Lesson 20: (More) Optimization Problems

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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

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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

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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

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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.

. . . . . .

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

.

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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?

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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,∞).

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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.