142PP8W2021-T2
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Dec 19, 2024
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MATH 142 – Calculus II for Management andEconomicsPractice Problems #8 with Full SolutionsPaul Tsopm´en´eEdited by Ben RobideauFebruary 3, 2022Topics: The Area Between Two Curves (applications) andNumerical Integration1.Suppose a company wants to introduce a new machine that will produce a rate ofannual savings (in dollars) given byS0(t) = 150-t2, wheretis the number ofyears of operation of the machine, while producing a rate of annual costs (in dollars)C0(t) =t2+114t.[Solution onpage 3](a)For how many years will the company realize savings?(b)What is the net total savings during this period?2.Find the consumers’ surplus if the demand function for grass seed is given (in dollars)byD(q) =200(3q+ 1)2. Assume supply and demand are in equilibrium atq= 3.[Solution onpage 5]3.Find the producers’ surplus if the supply functions for pork bellies is given (in dollars)byS(q) =q52+ 2q32+ 50. Assume supply and demand are in equilibrium atq= 16.[Solution onpage 6]4.Use the trapezoidal rule withn= 4 to approximate the integralZ10√3x2+ 5dx.[Solution onpage 7]5.Use Simpson’s rule withn= 4 to approximate the integralZ10√3x2+ 5dx.[Solution onpage 9]
26.Approximate the area under the curvey=√9-x2and above thex-axis by usingn= 6 with the following methods:[Solution onpage 10](a)The trapezoidal rule.(b)Simpson’s rule.(c)Compare the results with the area found by the formula for the area of circle.Which of the two approximation was more accurate?7.A restaurant’s marginal cost (in hundreds of dollars per month) were as follows overthe first five months of a specific year.Month,x12345Marginal Cost,f(x)77.37.67.49Approximate the total cost over the given period using Simpson’s rule.[Solution onpage 11]2022 Paul Tsopm´en´e All Rights Reserved
3Solutions to Practice Problems #81.[Exercise onpage 1](a)The company will realize savings if the marginal savings,S0(t), is above themarginal cost,C0(t).So, we need to determine the interval in whichS0(t)≥C0(t). To do this, we need to solve the equationC0(t) =S0(t). We havet2+114t= 150-t22t2+114t-150 = 08t2+ 11t-600 = 0Multiply by 4Using thequadratic formula, we gett=-11±p112-4(8)(-600)2(8)=-11±√1932116=-11±13916So,t=-11 + 13916= 8ort=-11-13916=-758Since the timetis a nonnegative number, we reject the solution-758. It followsthat the only meaningful solution to the equationC0(t) =S0(t) ist= 8. Sincet≥0, we have the intervals [0,8] and [8,∞).Interval[0,8]:We choose an arbitrary number, say 1. Substituting this intoC0(t) andS0(t),we getC0(1) = 12+114(1) = 1 + 2.75 = 3.75S0(1) = 150-12= 149SinceS0(1)> C0(1), it follows that marginal savings is above marginal cost onthe interval [0,8].Interval[8,∞):We choose 9. Substituting this intoC0(t) andS0(t), we getC0(9) = 92+114(9) = 105.75S0(9) = 150-92= 69SinceC0(9)> S0(9), marginal cost is above marginal savings on the interval2022 Paul Tsopm´en´e All Rights Reserved
4[8,∞).Thus, the company will realize savings for8 years.(b)The total savings over the 8-year period is given by the integralZ80S0(t)dtwhich can be viewed as the area under the curvey=S0(t) over the interval[0,8]. The total cost over the same period isZ80C0(t)dtwhich is the area under the curvey=C0(t) over the interval [0,8]. So, the nettotal savings during this period isZ80S0(t)dt-Z80C0(t)dt=Z80[S0(t)-C0(t)]dtwhich is nothing but the area between marginal cost and marginal savings curvesfromt= 0 tot= 8. We findNet total savings =Z80[S0(t)-C0(t)]dt=Z80(150-t2)-t2+114tdt=Z80-2t2-114t+ 150dt=-2t33-114t22+ 150t80Zxndx=xn+1n+ 1+C,Zk dx=kx+C,Zbaf(x)dx=F(x)ba=-23(8)3-118(8)2+ 150(8)-(0-0 + 0)F(x)ba=F(b)-F(a)=-10243-88 + 1200=-10243-2643+360032022 Paul Tsopm´en´e All Rights Reserved
5=23123≈770.67Therefore, the company will save a total of$770.67 over the 8-year period.2.[Exercise onpage 1]Consumer SurplusConsumer Surplus (CS)is the difference between the price that consumerswould be willing to pay and the price that they actually pay. Letp=D(q)be the demand function with equilibrium pricep0and equilibrium quantityq0.Then, the consumers’ surplus is given by the formulaCS =Zq00[D(q)-p0]dqGeometrically, this is the area between the curvesp=D(q) andp=p0overthe interval [0, q0].For the demand functionp=D(q) =200(3q+1)2with equilibrium quantityq0= 3, theequilibrium price isp0=D(q0) =D(3) =200(3(3) + 1)2=200100= 2We haveCS =Zq00[D(q)-p0]dq=Z30200(3q+ 1)2-2dqTo find this integral, we make the substitutionu= 3q+ 1. Then,du= 3dqso thatdu3=dq.Ifq= 0,u= 1Ifq= 3,u= 3(3) + 1 = 10Therefore,CS =Z30200(3q+ 1)2-2dq=Z101200u2-2du3Substitute, change bounds2022 Paul Tsopm´en´e All Rights Reserved
6=Z1012003u-2-23du=2003u-1-1-23u101Zxndx=xn+1n+ 1+C,Zk dx=kx+C,Zbaf(x)dx=F(x)ba=-2003u-23u101=-2003(10)-23(10)--2003-23F(x)ba=F(b)-F(a)=-203-203--2023=-403+2023=1623= 54Thus, the consumers’ surplus is$54 .3.[Exercise onpage 1]Producers’ SurplusSuppose a manufacturer would be willing to supply a product at a price lowerthan the equilibrium price.Then, the total of the differences between theequilibrium price and the lower prices at which the manufacturer would sellthe product is called theproducers’ surplus (PS). Letp=S(q) be thesupply function with equilibrium pricep0and equilibrium quantity (or supply)q0. Then, the producers’ surplus is given by the formulaPS =Zq00[p0-S(q)]dqGeometrically, this is the area between the curvesp=p0andp=S(q) overthe interval [0, q0].For the supply functionS(q) =q52+ 2q32+ 50 with equilibrium quantityq0= 16, the2022 Paul Tsopm´en´e All Rights Reserved
7equilibrium price isp0=S(q0) =S(16) = (16)52+ 2(16)32+ 50 = 1024 + 128 + 50 = 1202.We havePS =Zq00[p0-S(q)]dq=Z160h1202-q52+ 2q32+ 50idq=Z160-q52-2q32+ 1152dq=-q7272-2q5252+ 1152q!160Zxndx=xn+1n+ 1+C,Zk dx=kx+C,Zbaf(x)dx=F(x)ba=-27q72-45q52+ 1152q160abc=acb=-27(16)72-45(16)52+ 1152(16)-(-0-0 + 0)F(x)ba=F(b)-F(a)=-327687-40965+ 18432≈ -4681.14-819.2 + 18432= 12,931.66Hence, the producers’ surplus is approximately$12,931.66 .4.[Exercise onpage 1]Trapezoidal RuleLetfbe a continuous function on the interval [a, b].Suppose we subdivide[a, b] intonequal subintervals by pointsa=x0, x1, x2,· · ·, xn-1, xn=b. Thatis,x0=aandxi+1=xi+ Δxfor0≤i≤n-12022 Paul Tsopm´en´e All Rights Reserved
8where Δx=b-anis the width of each subinterval. So,x0=a,x3=x2+ Δxx1=a+ Δx...x2=x1+ Δxxn=xn-1+ Δx=bBy thetrapezoidal rule, the definite integralZbaf(x)dxcan be approximatedby the expression Δx12f(x0) +f(x1) +· · ·+f(xn-1) +12f(xn) . That is,Zbaf(x)dx≈Δx12f(x0) +f(x1) +· · ·+f(xn-1) +12f(xn)Letf(x) =√3x2+ 5. For the integralZ10√3x2+ 5dx,a= 0 andb= 1. Moreover,from the question,n= 4.The width is Δx=b-an=1-04= 0.25.So,x0=a= 0x3= 0.5 + 0.25 = 0.75x1= 0 + 0.25 = 0.25x4= 0.75 + 0.25 = 1 =bx2= 0.25 + 0.25 = 0.5Evaluating the function at eachxi, we getf(x0) =p3(0)2+ 5≈2.23606f(x3) =p3(0.75)2+ 5≈2.58602f(x1) =p3(0.25)2+ 5≈2.27760f(x4) =p3(1)2+ 5≈2.82842f(x2) =p3(0.5)2+ 5≈2.39791By thetrapezoidal rule,Z10√3x2+ 5dx≈Δx12f(x0) +f(x1) +f(x2) +f(x3) +12f(x4)= 0.2512(2.23606) + 2.27760 + 2.39791 + 2.58602 +12(2.82842)2022 Paul Tsopm´en´e All Rights Reserved
9= 0.25(9.79377) = 2.4484425Thus,Z10√3x2+ 5dx≈2.44844 .5.[Exercise onpage 1]Simpson’s RuleLetfbe a continuous function on the interval [a, b].Suppose we subdivide[a, b] into an even numbernof equal subintervals by pointsa=x0, x1, x2,· · ·, xn-1, xn=b.Let Δx=b-anbe the width of each subinterval. Then, bySimpson’s rule,Zbaf(x)dx≈Δx3[f(x0) + 4f(x1) + 2f(x2) + 4f(x3) +· · ·+ 2f(xn-2) + 4f(xn-1) +f(xn)]Note the pattern of coefficients:1,4,2,4,2,4,2,4,2,· · ·,4,2,4,1Examples:Ifn= 2, Δx=b-a2andZbaf(x)dx≈Δx3[f(x0) + 4f(x1) +f(x2)]Ifn= 4, Δx=b-a4andZbaf(x)dx≈Δx3[f(x0) + 4f(x1) + 2f(x2) + 4f(x3) +f(x4)]Ifn= 6, Δx=b-a6andZbaf(x)dx≈Δx3[f(x0) + 4f(x1) + 2f(x2) + 4f(x3) + 2f(x4) + 4f(x5) +f(x6)]Consider the integralZ10√3x2+ 5dx.In the previous question, we foundxiand2022 Paul Tsopm´en´e All Rights Reserved
10f(xi),0≤i≤4. Using these values andSimpson’s rulewithn= 4, we haveZ10√3x2+ 5dx≈Δx3[f(x0) + 4f(x1) + 2f(x2) + 4f(x3) +f(x4)]=0.253[2.23606 + 4(2.27760) + 2(2.39791) + 4(2.58602) + 2.82842]=0.253(29.31478)≈2.44289Aside:The integralZ10√3x2+ 5dcannot be evaluated by any technique we haveseen so far.Using a technique that we won’t discuss here, one can show that theexact value of this integral to five decimal places is 2.44291.6.[Exercise onpage 2](a)Letf(x) =√9-x2. We first need to find the points of intersection betweenthe curvey=√9-x2and thex-axis.This will give us the interval(s).Set√9-x2= 0 to find√9-x2= 09-x2= 9x2= 9x=±√9x=±3So, the interval is [-3,3], and the area under the curvey=f(x) and above thex-axis is given by the definite integralZ3-3√9-x2dx.We now approximate this using the trapezoidal rule withn= 6.Firstly, thewidth of each subinterval isΔx=b-an=3-(-3)6=66= 1.2022 Paul Tsopm´en´e All Rights Reserved
11Using this, we havex0=-3x1=-2x2=-1x3= 0x4= 1x5= 2x6= 3Evaluatingf(x) at eachxi, we getf(x0) = 0f(x1)≈2.23606f(x2)≈2.82842f(x3) = 3f(x4)≈2.82842f(x5)≈2.23606f(x6) = 0By thetrapezoidal rule,Z3-3√9-x2dx≈Δx12f(x0) +f(x1) +f(x2) +f(x3) +f(x4) +f(x5) +12f(x6)= 112(0) + 2.23606 + 2.82842 + 3 + 2.82842 + 2.23606 +12(0)=13.12896(b)We can use the same Δxandf(xi)’s from the previous question. However, withSimpson’s rule, we haveZ3-3√9-x2dx≈Δx3[f(x0) + 4f(x1) + 2f(x2) + 4f(x3) + 2f(x4) + 4f(x5) +f(x6)]=13[0 + 4(2.23606) + 2(2.82842) + 4(3) + 2(2.82842) + 4(2.23606) + 0]=13(41.20216)≈13.73405(c)Geometrically, the area of the region under the curvey=√9-x2above thex-axis is half of the area of the circle of radiusr= 3. That is,Z3-3√9-x2dx=πr22=π(3)22=9π2≈14.13716So,the result obtained using Simpson’s rule is more accurate.7.[Exercise onpage 2]The total cost over the given period is given by the definite integral of the marginal2022 Paul Tsopm´en´e All Rights Reserved
12cost from 1 to 5, that is,Z51C0(x)dxLetx0= 1x1= 2x2= 3x3= 4x4= 5These points determinen= 4 intervals:[1,2],[2,3],[3,4],and[4,5]The width of each interval is clearly Δx= 1. UsingSimpson’s rule, we findZ51C0(x)dx≈Δx3[C0(x0) + 4C0(x1) + 2C0(x2) + 4C0(x3) +C0(x4)]=13[7 + 4(7.3) + 2(7.6) + 4(7.4) + 9]From table=13(7 + 29.2 + 15.2 + 29.6 + 9)=13(90) = 30Therefore, the total cost over the given period is approximately$3000 (remember,the cost is in hundreds of dollars).2022 Paul Tsopm´en´e All Rights Reserved