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Att= 9, therefore,A(9) = $100e0.45.03410.0pointsA software company finds that the marginalprice atxunits of demand per week for itssoftware is proportional to the pricep. At aprice of $100 there is no demand, while thedemand is for 70 units at a price of $10. Howshould the company price the software if itwishes to sell 90 units per week?1.price = $6.172.price = $8.173.price = $7.174.price = $5.17correct5.price = $9.17Explanation:As a function of demandx, the marginalprice is given by the differential equationdpdx=−kp,since the marginal price is proportional to thepricep; note that the derivative will be neg-ative because demand will decrease as price
pacheco (jnp926) – Homework 5 – staron – (52840)18increases.After Integration this differentialequation becomesintegraldisplay1AdA=integraldisplayr100dt.Thuslnp=−kx+C.In turn, after exponentiation this becomesp(x) =eCe−kx=Ae−kxwithC, and henceA, an arbitrary constant.The value ofAis detemined by the conditionp(0) = 100, for thenp(x) = 100e−kx.On the other hand, the value ofkis deter-mined by the conditionp(70) = 10, for then10 = 100e−70k, sok=170lnparenleftBig10010parenrightBig.Thusp(x) = 100e−x70ln(100/10).Atx= 90, therefore,p(90) = 100e−9070ln(10)= $5.17.03510.0pointsScientists began studying the elk popula-tion in Yellowstone Park in 1990 when therewere 700 elk.They determined thattyearsafter the study began the population size,N(t), was increasing at a rate proportional to1300−N(t). If the population was 800 in year2000, estimate the size of the elk populationin year 2010?1.population size≈9042.population size≈9443.population size≈884correct4.population size≈9245.population size≈864Explanation:The population size,N(t),satisfies theequationsdNdt=k(1300−N),N(0) = 700withka constant. Thus−ln(1300−N) =kt+Cwhich after exponentiation becomesN(t) = 1300−Ae−ktwhereAis an arbitrary constant.NowAisdetermined by the initial condition onN(t)becauseN(0) = 700=⇒A= 600.Consequently,N(t) = 1300−600e−kt.On the other hand, the value ofkis deter-mined by the population size in year 2000whenN(10) = 800, for then500 = 600e−10k.Taking logs of both sides we thus see thatk=110lnparenleftBig65parenrightBig.In year 2010, therefore,N(20) = 1300−600e−2ln65.Hence, in year 2010,population size≈884.03610.0points
pacheco (jnp926) – Homework 5 – staron – (52840)19A population is modeled by the differentialequationdPdt= 1.9Pparenleftbigg1−P4000parenrightbigg.For what values ofPis the population in-creasing?1.P >20002.0< P <4000correct3.P >40004.P >1.95.0< P <1.9Explanation:1.7PparenleftBig1−P4500parenrightBigIn order to check for increase, the derivativemust be above 0.1.7PparenleftBig1−P4500parenrightBig>01.7P−1.7P24500>01.7P >1.7P245001.7P∗4500>1.7P24500∗45007650P1.7>1.7P21.74500PP>P2P4500> P.P cannot be negative because it is used torepresent population.03710.0pointsScientists can determine the age of ancient ob-jects by a method calledradiocarbondating.