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Section4.6OptimizationProblems
Math1aIntroductiontoCalculus
March19, 2008
Announcements
◮ ProblemSessionsSunday, Thursday, 7pm, SC 310◮ OfficehoursTues, Weds, 2–4pmSC 323
..Image: Flickruser LawrenceOP
. . . . . .
Announcements
◮ ProblemSessionsSunday, Thursday, 7pm, SC 310◮ OfficehoursTues, Weds, 2–4pmSC 323◮ Pleasereviewthepolicyonlatehomeworks
. . . . . .
Outline
LeadbyExample
TheTextintheBox
MoreExamplesTheBestFencingPlanTheShortestFenceNormanWindows
. . . . . .
LeadingbyExample
ExampleWhatistherectangleoffixedperimeterwithmaximumarea?
SolutionDrawarectangle.
.
.
.ℓ
.w
. . . . . .
LeadingbyExample
ExampleWhatistherectangleoffixedperimeterwithmaximumarea?
SolutionDrawarectangle.
.
.
.ℓ
.w
. . . . . .
LeadingbyExample
ExampleWhatistherectangleoffixedperimeterwithmaximumarea?
SolutionDrawarectangle.
.
.
.ℓ
.w
. . . . . .
LeadingbyExample
ExampleWhatistherectangleoffixedperimeterwithmaximumarea?
SolutionDrawarectangle.
.
.
.ℓ
.w
. . . . . .
Solution(Continued)Letitslengthbe ℓ anditswidthbe w. Theobjectivefunctionisarea A = ℓw.
Thisisafunctionoftwovariables, soweneedtousetheconstraintthattheperimeterisfixed. Since p = 2ℓ + 2w, wehave
ℓ =p− 2w
2,
so
A = ℓw =p− 2w
2·w =
12(p− 2w)(w) =
12pw−w2
Nowwehave A asafunctionof w alone(p isconstant). Thenaturaldomainofthisfunctionis [0,p/2]. Bothoftheseendpointswouldresultinadegeneraterectangle—alinesegment—ofzeroarea.
. . . . . .
Solution(Continued)Letitslengthbe ℓ anditswidthbe w. Theobjectivefunctionisarea A = ℓw.Thisisafunctionoftwovariables, soweneedtousetheconstraintthattheperimeterisfixed. Since p = 2ℓ + 2w, wehave
ℓ =p− 2w
2,
so
A = ℓw =p− 2w
2·w =
12(p− 2w)(w) =
12pw−w2
Nowwehave A asafunctionof w alone(p isconstant). Thenaturaldomainofthisfunctionis [0,p/2]. Bothoftheseendpointswouldresultinadegeneraterectangle—alinesegment—ofzeroarea.
. . . . . .
Solution(Continued)Letitslengthbe ℓ anditswidthbe w. Theobjectivefunctionisarea A = ℓw.Thisisafunctionoftwovariables, soweneedtousetheconstraintthattheperimeterisfixed. Since p = 2ℓ + 2w, wehave
ℓ =p− 2w
2,
so
A = ℓw =p− 2w
2·w =
12(p− 2w)(w) =
12pw−w2
Nowwehave A asafunctionof w alone(p isconstant). Thenaturaldomainofthisfunctionis [0,p/2]. Bothoftheseendpointswouldresultinadegeneraterectangle—alinesegment—ofzeroarea.
. . . . . .
Solution(Continued)Letitslengthbe ℓ anditswidthbe w. Theobjectivefunctionisarea A = ℓw.Thisisafunctionoftwovariables, soweneedtousetheconstraintthattheperimeterisfixed. Since p = 2ℓ + 2w, wehave
ℓ =p− 2w
2,
so
A = ℓw =p− 2w
2·w =
12(p− 2w)(w) =
12pw−w2
Nowwehave A asafunctionof w alone(p isconstant). Thenaturaldomainofthisfunctionis [0,p/2]. Bothoftheseendpointswouldresultinadegeneraterectangle—alinesegment—ofzeroarea.
. . . . . .
Solution(Concluded)Wehave
dAdr
=12p− 2w,
andthecriticalpointsarewhen
0 =12p− 2w =⇒ w =
p4
.
Sincethisistheonlycriticalpoint, itmustbethemaximum. In
thiscase ℓ =p4aswell. Wehaveasquare!
. . . . . .
Solution(Concluded)Wehave
dAdr
=12p− 2w,
andthecriticalpointsarewhen
0 =12p− 2w =⇒ w =
p4
.
Sincethisistheonlycriticalpoint, itmustbethemaximum. In
thiscase ℓ =p4aswell.
Wehaveasquare!
. . . . . .
Solution(Concluded)Wehave
dAdr
=12p− 2w,
andthecriticalpointsarewhen
0 =12p− 2w =⇒ w =
p4
.
Sincethisistheonlycriticalpoint, itmustbethemaximum. In
thiscase ℓ =p4aswell. Wehaveasquare!
. . . . . .
Outline
LeadbyExample
TheTextintheBox
MoreExamplesTheBestFencingPlanTheShortestFenceNormanWindows
. . . . . .
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.
. . . . . .
Outline
LeadbyExample
TheTextintheBox
MoreExamplesTheBestFencingPlanTheShortestFenceNormanWindows
. . . . . .
AnotherExample
Example(TheBestFencingPlan)A rectangularplotoffarmlandwillbeboundedononesidebyariverandontheotherthreesidesbyasingle-strandelectricfence. With800mofwireatyourdisposal, whatisthelargestareayoucanenclose, andwhatareitsdimensions?
. . . . . .
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.
. . . . . .
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?
AnswerThedimensionsofthefenceare 12m× 18m andamountoffencerequiredis 72m2.
. . . . . .
Yourturn
Example(Theshortestfence)A 216m2 rectangularpeapatchistobeenclosedbyafenceanddividedintotwoequalpartsbyanotherfenceparalleltooneofitssides. Whatdimensionsfortheouterrectanglewillrequirethesmallesttotallengthoffence? Howmuchfencewillbeneeded?
AnswerThedimensionsofthefenceare 12m× 18m andamountoffencerequiredis 72m2.
. . . . . .
SolutionLetthelengthandwidthofthepeapatchbe ℓ and w. Theamountoffenceneededis f = 2ℓ + 3w Since ℓw = A, aconstant,wehave
f(w) = 2Aw
+ 3w.
Sodfdw
= −2Aw2 + 3
whichiszerowhen w =
√2A3. Since lim
w→0+f(w) and lim
w→∞f(w)
areboth ∞, thecriticalpointisaminimum. Sotheareais
minimizedwhen w =
√2A3
= 12 and ℓ =Aw
=
√3A2
= 18. The
amountoffenceneededis
f
(√2A3
)= 2 ·
√2A2
+ 3
√2A3
= 2√6A = 2
√6 · 216 = 72 m.
. . . . . .
ExampleA Normanwindowhastheoutlineofasemicircleontopofarectangle. Supposethereis 8 + 4π feetofwoodtrimavailable.Discusswhyawindowdesignermightwanttomaximizetheareaofthewindow. Findthedimensionsoftherectangleandsemicirclethatwillmaximizetheareaofthewindow.
.
AnswerThedimensionsare2ftby4ft.
. . . . . .
ExampleA Normanwindowhastheoutlineofasemicircleontopofarectangle. Supposethereis 8 + 4π feetofwoodtrimavailable.Discusswhyawindowdesignermightwanttomaximizetheareaofthewindow. Findthedimensionsoftherectangleandsemicirclethatwillmaximizetheareaofthewindow.
.
AnswerThedimensionsare2ftby4ft.
. . . . . .
SolutionWehavetomaximize A = ℓw + (w/2)2π subjecttotheconstraintthat 2ℓ + w + πw = p. Solvingfor ℓ intermsof w gives
ℓ = 12(p−w− πw)
So A = 12w(p−w− πw) + 1
4πw2. Differentiatinggives
A′(w) =πw2
+12(−1− π)w +
12(p− πw−w)
whichiszerowhen w =p
2 + π. If p = 8 + 4π, w = 4. Itfollows
that ℓ = 2.
