. . . . . .
Section4.6MoreOptimizationProblems
Math1aIntroductiontoCalculus
March21, 2008
Announcements
◮ ProblemSessionsSunday, Thursday, 7pm, SC 310(notnextweek)
◮ OfficehoursTues, Weds, 2–4pmSC 323(notnextweek)
..Image: Flickruser glassbeat
. . . . . .
Announcements
◮ ProblemSessionsSunday, Thursday, 7pm, SC 310(notnextweek)
◮ OfficehoursTues, Weds, 2–4pmSC 323(notnextweek)
. . . . . .
Outline
Modeling
TheTextintheBox
MoreExamplesShortestFenceNormanWindows
WorksheetTwo-literbottlesTheStatueofLiberty
. . . . . .
TheModelingProcess
...Real-WorldProblems
..Mathematical
Model
..Real-WorldPredictions
..MathematicalConclusions
.formulate
.solve
.predict
.test
. . . . . .
Outline
Modeling
TheTextintheBox
MoreExamplesShortestFenceNormanWindows
WorksheetTwo-literbottlesTheStatueofLiberty
. . . . . .
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
Modeling
TheTextintheBox
MoreExamplesShortestFenceNormanWindows
WorksheetTwo-literbottlesTheStatueofLiberty
. . . . . .
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 + π feetofwoodtrimavailable.Discusswhyawindowdesignermightwanttomaximizetheareaofthewindow. Findthedimensionsoftherectangleandsemicirclethatwillmaximizetheareaofthewindow.
.
AnswerThedimensionsoftherectangularpartare2ftby4ft.
. . . . . .
ExampleA Normanwindowhastheoutlineofasemicircleontopofarectangle. Supposethereis 8 + π feetofwoodtrimavailable.Discusswhyawindowdesignermightwanttomaximizetheareaofthewindow. Findthedimensionsoftherectangleandsemicirclethatwillmaximizetheareaofthewindow.
.
AnswerThedimensionsoftherectangularpartare2ftby4ft.
. . . . . .
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.
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Outline
Modeling
TheTextintheBox
MoreExamplesShortestFenceNormanWindows
WorksheetTwo-literbottlesTheStatueofLiberty
. . . . . .
Worksheet
.
.Image: ErickCifuentes
. . . . . .
Two-literbottles
A two-litersodabottleisroughlyshapedlikeacylinderwithasphericalcap, andismadefromaplasticcalledpolyethyleneterephthalate(PET).Itsvolumeisfixedattwoliters.Whatdimensionsofthebottlewillminimizethecostofproduction?
. . . . . .
Solution I
Thevolumeofsuchabottleis
V = πr2h + 23πr
3,
whichisfixed. Thus
h =V− 2
3πr3
πr2=
Vπr2
− 23 r.
Theobjectivefunctionisthesurfacearea(sincethematerialisallthesame, costisproportionaltomaterialsused)
A = 2πrh + 3πr2 = 2πr
(V− 2
3πr2
πr2
)+ 3πr2 =
2Vr
+53πr2.
. . . . . .
Solution II
Thedomainofthisfunctionis (0,∞). Tofindthecriticalpointsweneedtofind
dAdr
= −2Vr2
+103
πr =−2V + 10
3 πr3
r2.
ThecriticalpointsarewhendAdr
= 0, or
0 = −2V +103
πr3
=⇒ r =
(3V5π
)1/3
.
. . . . . .
Solution III
Substitutingintoourexpressionfor h tellsus(afteralotofalgebra)that h = r. That’sdefinitelyamuchsquatterbottlethatweseeinthestores. Soit’snotmaterialcoststhatthey’reminimizing(unlessourshapeistoooff).
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TheStatueofLiberty
TheStatueofLibertystandsontopofapedestalwhichisontopofonoldfort. Thetopofthepedestalis47mabovegroundlevel.Thestatueitselfmeasures46mfromthetopofthepedestaltothetipofthetorch.Whatdistanceshouldonestandawayfromthestatueinordertomaximizetheviewofthestatue? Thatis, whatdistancewillmaximizetheportionoftheviewer’svisiontakenupbythestatue?
. . . . . .
Model I
Theanglesubtendedbythestatueintheviewer’seyecanbeexpressedas
θ = arctan(a + bx
)−arctan
(bx
).
a
bθ
x
. . . . . .
Solution I
Maximizing θ withrespectto x isasimplematterofdifferentiation:
dθ
dx=
1
1 +(a+bx
)2 · −(a + b)
x2− 1
1 +(bx
)2 · −bx2
=b
x2 + b2− a + b
x2 + (a + b)2
=
[x2 + (a + b)2
]b− (a + b)
[x2 + b2
](x2 + b2) [x2 + (a + b)2]
. . . . . .
Solution II
Thisderivativeiszeroifandonlyifthenumeratoriszero, soweseek x suchthat
0 =[x2 + (a + b)2
]b− (a + b)
[x2 + b2
]= a(ab + b2 − x2),
orx =
√b(a + b).
Usingthefirstderivativetest, weseethat dθ/dx > 0 if0 < x <
√b(a + b) and dθ/dx < 0 if x >
√b(a + b). Sothisis
definitelytheabsolutemaximumon (0,∞).
. . . . . .
AnalysisandDiscussion
Ifwesubstituteinthenumericaldimensionsgiven, wehave
x =√
(46)(93) ≈ 66.1 meters
Thisdistancewouldputyouprettyclosetothefrontoftheoldfortwhichliesatthebaseoftheisland. Unfortunately, you’renotallowedtowalkonthispartofthelawn.Thelength
√b(a + b) isthe geometricmean ofthetwodistances
measurefromtheground—tothetopofthepedestal(a)andthetopofthestatue(a+ b). Thegeometricmeanisoftwonumbersisalwaysbetweenthemandgreaterthanorequaltotheiraverage.
