Optimization Direct, Inc., Cplex & Transportation Optimization Problems(Advertisement)
Lesson 19: Optimization Problems
-
Upload
matthew-leingang -
Category
Technology
-
view
14.214 -
download
1
description
Transcript of Lesson 19: Optimization Problems
![Page 1: Lesson 19: Optimization Problems](https://reader034.fdocuments.us/reader034/viewer/2022051209/547cc421b4af9fce158b51bc/html5/thumbnails/1.jpg)
. . . . . .
Section4.6OptimizationProblems
Math1aIntroductiontoCalculus
March19, 2008
Announcements
◮ ProblemSessionsSunday, Thursday, 7pm, SC 310◮ OfficehoursTues, Weds, 2–4pmSC 323
..Image: Flickruser LawrenceOP
![Page 2: Lesson 19: Optimization Problems](https://reader034.fdocuments.us/reader034/viewer/2022051209/547cc421b4af9fce158b51bc/html5/thumbnails/2.jpg)
. . . . . .
Announcements
◮ ProblemSessionsSunday, Thursday, 7pm, SC 310◮ OfficehoursTues, Weds, 2–4pmSC 323◮ Pleasereviewthepolicyonlatehomeworks
![Page 3: Lesson 19: Optimization Problems](https://reader034.fdocuments.us/reader034/viewer/2022051209/547cc421b4af9fce158b51bc/html5/thumbnails/3.jpg)
. . . . . .
Outline
LeadbyExample
TheTextintheBox
MoreExamplesTheBestFencingPlanTheShortestFenceNormanWindows
![Page 4: Lesson 19: Optimization Problems](https://reader034.fdocuments.us/reader034/viewer/2022051209/547cc421b4af9fce158b51bc/html5/thumbnails/4.jpg)
. . . . . .
LeadingbyExample
ExampleWhatistherectangleoffixedperimeterwithmaximumarea?
SolutionDrawarectangle.
.
.
.ℓ
.w
![Page 5: Lesson 19: Optimization Problems](https://reader034.fdocuments.us/reader034/viewer/2022051209/547cc421b4af9fce158b51bc/html5/thumbnails/5.jpg)
. . . . . .
LeadingbyExample
ExampleWhatistherectangleoffixedperimeterwithmaximumarea?
SolutionDrawarectangle.
.
.
.ℓ
.w
![Page 6: Lesson 19: Optimization Problems](https://reader034.fdocuments.us/reader034/viewer/2022051209/547cc421b4af9fce158b51bc/html5/thumbnails/6.jpg)
. . . . . .
LeadingbyExample
ExampleWhatistherectangleoffixedperimeterwithmaximumarea?
SolutionDrawarectangle.
.
.
.ℓ
.w
![Page 7: Lesson 19: Optimization Problems](https://reader034.fdocuments.us/reader034/viewer/2022051209/547cc421b4af9fce158b51bc/html5/thumbnails/7.jpg)
. . . . . .
LeadingbyExample
ExampleWhatistherectangleoffixedperimeterwithmaximumarea?
SolutionDrawarectangle.
.
.
.ℓ
.w
![Page 8: Lesson 19: Optimization Problems](https://reader034.fdocuments.us/reader034/viewer/2022051209/547cc421b4af9fce158b51bc/html5/thumbnails/8.jpg)
. . . . . .
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.
![Page 9: Lesson 19: Optimization Problems](https://reader034.fdocuments.us/reader034/viewer/2022051209/547cc421b4af9fce158b51bc/html5/thumbnails/9.jpg)
. . . . . .
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.
![Page 10: Lesson 19: Optimization Problems](https://reader034.fdocuments.us/reader034/viewer/2022051209/547cc421b4af9fce158b51bc/html5/thumbnails/10.jpg)
. . . . . .
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.
![Page 11: Lesson 19: Optimization Problems](https://reader034.fdocuments.us/reader034/viewer/2022051209/547cc421b4af9fce158b51bc/html5/thumbnails/11.jpg)
. . . . . .
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.
![Page 12: Lesson 19: Optimization Problems](https://reader034.fdocuments.us/reader034/viewer/2022051209/547cc421b4af9fce158b51bc/html5/thumbnails/12.jpg)
. . . . . .
Solution(Concluded)Wehave
dAdr
=12p− 2w,
andthecriticalpointsarewhen
0 =12p− 2w =⇒ w =
p4
.
Sincethisistheonlycriticalpoint, itmustbethemaximum. In
thiscase ℓ =p4aswell. Wehaveasquare!
![Page 13: Lesson 19: Optimization Problems](https://reader034.fdocuments.us/reader034/viewer/2022051209/547cc421b4af9fce158b51bc/html5/thumbnails/13.jpg)
. . . . . .
Solution(Concluded)Wehave
dAdr
=12p− 2w,
andthecriticalpointsarewhen
0 =12p− 2w =⇒ w =
p4
.
Sincethisistheonlycriticalpoint, itmustbethemaximum. In
thiscase ℓ =p4aswell.
Wehaveasquare!
![Page 14: Lesson 19: Optimization Problems](https://reader034.fdocuments.us/reader034/viewer/2022051209/547cc421b4af9fce158b51bc/html5/thumbnails/14.jpg)
. . . . . .
Solution(Concluded)Wehave
dAdr
=12p− 2w,
andthecriticalpointsarewhen
0 =12p− 2w =⇒ w =
p4
.
Sincethisistheonlycriticalpoint, itmustbethemaximum. In
thiscase ℓ =p4aswell. Wehaveasquare!
![Page 15: Lesson 19: Optimization Problems](https://reader034.fdocuments.us/reader034/viewer/2022051209/547cc421b4af9fce158b51bc/html5/thumbnails/15.jpg)
. . . . . .
Outline
LeadbyExample
TheTextintheBox
MoreExamplesTheBestFencingPlanTheShortestFenceNormanWindows
![Page 16: Lesson 19: Optimization Problems](https://reader034.fdocuments.us/reader034/viewer/2022051209/547cc421b4af9fce158b51bc/html5/thumbnails/16.jpg)
. . . . . .
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.
![Page 17: Lesson 19: Optimization Problems](https://reader034.fdocuments.us/reader034/viewer/2022051209/547cc421b4af9fce158b51bc/html5/thumbnails/17.jpg)
. . . . . .
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.
![Page 18: Lesson 19: Optimization Problems](https://reader034.fdocuments.us/reader034/viewer/2022051209/547cc421b4af9fce158b51bc/html5/thumbnails/18.jpg)
. . . . . .
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.
![Page 19: Lesson 19: Optimization Problems](https://reader034.fdocuments.us/reader034/viewer/2022051209/547cc421b4af9fce158b51bc/html5/thumbnails/19.jpg)
. . . . . .
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.
![Page 20: Lesson 19: Optimization Problems](https://reader034.fdocuments.us/reader034/viewer/2022051209/547cc421b4af9fce158b51bc/html5/thumbnails/20.jpg)
. . . . . .
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.
![Page 21: Lesson 19: Optimization Problems](https://reader034.fdocuments.us/reader034/viewer/2022051209/547cc421b4af9fce158b51bc/html5/thumbnails/21.jpg)
. . . . . .
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.
![Page 22: Lesson 19: Optimization Problems](https://reader034.fdocuments.us/reader034/viewer/2022051209/547cc421b4af9fce158b51bc/html5/thumbnails/22.jpg)
. . . . . .
Outline
LeadbyExample
TheTextintheBox
MoreExamplesTheBestFencingPlanTheShortestFenceNormanWindows
![Page 23: Lesson 19: Optimization Problems](https://reader034.fdocuments.us/reader034/viewer/2022051209/547cc421b4af9fce158b51bc/html5/thumbnails/23.jpg)
. . . . . .
AnotherExample
Example(TheBestFencingPlan)A rectangularplotoffarmlandwillbeboundedononesidebyariverandontheotherthreesidesbyasingle-strandelectricfence. With800mofwireatyourdisposal, whatisthelargestareayoucanenclose, andwhatareitsdimensions?
![Page 24: Lesson 19: Optimization Problems](https://reader034.fdocuments.us/reader034/viewer/2022051209/547cc421b4af9fce158b51bc/html5/thumbnails/24.jpg)
. . . . . .
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.
![Page 25: Lesson 19: Optimization Problems](https://reader034.fdocuments.us/reader034/viewer/2022051209/547cc421b4af9fce158b51bc/html5/thumbnails/25.jpg)
. . . . . .
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.
![Page 26: Lesson 19: Optimization Problems](https://reader034.fdocuments.us/reader034/viewer/2022051209/547cc421b4af9fce158b51bc/html5/thumbnails/26.jpg)
. . . . . .
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.
![Page 27: Lesson 19: Optimization Problems](https://reader034.fdocuments.us/reader034/viewer/2022051209/547cc421b4af9fce158b51bc/html5/thumbnails/27.jpg)
. . . . . .
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.
![Page 28: Lesson 19: Optimization Problems](https://reader034.fdocuments.us/reader034/viewer/2022051209/547cc421b4af9fce158b51bc/html5/thumbnails/28.jpg)
. . . . . .
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.
![Page 29: Lesson 19: Optimization Problems](https://reader034.fdocuments.us/reader034/viewer/2022051209/547cc421b4af9fce158b51bc/html5/thumbnails/29.jpg)
. . . . . .
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.
![Page 30: Lesson 19: Optimization Problems](https://reader034.fdocuments.us/reader034/viewer/2022051209/547cc421b4af9fce158b51bc/html5/thumbnails/30.jpg)
. . . . . .
Yourturn
Example(Theshortestfence)A 216m2 rectangularpeapatchistobeenclosedbyafenceanddividedintotwoequalpartsbyanotherfenceparalleltooneofitssides. Whatdimensionsfortheouterrectanglewillrequirethesmallesttotallengthoffence? Howmuchfencewillbeneeded?
AnswerThedimensionsofthefenceare 12m× 18m andamountoffencerequiredis 72m2.
![Page 31: Lesson 19: Optimization Problems](https://reader034.fdocuments.us/reader034/viewer/2022051209/547cc421b4af9fce158b51bc/html5/thumbnails/31.jpg)
. . . . . .
Yourturn
Example(Theshortestfence)A 216m2 rectangularpeapatchistobeenclosedbyafenceanddividedintotwoequalpartsbyanotherfenceparalleltooneofitssides. Whatdimensionsfortheouterrectanglewillrequirethesmallesttotallengthoffence? Howmuchfencewillbeneeded?
AnswerThedimensionsofthefenceare 12m× 18m andamountoffencerequiredis 72m2.
![Page 32: Lesson 19: Optimization Problems](https://reader034.fdocuments.us/reader034/viewer/2022051209/547cc421b4af9fce158b51bc/html5/thumbnails/32.jpg)
. . . . . .
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.
![Page 33: Lesson 19: Optimization Problems](https://reader034.fdocuments.us/reader034/viewer/2022051209/547cc421b4af9fce158b51bc/html5/thumbnails/33.jpg)
. . . . . .
ExampleA Normanwindowhastheoutlineofasemicircleontopofarectangle. Supposethereis 8 + 4π feetofwoodtrimavailable.Discusswhyawindowdesignermightwanttomaximizetheareaofthewindow. Findthedimensionsoftherectangleandsemicirclethatwillmaximizetheareaofthewindow.
.
AnswerThedimensionsare2ftby4ft.
![Page 34: Lesson 19: Optimization Problems](https://reader034.fdocuments.us/reader034/viewer/2022051209/547cc421b4af9fce158b51bc/html5/thumbnails/34.jpg)
. . . . . .
ExampleA Normanwindowhastheoutlineofasemicircleontopofarectangle. Supposethereis 8 + 4π feetofwoodtrimavailable.Discusswhyawindowdesignermightwanttomaximizetheareaofthewindow. Findthedimensionsoftherectangleandsemicirclethatwillmaximizetheareaofthewindow.
.
AnswerThedimensionsare2ftby4ft.
![Page 35: Lesson 19: Optimization Problems](https://reader034.fdocuments.us/reader034/viewer/2022051209/547cc421b4af9fce158b51bc/html5/thumbnails/35.jpg)
. . . . . .
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.