. . . . . .

Section4.7ApplicationstoBusinessand

Economics

Math1aIntroductiontoCalculus

April2, 2008

Announcements

◮ ProblemSessionsSunday, Thursday, 7pm, SC 310◮ OfficehoursTues, Weds, 2–4pmSC 323◮ MidtermII:4/11inclass

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. . . . . .

Announcements

◮ ProblemSessionsSunday, Thursday, 7pm, SC 310◮ OfficehoursTues, Weds, 2–4pmSC 323◮ MidtermII:4/11inclass

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Outline

Lasttime

Cost

Marketing

UtilityMaximization

Nexttime

. . . . . .

Theorem(L’Hopital’sRule)Suppose f and g aredifferentiablefunctionsand g′(x) ̸= 0 near a(exceptpossiblyat a). Supposethat

limx→a

f(x) = 0 and limx→a

g(x) = 0

or

limx→a

f(x) = ±∞ and limx→a

g(x) = ±∞

Then

limx→a

f(x)g(x)

= limx→a

f′(x)g′(x)

,

ifthelimitontheright-handsideisfinite, ∞, or −∞.

L’Hôpital’srulealsoappliesforlimitsoftheform∞∞

.

. . . . . .

Theorem(L’Hopital’sRule)Suppose f and g aredifferentiablefunctionsand g′(x) ̸= 0 near a(exceptpossiblyat a). Supposethat

limx→a

f(x) = 0 and limx→a

g(x) = 0

or

limx→a

f(x) = ±∞ and limx→a

g(x) = ±∞

Then

limx→a

f(x)g(x)

= limx→a

f′(x)g′(x)

,

ifthelimitontheright-handsideisfinite, ∞, or −∞.

L’Hôpital’srulealsoappliesforlimitsoftheform∞∞

.

. . . . . .

SummaryForm Method

00 L’Hôpital’sruledirectly

∞∞ L’Hôpital’sruledirectly

0 · ∞ jiggletomake 00 or

∞∞ .

∞−∞ factortomakeanindeterminateproduct

00 take ln tomakeanindeterminateproduct

∞0 ditto

1∞ ditto

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Outline

Lasttime

Cost

Marketing

UtilityMaximization

Nexttime

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Someeconomicsterms

DefinitionThe (total)cost functionofagoodistheamountthatmustbespentto produce x itemsofagood.

Thisisincontrastto

DefinitionThe price functionofagoodistheamountthatmustbespenttopurchase anitemofagoodif x itemsaresuppliedtothemarket.

Theunitsofcostaredollars(orothercurrency), whiletheunitsofpricearedollarsperitem(orothermeasureofquantity).

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Cost

Let C(x) bethecostofproducing x unitsofaproduct. Whatshapeshouldthegraphof C have?

. . . . . .

Since C′ > 0 always, what’smoreinterestingisthe averagecostfunction

c(x) =C(x)x

.

Assumingthatwecanselleverythingwemake, wewanttominimizethecostsofproduction.Whenistheaveragecostminimized?

c′(x) =xC′(x) − C(x)

x2=

C′(x) − c(x)x

,

so c′(x) = 0 ifandonlyif C′(x) = c(x). Inotherwords,

TheoremAtthepointofminimalaveragecostofproduction, theaveragecostequalsthemarginalcost.

. . . . . .

Since C′ > 0 always, what’smoreinterestingisthe averagecostfunction

c(x) =C(x)x

.

Assumingthatwecanselleverythingwemake, wewanttominimizethecostsofproduction.

Whenistheaveragecostminimized?

c′(x) =xC′(x) − C(x)

x2=

C′(x) − c(x)x

,

so c′(x) = 0 ifandonlyif C′(x) = c(x). Inotherwords,

TheoremAtthepointofminimalaveragecostofproduction, theaveragecostequalsthemarginalcost.

. . . . . .

Since C′ > 0 always, what’smoreinterestingisthe averagecostfunction

c(x) =C(x)x

.

Assumingthatwecanselleverythingwemake, wewanttominimizethecostsofproduction.Whenistheaveragecostminimized?

c′(x) =xC′(x) − C(x)

x2=

C′(x) − c(x)x

,

so c′(x) = 0 ifandonlyif C′(x) = c(x). Inotherwords,

TheoremAtthepointofminimalaveragecostofproduction, theaveragecostequalsthemarginalcost.

. . . . . .

Since C′ > 0 always, what’smoreinterestingisthe averagecostfunction

c(x) =C(x)x

.

Assumingthatwecanselleverythingwemake, wewanttominimizethecostsofproduction.Whenistheaveragecostminimized?

c′(x) =xC′(x) − C(x)

x2=

C′(x) − c(x)x

,

so c′(x) = 0 ifandonlyif C′(x) = c(x). Inotherwords,

TheoremAtthepointofminimalaveragecostofproduction, theaveragecostequalsthemarginalcost.

. . . . . .

Since C′ > 0 always, what’smoreinterestingisthe averagecostfunction

c(x) =C(x)x

.

Assumingthatwecanselleverythingwemake, wewanttominimizethecostsofproduction.Whenistheaveragecostminimized?

c′(x) =xC′(x) − C(x)

x2=

C′(x) − c(x)x

,

so c′(x) = 0 ifandonlyif C′(x) = c(x). Inotherwords,

TheoremAtthepointofminimalaveragecostofproduction, theaveragecostequalsthemarginalcost.

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ExampleLet C(x) = 16, 000 + 200x + 4x3/2. Find

(a) thecost, averagecost, andmarginalcostataproductionlevelof1000units,

(b) theproductionlevelthatwillminimizeaveragecost

(c) theminimumaveragecost.

Solution

(a) 8000(27 + 5

√10

), 8

(27 + 5

√10

), 200 + 60

√10, or

342, 491, 342.491, 389.737

(b) 400

(c) 300

. . . . . .

ExampleLet C(x) = 16, 000 + 200x + 4x3/2. Find

(a) thecost, averagecost, andmarginalcostataproductionlevelof1000units,

(b) theproductionlevelthatwillminimizeaveragecost

(c) theminimumaveragecost.

Solution

(a) 8000(27 + 5

√10

), 8

(27 + 5

√10

), 200 + 60

√10, or

342, 491, 342.491, 389.737

(b) 400

(c) 300

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Outline

Lasttime

Cost

Marketing

UtilityMaximization

Nexttime

. . . . . .

Marketing

Let p(x) bethepricefunction. Whatshouldtheshapeof p be?

DefinitionThe revenue functionisthetotalamounttakeninfromthesaleofx itemsofagood.

WehaveR(x) = x · p(x)

andthe profitP(x) = R(x) − C(x).

Clearly, profitismaximizedwhen P′(x) = 0.

TheoremAtthepointofmaximalprice, themarginalrevenueequalsthemarginalcost.

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Marketing

Let p(x) bethepricefunction. Whatshouldtheshapeof p be?

DefinitionThe revenue functionisthetotalamounttakeninfromthesaleofx itemsofagood.

WehaveR(x) = x · p(x)

andthe profitP(x) = R(x) − C(x).

Clearly, profitismaximizedwhen P′(x) = 0.

TheoremAtthepointofmaximalprice, themarginalrevenueequalsthemarginalcost.

. . . . . .

Marketing

Let p(x) bethepricefunction. Whatshouldtheshapeof p be?

DefinitionThe revenue functionisthetotalamounttakeninfromthesaleofx itemsofagood.

WehaveR(x) = x · p(x)

andthe profitP(x) = R(x) − C(x).

Clearly, profitismaximizedwhen P′(x) = 0.

TheoremAtthepointofmaximalprice, themarginalrevenueequalsthemarginalcost.

. . . . . .

Marketing

Let p(x) bethepricefunction. Whatshouldtheshapeof p be?

DefinitionThe revenue functionisthetotalamounttakeninfromthesaleofx itemsofagood.

WehaveR(x) = x · p(x)

andthe profitP(x) = R(x) − C(x).

Clearly, profitismaximizedwhen P′(x) = 0.

TheoremAtthepointofmaximalprice, themarginalrevenueequalsthemarginalcost.

. . . . . .

Marketing

Let p(x) bethepricefunction. Whatshouldtheshapeof p be?

DefinitionThe revenue functionisthetotalamounttakeninfromthesaleofx itemsofagood.

WehaveR(x) = x · p(x)

andthe profitP(x) = R(x) − C(x).

Clearly, profitismaximizedwhen P′(x) = 0.

TheoremAtthepointofmaximalprice, themarginalrevenueequalsthemarginalcost.

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ExampleA manufacturerhasbeenselling1000televisionsetsaweekat$450each. A marketsurveyindicatesthatforeach$10rebateofferedtothebuyer, thenumberofsetssoldwillincreaseby100perweek.

(a) Findthedemandfunction

(b) Howlargearebateshouldthecompanyofferthebuyerinordertomaximizerevenue?

(c) Ifitsweeklycostfunctionis C(x) = 68, 000 + 150x, howshouldthemanufacturersetthesizeoftherebateinordertomaximizeitsprofit?

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Solution

(a) Tofind p(x) wehave p(1000) = 450, p(1100) = 440. So

p(x) =p(1100) − p(1000)

1100− 1000(x− 1000) + p(1000)

= − 110

(x− 1000) + 450 = − 110

x + 550

(b) Thesalesfunctionis R(x) = xp(x) = − 110

x2 + 550x. Marginal

revenueistherefore

R′(x) = −15x + 550.

Revenueismaximizedwhen R′(x) = 0, orx = 550× 5 = 2750. Now p(2750) = −275 + 550 = 275,sotherevenue-optimizingrebateis$275.

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Solution

(a) Tofind p(x) wehave p(1000) = 450, p(1100) = 440. So

p(x) =p(1100) − p(1000)

1100− 1000(x− 1000) + p(1000)

= − 110

(x− 1000) + 450 = − 110

x + 550

(b) Thesalesfunctionis R(x) = xp(x) = − 110

x2 + 550x. Marginal

revenueistherefore

R′(x) = −15x + 550.

Revenueismaximizedwhen R′(x) = 0, orx = 550× 5 = 2750. Now p(2750) = −275 + 550 = 275,sotherevenue-optimizingrebateis$275.

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Solution(Continued)

(c) Nowwehave

P(x) = R(x) − C(x) = − 110

x2 + 550x− (68, 000 + 150x)

= − 110

x2 + 400x− 68, 000

Marginalprofitis

P′(x) = −15x + 400

whichiszerowhen x = 2000. Thiscorrespondstoarebateof$100.

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Outline

Lasttime

Cost

Marketing

UtilityMaximization

Nexttime

. . . . . .

UtilityMaximization

A studentderivesutility(happiness)fromconsumingburritosandchipsaccordingtothefunction

u(x, y) = 10x1/2y2/5

where x isthenumberofburritosconsumedand y isthenumberofbasketsofchipsconsumedeveryweek. Burritoscost$5apieceandbasketsofchipscost$2apiece.Ifthestudenthasafixedbudgetof$18withwhichtobuyhisMexicanfood, whatquantitiesofburritosandchipsshouldhebuytomaximizehisutility?

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SolutionWeneedtomaximize u subjecttotheconstraintthat5x + 2y = 60. So

u(x) = 10x1/2(18− 5x

2

)2/5

dudx

= 1022/5

{12x

−1/2(18− 5x)2/5 + x1/2 25(18− 5x)−3/5(−5)

}=

10

22/5x−1/2(18− 5x)−3/5 {1

2(18− 5x) − 2x}

=10

22/5x−1/2(18− 5x)−3/5 (

9− 92x

)So u′(x) = 0 when x = 2. Then y = 4.

. . . . . .

SolutionWeneedtomaximize u subjecttotheconstraintthat5x + 2y = 60. So

u(x) = 10x1/2(18− 5x

2

)2/5

dudx

= 1022/5

{12x

−1/2(18− 5x)2/5 + x1/2 25(18− 5x)−3/5(−5)

}=

10

22/5x−1/2(18− 5x)−3/5 {1

2(18− 5x) − 2x}

=10

22/5x−1/2(18− 5x)−3/5 (

9− 92x

)So u′(x) = 0 when x = 2. Then y = 4.

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Outline

Lasttime

Cost

Marketing

UtilityMaximization

Nexttime

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NexttimeNewton’sMethod