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We know the procedure for optimization. When applied to more real-world problems there is a wealth of questions to answer.
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. . . . . .
Section4.5OptimizationProblems
V63.0121, CalculusI
April5, 2009
Announcements
I Quiz5isnextweek, coveringSections4.1–4.4I I ammovingtoWWH 624sometimenextweek(April13th)I HappyOpeningDay!
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. . . . . .
OfficeHoursandotherhelpInadditiontorecitation
Day Time Who/What WhereinWWHM 1:00–2:00 LeingangOH 718/624
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F 2:00–4:00 CurtoOH 1310I ammovingtoWWH 624sometimenextweek(April13th)
. . . . . .
Outline
LeadingbyExample
TheTextintheBox
MoreExamples
. . . . . .
LeadingbyExample
ExampleWhatistherectangleoffixedperimeterwithmaximumarea?
Solution
I Drawarectangle.
.
.
.ℓ
.w
. . . . . .
LeadingbyExample
ExampleWhatistherectangleoffixedperimeterwithmaximumarea?
Solution
I Drawarectangle.
.
.
.ℓ
.w
. . . . . .
LeadingbyExample
ExampleWhatistherectangleoffixedperimeterwithmaximumarea?
Solution
I Drawarectangle.
.
.
.ℓ
.w
. . . . . .
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)
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)
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)
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)
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] (wewanttomakesure A(w) ≥ 0).
. . . . . .
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(Concluded)WeusetheClosedIntervalMethodfor A(w) =
12pw−w2 on
[0,p/2].I Attheendpoints, A(0) = A(p/2) = 0.
I Tofindthecriticalpoints, wefinddAdr
=12p− 2w.
I Thecriticalpointsarewhen
0 =12p− 2w =⇒ w =
p4
I Sincethisistheonlycriticalpoint, itmustbethemaximum.
Inthiscase ℓ =p4aswell.
I Wehaveasquare! Themaximalareais A(p/4) = p2/16.
. . . . . .
Solution(Concluded)WeusetheClosedIntervalMethodfor A(w) =
12pw−w2 on
[0,p/2].I Attheendpoints, A(0) = A(p/2) = 0.
I Tofindthecriticalpoints, wefinddAdr
=12p− 2w.
I Thecriticalpointsarewhen
0 =12p− 2w =⇒ w =
p4
I Sincethisistheonlycriticalpoint, itmustbethemaximum.
Inthiscase ℓ =p4aswell.
I Wehaveasquare! Themaximalareais A(p/4) = p2/16.
. . . . . .
Solution(Concluded)WeusetheClosedIntervalMethodfor A(w) =
12pw−w2 on
[0,p/2].I Attheendpoints, A(0) = A(p/2) = 0.
I Tofindthecriticalpoints, wefinddAdr
=12p− 2w.
I Thecriticalpointsarewhen
0 =12p− 2w =⇒ w =
p4
I Sincethisistheonlycriticalpoint, itmustbethemaximum.
Inthiscase ℓ =p4aswell.
I Wehaveasquare! Themaximalareais A(p/4) = p2/16.
. . . . . .
Solution(Concluded)WeusetheClosedIntervalMethodfor A(w) =
12pw−w2 on
[0,p/2].I Attheendpoints, A(0) = A(p/2) = 0.
I Tofindthecriticalpoints, wefinddAdr
=12p− 2w.
I Thecriticalpointsarewhen
0 =12p− 2w =⇒ w =
p4
I Sincethisistheonlycriticalpoint, itmustbethemaximum.
Inthiscase ℓ =p4aswell.
I Wehaveasquare! Themaximalareais A(p/4) = p2/16.
. . . . . .
Solution(Concluded)WeusetheClosedIntervalMethodfor A(w) =
12pw−w2 on
[0,p/2].I Attheendpoints, A(0) = A(p/2) = 0.
I Tofindthecriticalpoints, wefinddAdr
=12p− 2w.
I Thecriticalpointsarewhen
0 =12p− 2w =⇒ w =
p4
I Sincethisistheonlycriticalpoint, itmustbethemaximum.
Inthiscase ℓ =p4aswell.
I Wehaveasquare! Themaximalareais A(p/4) = p2/16.
. . . . . .
Outline
LeadingbyExample
TheTextintheBox
MoreExamples
. . . . . .
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.
. . . . . .
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.
. . . . . .
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.
. . . . . .
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.
. . . . . .
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.
. . . . . .
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.
. . . . . .
TheClosedIntervalMethod
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.
. . . . . .
TheFirstDerivativeTest
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.
. . . . . .
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).
. . . . . .
Whichtousewhen?
CIM 1DT 2DTPro no need for in-
equalitiesgets global ex-trema automati-cally
w o r k s o nnon-closed,non-boundedintervalsonlyonederiva-tive
w o r k s o nnon-closed,non-boundedintervalsno need for in-equalities
Con only for closedbounded inter-vals
UsesinequalitiesMore work atboundary thanCIM
Morederivativesless conclusivethan1DTmore work atboundary thanCIM
. . . . . .
Whichtousewhen? Thebottomline
I UseCIM ifitapplies: thedomainisaclosed, boundedinterval
I Ifdomainisnotclosedornotbounded, use2DT ifyouliketotakederivatives, or1DT ifyouliketocomparesigns.
. . . . . .
Outline
LeadingbyExample
TheTextintheBox
MoreExamples
. . . . . .
AnotherExample
Example(TheBestFencingPlan)A rectangularplotoffarmlandwillbeboundedononesidebyariverandontheotherthreesidesbyasingle-strandelectricfence. With800mofwireatyourdisposal, whatisthelargestareayoucanenclose, andwhatareitsdimensions?
I Known: amountoffenceusedI Unknown: areaenclosedI Objective: maximizeareaI Constraint: fixedfencelength
. . . . . .
Solution
1. 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
= p− 4w, whichiszerowhen w =p4.
Q(p4
)= p · p
4− 2 · p
2
16=
p2
8= 80000m2
Since Q(0) = Q(p/2) = 0, thiscriticalpointisamaximum.
. . . . . .
AnotherExample
Example(TheBestFencingPlan)A rectangularplotoffarmlandwillbeboundedononesidebyariverandontheotherthreesidesbyasingle-strandelectricfence. With800mofwireatyourdisposal, whatisthelargestareayoucanenclose, andwhatareitsdimensions?
I Known: amountoffenceusedI Unknown: areaenclosedI Objective: maximizeareaI Constraint: fixedfencelength
. . . . . .
AnotherExample
Example(TheBestFencingPlan)A rectangularplotoffarmlandwillbeboundedononesidebyariverandontheotherthreesidesbyasingle-strandelectricfence. With800mofwireatyourdisposal, whatisthelargestareayoucanenclose, andwhatareitsdimensions?
I Known: amountoffenceusedI Unknown: areaenclosed
I Objective: maximizeareaI Constraint: fixedfencelength
. . . . . .
AnotherExample
Example(TheBestFencingPlan)A rectangularplotoffarmlandwillbeboundedononesidebyariverandontheotherthreesidesbyasingle-strandelectricfence. With800mofwireatyourdisposal, whatisthelargestareayoucanenclose, andwhatareitsdimensions?
I Known: amountoffenceusedI Unknown: areaenclosedI Objective: maximizeareaI Constraint: fixedfencelength
. . . . . .
Solution
1. 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
= p− 4w, whichiszerowhen w =p4.
Q(p4
)= p · p
4− 2 · p
2
16=
p2
8= 80000m2
Since Q(0) = Q(p/2) = 0, thiscriticalpointisamaximum.
. . . . . .
Solution
1. 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
= p− 4w, whichiszerowhen w =p4.
Q(p4
)= p · p
4− 2 · p
2
16=
p2
8= 80000m2
Since Q(0) = Q(p/2) = 0, thiscriticalpointisamaximum.
. . . . . .
Diagram
.
.
.
.
.w
.ℓ
. . . . . .
Solution
1. 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
= p− 4w, whichiszerowhen w =p4.
Q(p4
)= p · p
4− 2 · p
2
16=
p2
8= 80000m2
Since Q(0) = Q(p/2) = 0, thiscriticalpointisamaximum.
. . . . . .
Solution
1. 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
= p− 4w, whichiszerowhen w =p4.
Q(p4
)= p · p
4− 2 · p
2
16=
p2
8= 80000m2
Since Q(0) = Q(p/2) = 0, thiscriticalpointisamaximum.
. . . . . .
Diagram
.
.
.
.
.w
.ℓ
. . . . . .
Diagram
.
.
.
.
.w
.ℓ
. . . . . .
Solution
1. 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
= p− 4w, whichiszerowhen w =p4.
Q(p4
)= p · p
4− 2 · p
2
16=
p2
8= 80000m2
Since Q(0) = Q(p/2) = 0, thiscriticalpointisamaximum.
. . . . . .
Solution
1. 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
= p− 4w, whichiszerowhen w =p4.
Q(p4
)= p · p
4− 2 · p
2
16=
p2
8= 80000m2
Since Q(0) = Q(p/2) = 0, thiscriticalpointisamaximum.
. . . . . .
Solution
1. 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
= p− 4w, whichiszerowhen w =p4.
Q(p4
)= p · p
4− 2 · p
2
16=
p2
8= 80000m2
Since Q(0) = Q(p/2) = 0, thiscriticalpointisamaximum.
. . . . . .
Solution
1. 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
= p− 4w, whichiszerowhen w =p4.
Q(p4
)= p · p
4− 2 · p
2
16=
p2
8= 80000m2
Since Q(0) = Q(p/2) = 0, thiscriticalpointisamaximum.
. . . . . .
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)So
dfdw
= −2Aw2 + 3
whichiszerowhen w =
√2A3.
Since f′′(w) = 4Aw−3, whichispositiveforallpositive w, thecriticalpointisaminimum, infacttheglobalminimum.
Sotheareaisminimizedwhen w =
√2A3
= 12 and
ℓ =Aw
=
√3A2
= 18. Theamountoffenceneededis
f
(√2A3
)= 2 ·
√2A2
+ 3
√2A3
= 2√6A = 2
√6 · 216 = 72m
. . . . . .
ExampleA Normanwindowhastheoutlineofasemicircleontopofarectangle. Supposethereis 8 + 2π feetofwoodtrimavailable.Findthedimensionsoftherectangleandsemicirclethatwillmaximizetheareaofthewindow.
.
AnswerThedimensionsare4ftby2ft.
. . . . . .
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.
. . . . . .
Summary
I RememberthechecklistI Askyourself: whatistheobjective?I Rememberyourgeometry:
I similartrianglesI righttrianglesI trigonometricfunctions