Lesson 22: Optimization II (Section 10 version)
-
Upload
matthew-leingang -
Category
Technology
-
view
499 -
download
1
description
Transcript of Lesson 22: Optimization II (Section 10 version)
. . . . . .
Section4.5OptimizationProblems, PartDeux
V63.0121, CalculusI
April8, 2009
Announcements
I Quiz5isnextweek, coveringSections4.1–4.4I I ammovingtoWWH 624sometimenextweek(April13th)I HappyPassover/Easter
. . . . . .
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′′ becontinuouson [a,b]. Let c bebeapointin(a,b) with f′(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) > 1 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) > 1 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) > 1 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) > 1 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) > 1 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) > 1 if x > 1. So1 istheglobalminimum.
I Thepointontheparabolaclosestto (3, 0) is (1, 1). Theminimumdistanceis
√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
. . . . . .
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 surveyrevelasthatallunitscanberentedwhenr = 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 surveyrevelasthatallunitscanberentedwhenr = 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 0 ≤ r ≤ 1900.I A(0) = −19000, A(1900) = 0, sotheseareprobablynotmaximal
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 0 ≤ r ≤ 1900.
I A(0) = −19000, A(1900) = 0, sotheseareprobablynotmaximal
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 0 ≤ r ≤ 1900.I A(0) = −19000, A(1900) = 0, sotheseareprobablynotmaximal
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 0 ≤ r ≤ 1900.I A(0) = −19000, A(1900) = 0, sotheseareprobablynotmaximal
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 0 ≤ r ≤ 1900.I A(0) = −19000, A(1900) = 0, sotheseareprobablynotmaximal
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
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]
. . . . . .
TheStatueofLibertyFindingthecriticalpoints
dθ
dx=
[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
dθ
dx=
[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
dθ
dx=
[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
dθ
dx=
[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 ofthetwodistancesmeasurefromtheground—tothetopofthepedestal(a)andthetopofthestatue(a + b).
I Thegeometricmeanisoftwonumbersisalwaysbetweenthemandgreaterthanorequaltotheiraverage.
. . . . . .
Summary
I RememberthechecklistI Askyourself: whatistheobjective?I Rememberyourgeometry:
I similartrianglesI righttrianglesI trigonometricfunctions