# At t 9 therefore a 9 100 e 45 034 100points a

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At t = 9, therefore, A (9) = \$100 e 0 . 45 . 034 10.0points A software company finds that the marginal price at x units of demand per week for its software is proportional to the price p . At a price of \$100 there is no demand, while the demand is for 70 units at a price of \$10. How should the company price the software if it wishes to sell 90 units per week? 1. price = \$6 . 17 2. price = \$8 . 17 3. price = \$7 . 17 4. price = \$5 . 17 correct 5. price = \$9 . 17 Explanation: As a function of demand x , the marginal price is given by the differential equation dp dx = kp, since the marginal price is proportional to the price p ; note that the derivative will be neg- ative because demand will decrease as price
pacheco (jnp926) – Homework 5 – staron – (52840) 18 increases. After Integration this differential equation becomes integraldisplay 1 A dA = integraldisplay r 100 dt. Thus ln p = kx + C. In turn, after exponentiation this becomes p ( x ) = e C e kx = Ae kx with C , and hence A , an arbitrary constant. The value of A is detemined by the condition p (0) = 100, for then p ( x ) = 100 e kx . On the other hand, the value of k is deter- mined by the condition p (70) = 10, for then 10 = 100 e 70 k , so k = 1 70 ln parenleftBig 100 10 parenrightBig . Thus p ( x ) = 100 e x 70 ln(100 / 10) . At x = 90, therefore, p (90) = 100 e 90 70 ln(10) = \$5 . 17 . 035 10.0points Scientists began studying the elk popula- tion in Yellowstone Park in 1990 when there were 700 elk. They determined that t years after the study began the population size, N ( t ), was increasing at a rate proportional to 1300 N ( t ). If the population was 800 in year 2000, estimate the size of the elk population in year 2010? 1. population size 904 2. population size 944 3. population size 884 correct 4. population size 924 5. population size 864 Explanation: The population size, N ( t ), satisfies the equations dN dt = k (1300 N ) , N (0) = 700 with k a constant. Thus ln(1300 N ) = kt + C which after exponentiation becomes N ( t ) = 1300 Ae kt where A is an arbitrary constant. Now A is determined by the initial condition on N ( t ) because N (0) = 700 = A = 600 . Consequently, N ( t ) = 1300 600 e kt . On the other hand, the value of k is deter- mined by the population size in year 2000 when N (10) = 800, for then 500 = 600 e 10 k . Taking logs of both sides we thus see that k = 1 10 ln parenleftBig 6 5 parenrightBig . In year 2010, therefore, N (20) = 1300 600 e 2ln 6 5 . Hence, in year 2010, population size 884 . 036 10.0points
pacheco (jnp926) – Homework 5 – staron – (52840) 19 A population is modeled by the differential equation dP dt = 1 . 9 P parenleftbigg 1 P 4000 parenrightbigg . For what values of P is the population in- creasing? 1. P > 2000 2. 0 < P < 4000 correct 3. P > 4000 4. P > 1 . 9 5. 0 < P < 1 . 9 Explanation: 1 . 7 P parenleftBig 1 P 4500 parenrightBig In order to check for increase, the derivative must be above 0. 1 . 7 P parenleftBig 1 P 4500 parenrightBig > 0 1 . 7 P 1 . 7 P 2 4500 > 0 1 . 7 P > 1 . 7 P 2 4500 1 . 7 P 4500 > 1 . 7 P 2 4500 4500 7650 P 1 . 7 > 1 . 7 P 2 1 . 7 4500 P P > P 2 P 4500 > P . P cannot be negative because it is used to represent population. 037 10.0points Scientists can determine the age of ancient ob- jects by a method called radiocarbondating .