalocalminimum.

I If f′(x) hasthesamesignon (a, c) and (c,b), then c isnotalocalextremum.

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Recall: TheSecondDerivativeTestSeeSection4.3

Theorem(TheSecondDerivativeTest)Let f, f′, and f′′ becontinuouson [a,b]. Let c bebeapointin(a,b) with f′(c) = 0.

I If f′′(c) < 0, then f(c) isalocalmaximum.I If f′′(c) > 0, then f(c) isalocalminimum.

If f′′(c) = 0, thesecondderivativetestisinconclusive(thisdoesnotmean c isneither; wejustdon’tknowyet).

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Whichtousewhen?

CIM 1DT 2DTPro –noneedfor

inequalities–getsglobalextremaautomatically

–worksonnon-closed,non-boundedintervals–onlyonederivative

–worksonnon-closed,non-boundedintervals–noneedforinequalities

Con –onlyforclosedboundedintervals

–Usesinequalities–MoreworkatboundarythanCIM

–Morederivatives–lessconclusivethan1DT–moreworkatboundarythanCIM

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Whichtousewhen? Thebottomline

I UseCIM ifitapplies: thedomainisaclosed, boundedinterval

I Ifdomainisnotclosedornotbounded, use2DT ifyouliketotakederivatives, or1DT ifyouliketocomparesigns.

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Outline

LeadingbyExample

TheTextintheBox

MoreExamples

. . . . . .

AnotherExample

Example(TheBestFencingPlan)A rectangularplotoffarmlandwillbeboundedononesidebyariverandontheotherthreesidesbyasingle-strandelectricfence. With800mofwireatyourdisposal, whatisthelargestareayoucanenclose, andwhatareitsdimensions?

I Known: amountoffenceusedI Unknown: areaenclosedI Objective: maximizeareaI Constraint: fixedfencelength

. . . . . .

Solution1. Everybodyunderstand?

2. Drawadiagram.3. Introducenotation: Lengthandwidthare ℓ and w. Lengthof

wireusedis p.4. Q = area = ℓw.5. Since p = ℓ+ 2w, wehave ℓ = p− 2w andso

Q(w) = (p− 2w)(w) = pw− 2w2

Thedomainof Q is [0,p/2]

6.dQdw

= p− 4w, whichiszerowhen w =p4.

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

Q(p4

)= p · p

4− 2 · p

2

16=

p2

8= 80, 000m2

sothecriticalpointistheabsolutemaximum.

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AnotherExample



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AnotherExample


I Known: amountoffenceusedI Unknown: areaenclosed

I Objective: maximizeareaI Constraint: fixedfencelength

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AnotherExample



. . . . . .

Solution1. Everybodyunderstand?

2. Drawadiagram.3. Introducenotation: Lengthandwidthare ℓ and w. Lengthof


Q(w) = (p− 2w)(w) = pw− 2w2


6.dQdw


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

Q(p4

)= p · p

4− 2 · p

2

16=

p2

8= 80, 000m2


. . . . . .

Solution1. Everybodyunderstand?2. Drawadiagram.

3. Introducenotation: Lengthandwidthare ℓ and w. Lengthofwireusedis p.

4. Q = area = ℓw.5. Since p = ℓ+ 2w, wehave ℓ = p− 2w andso

Q(w) = (p− 2w)(w) = pw− 2w2


6.dQdw


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

Q(p4

)= p · p

4− 2 · p

2

16=

p2

8= 80, 000m2


. . . . . .

Diagram

A rectangularplotoffarmlandwillbeboundedononesidebyariverandontheotherthreesidesbyasingle-strandelectricfence. With800mofwireatyourdisposal, whatisthelargestareayoucanenclose, andwhatareitsdimensions?

.

.

.

.

.w

.ℓ

. . . . . .

Solution1. Everybodyunderstand?2. Drawadiagram.

3. Introducenotation: Lengthandwidthare ℓ and w. Lengthofwireusedis p.


Q(w) = (p− 2w)(w) = pw− 2w2


6.dQdw


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

Q(p4

)= p · p

4− 2 · p

2

16=

p2

8= 80, 000m2


. . . . . .

Solution1. Everybodyunderstand?2. Drawadiagram.3. Introducenotation: Lengthandwidthare ℓ and w. Lengthof

wireusedis p.


Q(w) = (p− 2w)(w) = pw− 2w2


6.dQdw


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

Q(p4

)= p · p

4− 2 · p

2

16=

p2

8= 80, 000m2


. . . . . .

Diagram

A rectangularplotoffarmlandwillbeboundedononesidebyariverandontheotherthreesidesbyasingle-strandelectricfence. With800mofwireatyourdisposal, whatisthelargestareayoucanenclose, andwhatareitsdimensions?

.

.

.

.

.w

.ℓ

. . . . . .


wireusedis p.


Q(w) = (p− 2w)(w) = pw− 2w2


6.dQdw


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

Q(p4

)= p · p

4− 2 · p

2

16=

p2

8= 80, 000m2


. . . . . .


wireusedis p.4. Q = area = ℓw.

5. Since p = ℓ+ 2w, wehave ℓ = p− 2w andso

Q(w) = (p− 2w)(w) = pw− 2w2


6.dQdw


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

Q(p4

)= p · p

4− 2 · p

2

16=

p2

8= 80, 000m2


. . . . . .



Q(w) = (p− 2w)(w) = pw− 2w2


6.dQdw


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

Q(p4

)= p · p

4− 2 · p

2

16=

p2

8= 80, 000m2


. . . . . .

Yourturn

Example(Theshortestfence)A 216m2 rectangularpeapatchistobeenclosedbyafenceanddividedintotwoequalpartsbyanotherfenceparalleltooneofitssides. Whatdimensionsfortheouterrectanglewillrequirethesmallesttotallengthoffence? Howmuchfencewillbeneeded?

SolutionLetthelengthandwidthofthepeapatchbe ℓ and w. Theamountoffenceneededis f = 2ℓ+ 3w. Since ℓw = A, aconstant, wehave

f(w) = 2Aw

+ 3w.

Thedomainisallpositivenumbers.

. . . . . .

.

. .

.ℓ

.w

f = 2ℓ+ 3w A = ℓw ≡ 216

. . . . . .

Solution(Continued)Weneedtofindtheminimumvalueof f(w) =

2Aw

+ 3w on

(0,∞).

I Wehavedfdw

= −2Aw2 + 3

whichiszerowhen w =

√2A3.

I Since f′′(w) = 4Aw−3, whichispositiveforallpositive w, thecriticalpointisaminimum, infacttheglobalminimum.

I Sotheareaisminimizedwhen w =

√2A3

= 12 and

ℓ =Aw

=

√3A2

= 18. Theamountoffenceneededis

f

(√2A3

)= 2·

√3A2

+3

√2A3

= 2√6A = 2

√6 · 216 = 72m

. . . . . .

Solution(Continued)Weneedtofindtheminimumvalueof f(w) =

2Aw

+ 3w on

(0,∞).I Wehave

dfdw

= −2Aw2 + 3

whichiszerowhen w =

√2A3.

I Since f′′(w) = 4Aw−3, whichispositiveforallpositive w, thecriticalpointisaminimum, infacttheglobalminimum.

I Sotheareaisminimizedwhen w =

√2A3

= 12 and

ℓ =Aw

=

√3A2

= 18. Theamountoffenceneededis

f

(√2A3

)= 2·

√3A2

+3

√2A3

= 2√6A = 2

√6 · 216 = 72m

. . . . . .

ExampleA Normanwindowhastheoutlineofasemicircleontopofarectangle. Supposethereis 8+ 2π feetofwoodtrimavailable.Findthedimensionsoftherectangleandsemicirclethatwillmaximizetheareaofthewindow.

.

AnswerThedimensionsare4ftby2ft.

. . . . . .

SolutionLet h and w betheheightandwidthofthewindow. Wehave

L = 2h+w+π

2w A = wh+

π

2

(w2

)2If L isfixedtobe 8+ 2π, wehave

h =16+ 4π − 2w− πw

4,

so

A =w4(16+ 4π − 2w− πw)+

π

8w2 = (π+ 4)w−

(12+

π

8

)w2.

So A′ = (π + 4)w−(1+

π

4

), whichiszerowhen

w =π + 41+ π

2= 4 ft. Thedimensionsare4ftby2ft.

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Summary

I RememberthechecklistI Askyourself: whatistheobjective?I Rememberyourgeometry:

I similartrianglesI righttrianglesI trigonometricfunctions

lesson 24: optimization

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