OMATLecture5
.pdf
School
Vietnamese-German University, Ho Chi Minh City**We aren't endorsed by this school
Course
FIN 101
Subject
Mathematics
Date
Dec 18, 2024
Pages
23
Uploaded by ChefQuail4848
Duong. T. PHAMChapter 5: Optimization of Functions of one variableDuong T. PHAMOMATDuong T. PHAM1 / 23
Duong. T. PHAMThe goals for this lectureApply the first and the second derivatives to find Maximum andMinimumFind Maximum and Minimum over an IntervalCalculate Maximum-Minimum values in Business and EconomicsApplicationsDuong T. PHAM2 / 23
Duong. T. PHAMDerivative and shape of a graphIncreasing and decreasing Test:Letf: (a,b)→Rbe a differentiablefunction.1Iff0(x)>0,∀x∈(a,b), thenfis increasing in (a,b).2Iff0(x)<0,∀x∈(a,b), thenfis decreasing in (a,b).xyy=f(x)abcdeDuong T. PHAM3 / 23
Duong. T. PHAMIncreasing and decreasing TestEx:Determine when the functionf(x) = 3x4-4x3-12x2+ 5is increas-ing and decreasing.Ans:f0(x) = 12x3-12x2-24x= 12x(x-2)(x+ 1)=⇒f0(x) = 0⇐⇒x=-1∨x= 0∨x= 2x-∞-102∞x-2-|-|-0+x-|-0+|+x+ 1-0+|+|+f0(x)-0+0-0+f(x)&0%5&-27%Duong T. PHAM4 / 23
Duong. T. PHAMIncreasing and decreasing Testxy-12y= 3x4-4x3-12x2+ 50Duong T. PHAM5 / 23
Duong. T. PHAMThe First derivative Test:Suppose thatcis a critical number of acontinuous functionf.(i)Iff0changes from positive to negative atc, thenfhas a localmaximum atc.(ii)Iff0changes from negative to positive atc, thenfhas a localminimum atc.(iii)Iff0does not change sign atc(for example, iff0is positive onboth sides ofcor negative on both sides), thenfhas no localmaximum or minimum atc.yxcf0<0f0>0local minimumyxcf0>0f0<0local maximumyxcf0>0f0>0no local min. or max.yxcf0<0f0<0no local min. or maxDuong T. PHAM6 / 23
Duong. T. PHAMThe first derivative TestEx:Find the local minimum and maximum values of the functionf(x) =3x4-4x3-12x2+ 5Ans:f0(x) = 12x3-12x2-24x= 12x(x-2)(x+ 1)=⇒f0(x) = 0⇐⇒x=-1∨x= 0∨x= 2x-∞-102∞x-2-|-|-0+x-|-0+|+x+ 1-0+|+|+f0(x)-0+0-0+f(x)&0%5&-27%fattains local minimum at-1 and 2; and attains local maximum at0.Duong T. PHAM7 / 23
Duong. T. PHAMThe first derivative Testxy-12y= 3x4-4x3-12x2+ 50Duong T. PHAM8 / 23
Duong. T. PHAMDefinition.Letf:I→R.If the graph offlies above all of its tangent lines, thenfis said to beconvexonIIf the graph offlies below all of its tangent lines, thenfis said to beconcaveonIyxconvexyxconcaveDuong T. PHAM9 / 23
Duong. T. PHAMConcavity TestConcavity Test:Letf:I→R. Then(a)Iff00(x)>0∀x∈I, then the graph offis convex onI.(b)Iff00(x)<0∀x∈I, then the graph offis concave onI.Discussion:f00(x)>0 for allx∈I=⇒f0(x) is increasing onIf00(x<0) for allx∈I=⇒f0(x) is decreasing onIyxf0increases→convexyxf0decreases→concaveDuong T. PHAM10 / 23
Duong. T. PHAMInflection PointDefinition.A pointPon a curvey=f(x)is called an inflection point iffiscontinuous there and the curve changes from convex to concave or fromconcave to convex atPyx(c,f(c))(c,f(c)) is inflection pointyx(c,f(c))(c,f(c)) is inflection pointDuong T. PHAM11 / 23
Duong. T. PHAMThe second derivative TestThe second derivative Test:Letfbe a function such thatf00is contin-uous nearc. Then(a)Iff0(c) = 0andf00(c)>0, thenfhas a local minimum atc,(b)Iff0(c) = 0andf00(c)<0, thenfhas a local maximum atc.Duong T. PHAM12 / 23
Duong. T. PHAMThe second derivative TestEx:Discuss the concavity, inflection points, local maxima and local min-ima of the curvey=x4-4x3.Ans:Denotef(x) =x4-4x3. Thenf0(x) = 4x3-12x2= 4x2(x-3)f00(x) = 12x2-24x= 12x(x-2)f0(x) = 0⇐⇒x= 0∨x= 3 andf00(x) = 0⇐⇒x= 0∨x= 2x023x2+0+|+|+x-3-|-|-0+x-2-|-0+|+f0(x)-0-|-0+f00(x)+0-0+|+f(x)&0&-16&-27%convexinfl.p.concaveinfl.p.convexlocal min.convexDuong T. PHAM13 / 23
Duong. T. PHAMThe second derivative Testxy2inflection point-163local min.-270inflection pointDuong T. PHAM14 / 23
Duong. T. PHAMFinding absolute minimum and absolute maximumThe closed interval method:Letf: [a,b]→Rbe a continuous function. Tofind the absolute max. and absolute min. off, we follows the steps:1Find the values offat critical numbers offin (a,b);2Find the valuesf(a)andf(b);3The largest number in steps 1 and 2 is the absolute maximum value off,and the smallest number in steps 1 and 2 is the absolute minimum off.Ex:Find abs. max. and abs. min. off(x) = 2x3-9x2+ 12x+ 2in[0,3]Ans:f0(x) = 6x2-18x+ 12 = 6(x2-3x+ 2)1f0(x) = 0⇐⇒x= 1 orx= 2, andf(1) = 7,f(2) = 62f(0) = 2andf(3) = 113Comparing the 4 values in Steps 1 and 2, we concludemax[0,3]f=f(3) = 11andmin[0,3]f=f(0) = 2Duong T. PHAM15 / 23
Duong. T. PHAMThe closed interval methodxyy= 2x3-9x2+ 12x+ 5027162113Duong T. PHAM16 / 23
Duong. T. PHAMCost, Revenue & ProfitExample.A stereo manufacturer determines that in order to sell x unitsof a new stereo, the price per unit, in dollars, must bep(x) = 1,000-xThe manufacturer also determines that the total cost of producingxunitsis given byC(x) = 3,000 + 20x.a) Find the total revenueR(x).b) Find the total profitP(x).c) How many units must the company produce and sell in order tomaximize profit?d) What is the maximum profit?e) What price per unit must be charged in order to make this maximumprofit?Duong T. PHAM17 / 23
Duong. T. PHAMCost, Revenue & ProfitExample.Price:p(x) = 1,000-x; Cost:C(x) = 3,000 + 2xa) Revenue = quantity·priceR(x) =x·p(x) =x(1000-x) = 1000x-x2.b) Profit = Total Revenue - Total CostP(x) =R(x)-C(x) = 1000x-x2-(3000 + 20x) =-x2+ 980x-3000c)Maximize the profit.We haveP0(x) =-2x+ 980.P0(x) = 0⇐⇒ -2x+ 980 = 0⇐⇒x= 490x-∞490∞P0(x)+0-P(x)%237100&Thus, profit is maximized when 490 units are bought and sold.Duong T. PHAM18 / 23
Duong. T. PHAMCost, Revenue & Profitx-∞490∞P0(x)+0-P(x)%237100&Example (continued).d) The maximum profit is given byP(490) =-(490)2+ 980(490)-3000 = $237,100.Thus, the stereo manufacturer makes a maximum profit of $237,100 when490 units are bought and solde) What price per unit must be charged in order to make this maximumprofit?The price per unit to achieve this maximum profit isp(490) = 1000-490 = $510.Duong T. PHAM19 / 23
Duong. T. PHAMMaximizing RevenueExample.Promoters of international fund-raising concerts must walk afine line between profit and loss, especially when determining the price tocharge for admission to closed-circuit TV showings in local theaters . Bykeeping records, a theater determines that, at an admission price of $26, itaverages 1000 people in attendance. For every drop in price of $1, it gains50 customers. Each customer spends average of $4 on concessions. Whatadmission price should the theater charge in order to maximize totalrevenue?Ans.Denotex= the number of dollars by which the price of $26 shouldbe decreased (ifxis negative, the price should be increased).Revenue = Revenue from tickets + Revenue from concessionsR(x) = #of people·ticket price + #of people·4R(x) = (1000 + 50x)(26-x) + (1000 + 50x)·4R(x) =-50x2+ 500x+ 30,000Duong T. PHAM20 / 23
Duong. T. PHAMMaximizing RevenueExample (continued).R(x) =-50x2+ 500x+ 30,000. To maximizeR(x), we findR0(x) =-100x+ 500. We haveR0(x) = 0⇐⇒ -100x+ 500 = 0⇐⇒x= 5x-∞5∞R0(x)+0-R(x)%31250&Thus,x= 5 yields a maximum revenue. The theater should charge$26-$5 = $21 per ticket.Duong T. PHAM21 / 23
Duong. T. PHAMDemand & RevenueExample.A swimming club offers memberships at the rate of $200,provided that a minimum of 100 people join. For each member in excess of100, the membership fee will be reduced $1 per person (for each member).At most 160 memberships will be sold. How many memberships shouldthe club try to sell in order to maximize its revenue?Ans.We consider the price functionp=p(x). Since a minimum of 100people is supposed to join, we only consider whenx≥100. At most 160memberships will be sold thenx≤160. Hence,100≤x≤160.We havep(100) =200.For each member in excess of 100, the membership fee will bereduced $1 per person (for each member) =⇒xincreases 1 unit,ydecreases 1 unit. =⇒p(160) = 200-60 = $140.Suppose thatp(x) =ax+b.We have(100a+b= 200160a+b= 140⇐⇒(a=-1b= 300=⇒p(x) =-x+ 300Duong T. PHAM22 / 23
Duong. T. PHAMDemand & RevenueExample (continued).Recall:p(x) =-x+ 300and our task is tomaximize the revenue function. The revenue function isR(x) =x·p(x) =x(-x+ 300) =-x2+ 300x.=⇒R0(x) =-2x+ 300.ConsiderR0(x) = 0⇐⇒ -2x+ 300 = 0⇐⇒x= 150x-∞150∞R0(x)+0-R(x)%22500&The revenue is maximized whenx= 150.Duong T. PHAM23 / 23