. . . . . .

Section4.5OptimizationII

V63.0121.034, CalculusI

November25, 2009

Announcements

I FinalExam, December18, 2:00–3:50pm

. . . . . .

Outline

Recall

MoreexamplesAdditionDistanceTrianglesEconomicsTheStatueofLiberty

. . . . . .

Checklistforoptimizationproblems

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.

. . . . . .

Recall: TheClosedIntervalMethodSeeSection4.1

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.

. . . . . .

Recall: TheFirstDerivativeTestSeeSection4.3

Theorem(TheFirstDerivativeTest)Let f beacontinuousfunctionand c acriticalpointof f in (a,b).

I If f′(x) > 0 on (a, c) (i.e., “before c”)and f′(x) < 0 on (c,b)(i.e., “after c”, then c isalocalmaximumfor f.

I If f′(x) < 0 on (a, c) and f′(x) > 0 on (c,b), then c isalocalminimumfor f.

I If f′(x) hasthesamesignon (a, c) and (c,b), then c isnotalocalextremum.

. . . . . .

Recall: TheSecondDerivativeTestSeeSection4.3

Theorem(TheSecondDerivativeTest)Let f, f′, and f′′ becontinuous. Let c bebeapointin (a,b) withf′(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? Thebottomline

I UseCIM ifitapplies: thedomainisaclosed, boundedinterval

I Ifdomainisnotclosedornotbounded, use2DT ifyouliketotakederivatives, or1DT ifyouliketocomparesigns.

. . . . . .

Outline

Recall

MoreexamplesAdditionDistanceTrianglesEconomicsTheStatueofLiberty

. . . . . .

Additionwithaconstraint

ExampleFindtwopositivenumbers x and y with xy = 16 and x+ y assmallaspossible.

Solution

I Objective: minimize S = x+ y subjecttotheconstraintthatxy = 16

I Eliminate y: y = 16/x so S = x+ 16/x. Thedomainofconsiderationis (0,∞).

I Findthecriticalpoints: S′(x) = 1− 16/x2, whichis 0 whenx = 4.

I Classifythecriticalpoints: S′′(x) = 32/x3, whichisalwayspositive. Sothegraphisalwaysconcaveup, 4 isalocalmin,andthereforetheglobalmin.

I Sothenumbersare x = y = 4, Smin = 8.

. . . . . .

Additionwithaconstraint

ExampleFindtwopositivenumbers x and y with xy = 16 and x+ y assmallaspossible.

Solution

I Objective: minimize S = x+ y subjecttotheconstraintthatxy = 16

I Eliminate y: y = 16/x so S = x+ 16/x. Thedomainofconsiderationis (0,∞).

I Findthecriticalpoints: S′(x) = 1− 16/x2, whichis 0 whenx = 4.

I Classifythecriticalpoints: S′′(x) = 32/x3, whichisalwayspositive. Sothegraphisalwaysconcaveup, 4 isalocalmin,andthereforetheglobalmin.

I Sothenumbersare x = y = 4, Smin = 8.

. . . . . .

Additionwithaconstraint

ExampleFindtwopositivenumbers x and y with xy = 16 and x+ y assmallaspossible.

Solution

I Objective: minimize S = x+ y subjecttotheconstraintthatxy = 16

I Eliminate y: y = 16/x so S = x+ 16/x. Thedomainofconsiderationis (0,∞).

I Findthecriticalpoints: S′(x) = 1− 16/x2, whichis 0 whenx = 4.

I Classifythecriticalpoints: S′′(x) = 32/x3, whichisalwayspositive. Sothegraphisalwaysconcaveup, 4 isalocalmin,andthereforetheglobalmin.

I Sothenumbersare x = y = 4, Smin = 8.

. . . . . .

Additionwithaconstraint

ExampleFindtwopositivenumbers x and y with xy = 16 and x+ y assmallaspossible.

Solution

I Objective: minimize S = x+ y subjecttotheconstraintthatxy = 16

I Eliminate y: y = 16/x so S = x+ 16/x. Thedomainofconsiderationis (0,∞).

I Findthecriticalpoints: S′(x) = 1− 16/x2, whichis 0 whenx = 4.

I Classifythecriticalpoints: S′′(x) = 32/x3, whichisalwayspositive. Sothegraphisalwaysconcaveup, 4 isalocalmin,andthereforetheglobalmin.

I Sothenumbersare x = y = 4, Smin = 8.

. . . . . .

Additionwithaconstraint

ExampleFindtwopositivenumbers x and y with xy = 16 and x+ y assmallaspossible.

Solution

I Objective: minimize S = x+ y subjecttotheconstraintthatxy = 16

I Eliminate y: y = 16/x so S = x+ 16/x. Thedomainofconsiderationis (0,∞).

I Findthecriticalpoints: S′(x) = 1− 16/x2, whichis 0 whenx = 4.

I Classifythecriticalpoints: S′′(x) = 32/x3, whichisalwayspositive. Sothegraphisalwaysconcaveup, 4 isalocalmin,andthereforetheglobalmin.

I Sothenumbersare x = y = 4, Smin = 8.

. . . . . .

Additionwithaconstraint

ExampleFindtwopositivenumbers x and y with xy = 16 and x+ y assmallaspossible.

Solution

I Objective: minimize S = x+ y subjecttotheconstraintthatxy = 16

I Eliminate y: y = 16/x so S = x+ 16/x. Thedomainofconsiderationis (0,∞).

I Findthecriticalpoints: S′(x) = 1− 16/x2, whichis 0 whenx = 4.

I Classifythecriticalpoints: S′′(x) = 32/x3, whichisalwayspositive. Sothegraphisalwaysconcaveup, 4 isalocalmin,andthereforetheglobalmin.

I Sothenumbersare x = y = 4, Smin = 8.

. . . . . .

Distance

ExampleFindthepoint P ontheparabola y = x2 closesttothepoint (3, 0).

Solution

Thedistancebetween (x, x2)and (3, 0) isgivenby

f(x) =√

(x− 3)2 + (x2 − 0)2

Wemayinsteadminimizethe square of f:

g(x) = f(x)2 = (x− 3)2 + x4

Thedomainis (−∞,∞).

. .x

.y

..(x, x2)

..3

. . . . . .

Distance

ExampleFindthepoint P ontheparabola y = x2 closesttothepoint (3, 0).

Solution

Thedistancebetween (x, x2)and (3, 0) isgivenby

f(x) =√

(x− 3)2 + (x2 − 0)2

Wemayinsteadminimizethe square of f:

g(x) = f(x)2 = (x− 3)2 + x4

Thedomainis (−∞,∞).

. .x

.y

..(x, x2)

..3

. . . . . .

Distance

ExampleFindthepoint P ontheparabola y = x2 closesttothepoint (3, 0).

SolutionThedistancebetween (x, x2)and (3, 0) isgivenby

f(x) =√

(x− 3)2 + (x2 − 0)2

Wemayinsteadminimizethe square of f:

g(x) = f(x)2 = (x− 3)2 + x4

Thedomainis (−∞,∞).

. .x

.y

..(x, x2)

..3

. . . . . .

Distance

ExampleFindthepoint P ontheparabola y = x2 closesttothepoint (3, 0).

SolutionThedistancebetween (x, x2)and (3, 0) isgivenby

f(x) =√

(x− 3)2 + (x2 − 0)2

Wemayinsteadminimizethe square of f:

g(x) = f(x)2 = (x− 3)2 + x4

Thedomainis (−∞,∞).

. .x

.y

..(x, x2)

..3

. . . . . .

Distance

ExampleFindthepoint P ontheparabola y = x2 closesttothepoint (3, 0).

SolutionThedistancebetween (x, x2)and (3, 0) isgivenby

f(x) =√

(x− 3)2 + (x2 − 0)2

Wemayinsteadminimizethe square of f:

g(x) = f(x)2 = (x− 3)2 + x4

Thedomainis (−∞,∞).. .x

.y

..(x, x2)

..3

. . . . . .

Distanceproblemminimizationstep

Wewanttofindtheglobalminimumof g(x) = (x− 3)2 + x4.

I g′(x) = 2(x− 3) + 4x3 = 4x3 + 2x− 6 = 2(2x3 + x− 3)I Ifapolynomialhasintegerroots, theyarefactorsoftheconstantterm(Euler)

I 1 isaroot, so 2x3 + x− 3 isdivisibleby x− 1:

f′(x) = 2(2x3 + x− 3) = 2(x− 1)(2x2 + 2x+ 3)

Thequadratichasnorealroots(thediscriminantb2 − 4ac < 0)

I Wesee f′(1) = 0, f′(x) < 0 if x < 1, and f′(x) > 0 if x > 1. So1 istheglobalminimum.

I Thepointontheparabolaclosestto (3, 0) is (1, 1). Theminimumdistanceis

√5.

. . . . . .

Distanceproblemminimizationstep

Wewanttofindtheglobalminimumof g(x) = (x− 3)2 + x4.I g′(x) = 2(x− 3) + 4x3 = 4x3 + 2x− 6 = 2(2x3 + x− 3)

I Ifapolynomialhasintegerroots, theyarefactorsoftheconstantterm(Euler)

I 1 isaroot, so 2x3 + x− 3 isdivisibleby x− 1:

f′(x) = 2(2x3 + x− 3) = 2(x− 1)(2x2 + 2x+ 3)

Thequadratichasnorealroots(thediscriminantb2 − 4ac < 0)

I Wesee f′(1) = 0, f′(x) < 0 if x < 1, and f′(x) > 0 if x > 1. So1 istheglobalminimum.

I Thepointontheparabolaclosestto (3, 0) is (1, 1). Theminimumdistanceis

√5.

. . . . . .

Distanceproblemminimizationstep

Wewanttofindtheglobalminimumof g(x) = (x− 3)2 + x4.I g′(x) = 2(x− 3) + 4x3 = 4x3 + 2x− 6 = 2(2x3 + x− 3)I Ifapolynomialhasintegerroots, theyarefactorsoftheconstantterm(Euler)

I 1 isaroot, so 2x3 + x− 3 isdivisibleby x− 1:

f′(x) = 2(2x3 + x− 3) = 2(x− 1)(2x2 + 2x+ 3)

Thequadratichasnorealroots(thediscriminantb2 − 4ac < 0)

I Wesee f′(1) = 0, f′(x) < 0 if x < 1, and f′(x) > 0 if x > 1. So1 istheglobalminimum.

I Thepointontheparabolaclosestto (3, 0) is (1, 1). Theminimumdistanceis

√5.

. . . . . .

Distanceproblemminimizationstep

Wewanttofindtheglobalminimumof g(x) = (x− 3)2 + x4.I g′(x) = 2(x− 3) + 4x3 = 4x3 + 2x− 6 = 2(2x3 + x− 3)I Ifapolynomialhasintegerroots, theyarefactorsoftheconstantterm(Euler)

I 1 isaroot, so 2x3 + x− 3 isdivisibleby x− 1:

f′(x) = 2(2x3 + x− 3) = 2(x− 1)(2x2 + 2x+ 3)

Thequadratichasnorealroots(thediscriminantb2 − 4ac < 0)

I Wesee f′(1) = 0, f′(x) < 0 if x < 1, and f′(x) > 0 if x > 1. So1 istheglobalminimum.

I Thepointontheparabolaclosestto (3, 0) is (1, 1). Theminimumdistanceis

√5.

. . . . . .

Distanceproblemminimizationstep

Wewanttofindtheglobalminimumof g(x) = (x− 3)2 + x4.I g′(x) = 2(x− 3) + 4x3 = 4x3 + 2x− 6 = 2(2x3 + x− 3)I Ifapolynomialhasintegerroots, theyarefactorsoftheconstantterm(Euler)

I 1 isaroot, so 2x3 + x− 3 isdivisibleby x− 1:

f′(x) = 2(2x3 + x− 3) = 2(x− 1)(2x2 + 2x+ 3)

Thequadratichasnorealroots(thediscriminantb2 − 4ac < 0)

I Wesee f′(1) = 0, f′(x) < 0 if x < 1, and f′(x) > 0 if x > 1. So1 istheglobalminimum.

I Thepointontheparabolaclosestto (3, 0) is (1, 1). Theminimumdistanceis

√5.

. . . . . .

Distanceproblemminimizationstep

Wewanttofindtheglobalminimumof g(x) = (x− 3)2 + x4.I g′(x) = 2(x− 3) + 4x3 = 4x3 + 2x− 6 = 2(2x3 + x− 3)I Ifapolynomialhasintegerroots, theyarefactorsoftheconstantterm(Euler)

I 1 isaroot, so 2x3 + x− 3 isdivisibleby x− 1:

f′(x) = 2(2x3 + x− 3) = 2(x− 1)(2x2 + 2x+ 3)

Thequadratichasnorealroots(thediscriminantb2 − 4ac < 0)

I Wesee f′(1) = 0, f′(x) < 0 if x < 1, and f′(x) > 0 if x > 1. So1 istheglobalminimum.

I Thepointontheparabolaclosestto (3, 0) is (1, 1). Theminimumdistanceis

√5.

. . . . . .

Remark

I We’veusedeachofthemethods(CIM,1DT,2DT) sofar.I Noticehowwearguedthatthecriticalpointswereabsoluteextremeseventhough1DT and2DT onlytellyourelative/localextremes.

. . . . . .

A problemwithatriangleExampleFindtherectangleofmaximalareainscribedina3-4-5righttriangleasshown.

Solution

I Letthedimensionsoftherectanglebe x and y.

I Similartrianglesgive

y3− x

=43

=⇒ 3y = 4(3−x)

I So y = 4− 43x and

A(x) = x(4− 4

3x)

= 4x−43x2

..3

.4.5

. . . . . .

A problemwithatriangleExampleFindtherectangleofmaximalareainscribedina3-4-5righttriangleasshown.

Solution

I Letthedimensionsoftherectanglebe x and y.

I Similartrianglesgive

y3− x

=43

=⇒ 3y = 4(3−x)

I So y = 4− 43x and

A(x) = x(4− 4

3x)

= 4x−43x2

..3

.4.5

.y

.x

. . . . . .

A problemwithatriangleExampleFindtherectangleofmaximalareainscribedina3-4-5righttriangleasshown.

Solution

I Letthedimensionsoftherectanglebe x and y.

I Similartrianglesgive

y3− x

=43

=⇒ 3y = 4(3−x)

I So y = 4− 43x and

A(x) = x(4− 4

3x)

= 4x−43x2

..3

.4.5

.y

.x

. . . . . .

A problemwithatriangleExampleFindtherectangleofmaximalareainscribedina3-4-5righttriangleasshown.

Solution

I Letthedimensionsoftherectanglebe x and y.

I Similartrianglesgive

y3− x

=43

=⇒ 3y = 4(3−x)

I So y = 4− 43x and

A(x) = x(4− 4

3x)

= 4x−43x2

..3

.4.5

.y

.x

. . . . . .

TriangleProblemmaximizationstep

Wewanttofindtheabsolutemaximumof A(x) = 4x− 43x2 on

theinterval [0, 3].

I A(0) = A(3) = 0

I A′(x) = 4− 83x, whichiszerowhen x =

128

= 1.5.

I Since A(1.5) = 3, thisistheabsolutemaximum.I Sothedimensionsoftherectangleofmaximalareaare1.5× 2.

. . . . . .

TriangleProblemmaximizationstep

Wewanttofindtheabsolutemaximumof A(x) = 4x− 43x2 on

theinterval [0, 3].I A(0) = A(3) = 0

I A′(x) = 4− 83x, whichiszerowhen x =

128

= 1.5.

I Since A(1.5) = 3, thisistheabsolutemaximum.I Sothedimensionsoftherectangleofmaximalareaare1.5× 2.

. . . . . .

TriangleProblemmaximizationstep

Wewanttofindtheabsolutemaximumof A(x) = 4x− 43x2 on

theinterval [0, 3].I A(0) = A(3) = 0

I A′(x) = 4− 83x, whichiszerowhen x =

128

= 1.5.

I Since A(1.5) = 3, thisistheabsolutemaximum.I Sothedimensionsoftherectangleofmaximalareaare1.5× 2.

. . . . . .

TriangleProblemmaximizationstep

Wewanttofindtheabsolutemaximumof A(x) = 4x− 43x2 on

theinterval [0, 3].I A(0) = A(3) = 0

I A′(x) = 4− 83x, whichiszerowhen x =

128

= 1.5.

I Since A(1.5) = 3, thisistheabsolutemaximum.

I Sothedimensionsoftherectangleofmaximalareaare1.5× 2.

. . . . . .

TriangleProblemmaximizationstep

Wewanttofindtheabsolutemaximumof A(x) = 4x− 43x2 on

theinterval [0, 3].I A(0) = A(3) = 0

I A′(x) = 4− 83x, whichiszerowhen x =

128

= 1.5.

I Since A(1.5) = 3, thisistheabsolutemaximum.I Sothedimensionsoftherectangleofmaximalareaare1.5× 2.

. . . . . .

AnEconomicsproblem

ExampleLet r bethemonthlyrentperunitinanapartmentbuildingwith100 units. A surveyrevealsthatallunitscanberentedwhenr = 900 andthatoneunitbecomesvacantwitheach 10 increaseinrent. Supposetheaveragemonthlymaintenancecostsperoccupiedunitis$100/month. Whatrentshouldbechargedtomaximizeprofit?

Solution

I Let n bethenumberofunitsrented, dependingonprice(thedemandfunction).

I Wehave n(900) = 100 and∆n∆r

= − 110

. So

n− 100 = − 110

(r− 900) =⇒ n(r) = − 110

r+ 190

. . . . . .

AnEconomicsproblem

ExampleLet r bethemonthlyrentperunitinanapartmentbuildingwith100 units. A surveyrevealsthatallunitscanberentedwhenr = 900 andthatoneunitbecomesvacantwitheach 10 increaseinrent. Supposetheaveragemonthlymaintenancecostsperoccupiedunitis$100/month. Whatrentshouldbechargedtomaximizeprofit?

Solution

I Let n bethenumberofunitsrented, dependingonprice(thedemandfunction).

I Wehave n(900) = 100 and∆n∆r

= − 110

. So

n− 100 = − 110

(r− 900) =⇒ n(r) = − 110

r+ 190

. . . . . .

EconomicsProblemFinishingthemodelandmaximizing

I Theprofitperunitrentedis r− 100, so

P(r) = (r− 100)n(r) = (r− 100)(− 110

r+ 190)

= − 110

r2 + 200r− 19000

I Wewanttomaximize P ontheinterval 900 ≤ r ≤ 1900.I A(900) = $800× 100 = $80, 000, A(1900) = 0

I A′(x) = −15r+ 200, whichiszerowhen r = 1000.

I n(1000) = 90, so P(r) = $900× 90 = $81, 000. Thisisthemaximumintake.

. . . . . .

EconomicsProblemFinishingthemodelandmaximizing

I Theprofitperunitrentedis r− 100, so

P(r) = (r− 100)n(r) = (r− 100)(− 110

r+ 190)

= − 110

r2 + 200r− 19000

I Wewanttomaximize P ontheinterval 900 ≤ r ≤ 1900.

I A(900) = $800× 100 = $80, 000, A(1900) = 0

I A′(x) = −15r+ 200, whichiszerowhen r = 1000.

I n(1000) = 90, so P(r) = $900× 90 = $81, 000. Thisisthemaximumintake.

. . . . . .

EconomicsProblemFinishingthemodelandmaximizing

I Theprofitperunitrentedis r− 100, so

P(r) = (r− 100)n(r) = (r− 100)(− 110

r+ 190)

= − 110

r2 + 200r− 19000

I Wewanttomaximize P ontheinterval 900 ≤ r ≤ 1900.I A(900) = $800× 100 = $80, 000, A(1900) = 0

I A′(x) = −15r+ 200, whichiszerowhen r = 1000.

I n(1000) = 90, so P(r) = $900× 90 = $81, 000. Thisisthemaximumintake.

. . . . . .

EconomicsProblemFinishingthemodelandmaximizing

I Theprofitperunitrentedis r− 100, so

P(r) = (r− 100)n(r) = (r− 100)(− 110

r+ 190)

= − 110

r2 + 200r− 19000

I Wewanttomaximize P ontheinterval 900 ≤ r ≤ 1900.I A(900) = $800× 100 = $80, 000, A(1900) = 0

I A′(x) = −15r+ 200, whichiszerowhen r = 1000.

I n(1000) = 90, so P(r) = $900× 90 = $81, 000. Thisisthemaximumintake.

. . . . . .

EconomicsProblemFinishingthemodelandmaximizing

I Theprofitperunitrentedis r− 100, so

P(r) = (r− 100)n(r) = (r− 100)(− 110

r+ 190)

= − 110

r2 + 200r− 19000

I Wewanttomaximize P ontheinterval 900 ≤ r ≤ 1900.I A(900) = $800× 100 = $80, 000, A(1900) = 0

I A′(x) = −15r+ 200, whichiszerowhen r = 1000.

I n(1000) = 90, so P(r) = $900× 90 = $81, 000. Thisisthemaximumintake.

. . . . . .

TheStatueofLiberty

TheStatueofLibertystandsontopofapedestalwhichisontopofonoldfort. Thetopofthepedestalis47mabovegroundlevel.Thestatueitselfmeasures46mfromthetopofthepedestaltothetipofthetorch.

Whatdistanceshouldonestandawayfromthestatueinordertomaximizetheviewofthestatue? Thatis, whatdistancewillmaximizetheportionoftheviewer’svisiontakenupbythestatue?

. . . . . .

TheStatueofLibertySetingupthemodel

Theanglesubtendedbythestatueintheviewer’seyecanbeexpressedas

θ = arctan(a+ bx

)−arctan

(bx

).

a

bθ

x

Thedomainof θ isallpositiverealnumbers x.

. . . . . .

TheStatueofLibertyFindingthederivative

θ = arctan(a+ bx

)− arctan

(bx

)So

dθdx

=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]

. . . . . .

TheStatueofLibertyFindingthecriticalpoints

dθdx

=

[x2 + (a+ b)2

]b− (a+ b)

[x2 + b2

](x2 + b2) [x2 + (a+ b)2]

I Thisderivativeiszeroifandonlyifthenumeratoriszero, soweseek x suchthat

0 =[x2 + (a+ b)2

]b− (a+ b)

[x2 + b2

]= a(ab+ b2 − x2)

I Theonlypositivesolutionis x =√

b(a+ b).I Usingthefirstderivativetest, weseethat dθ/dx > 0 if0 < x <

√b(a+ b) and dθ/dx < 0 if x >

√b(a+ b).

I Sothisisdefinitelytheabsolutemaximumon (0,∞).

. . . . . .

TheStatueofLibertyFindingthecriticalpoints

dθdx

=

[x2 + (a+ b)2

]b− (a+ b)

[x2 + b2

](x2 + b2) [x2 + (a+ b)2]

I Thisderivativeiszeroifandonlyifthenumeratoriszero, soweseek x suchthat

0 =[x2 + (a+ b)2

]b− (a+ b)

[x2 + b2

]= a(ab+ b2 − x2)

I Theonlypositivesolutionis x =√

b(a+ b).I Usingthefirstderivativetest, weseethat dθ/dx > 0 if0 < x <

√b(a+ b) and dθ/dx < 0 if x >

√b(a+ b).

I Sothisisdefinitelytheabsolutemaximumon (0,∞).

. . . . . .

TheStatueofLibertyFindingthecriticalpoints

dθdx

=

[x2 + (a+ b)2

]b− (a+ b)

[x2 + b2

](x2 + b2) [x2 + (a+ b)2]

I Thisderivativeiszeroifandonlyifthenumeratoriszero, soweseek x suchthat

0 =[x2 + (a+ b)2

]b− (a+ b)

[x2 + b2

]= a(ab+ b2 − x2)

I Theonlypositivesolutionis x =√

b(a+ b).

I Usingthefirstderivativetest, weseethat dθ/dx > 0 if0 < x <

√b(a+ b) and dθ/dx < 0 if x >

√b(a+ b).

I Sothisisdefinitelytheabsolutemaximumon (0,∞).

. . . . . .

TheStatueofLibertyFindingthecriticalpoints

dθdx

=

[x2 + (a+ b)2

]b− (a+ b)

[x2 + b2

](x2 + b2) [x2 + (a+ b)2]

I Thisderivativeiszeroifandonlyifthenumeratoriszero, soweseek x suchthat

0 =[x2 + (a+ b)2

]b− (a+ b)

[x2 + b2

]= a(ab+ b2 − x2)

I Theonlypositivesolutionis x =√

b(a+ b).I Usingthefirstderivativetest, weseethat dθ/dx > 0 if0 < x <

√b(a+ b) and dθ/dx < 0 if x >

√b(a+ b).

I Sothisisdefinitelytheabsolutemaximumon (0,∞).

. . . . . .

TheStatueofLibertyFinalanswer

Ifwesubstituteinthenumericaldimensionsgiven, wehave

x =√

(46)(93) ≈ 66.1 meters

Thisdistancewouldputyouprettyclosetothefrontoftheoldfortwhichliesatthebaseoftheisland.

Unfortunately, you’renotallowedtowalkonthispartofthelawn.

. . . . . .

TheStatueofLibertyDiscussion

I Thelength√

b(a+ b) isthe geometricmean ofthetwodistancesmeasuredfromtheground—tothetopofthepedestal(a)andthetopofthestatue(a+ b).

I Thegeometricmeanisoftwonumbersisalwaysbetweenthemandgreaterthanorequaltotheiraverage.