Q 7 I need help with calculus Growth of Managed Srvices Almost half of companies let other firms manage some of their Web operations-a practice called Webhosting. Managed services-monitoring a customer's technology services-is the fastest growing part of Web hosting. Managed services salesare expected to grow in accordance by the functionf(t) = 0.469t? + 0.758t + 0.44 (0 st s 10)where f(t) is measured in billions of dollars and t is measured in years, with t = 0 corresponding to 1999.1(a) Find the interval where f is increasing. (Enter your answer using interval notation.)(b) What does your result tell you about sales in managed services from 1999 through 2009?In 1999 managed service sales were much bigger than in 2009.Managed service sales were increasing during this period.O Managed service sales were decreasing during this period.First managed service sales were increasing and then became a constant.Managed service sales remained the same during this period.

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Functions and Change: A Modeling Approach to College Algebra (MindTap Course List)

6th Edition

ISBN:9781337111348

Author:Bruce Crauder, Benny Evans, Alan Noell

Publisher:Bruce Crauder, Benny Evans, Alan Noell

Chapter2: Graphical And Tabular Analysis

Section2.4: Solving Nonlinear Equations

Problem 21E: Grazing Kangaroos The amount of vegetation eaten in a day by a grazing animal V of food available...

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Grazing Kangaroos The amount of vegetation eaten in a day by a grazing animal V of food available measured as biomass, in units such as pounds per acre. This relationship is called the functional response. If there is little vegetation available, the daily intake will be small, since the animal will have difficulty finding and eating the food. As the amount of food biomass increases, so does the daily intake. Clearly, though, there is a limit to the amount the animal will eat, regardless of the amount of food available. This maximum amount eaten is the satiation level. a.For the western grey kangaroo of Australia, the functional response is G=2.54.8e0.004V, where G=G(V) is the daily intake measured in pounds and V is the vegetation biomass measured in pounds per acre. i. Draw a graph of G against V. Include vegetation biomass levels up to 2000 pounds per acre. ii. Is the graph you found in part i concave up or concave down? Explain in practical terms what your answer means about how this kangaroo feeds. iii. There is a minimal vegetation biomass level below which the western grey kangaroo will eat nothing. Another way of expressing this is to say that the animal cannot reduce the food biomass below this level. Find this minimal level. iv. Find the satiation level for the western grey kangaroo. b. For the red kangaroo of Australia, the functional response is R=1.91.9e0.033V, Where R is the daily intake measured in pounds and V is the vegetation biomass measured in pounds per acre. i. Add the graph of R against V to the graph of G you drew in part a. ii. A simple measure of the grazing efficiency of an animal involves the minimal vegetation biomass level described above: The lower the minimal level for an animal, the more efficient it is at grazing. Which is more efficient at grazing, the western grey kangaroo or the red kangaroo?
Sales Growth In this exercise, we develop a model for the growth rate G, in thousands of dollars per year, in sales of the product as a function of the sales level s, in thousands of dollars. The model assumes that there is a limit to the total amount of sales that can be attained. In this situation, we use the term unattained sales for difference this limit and the current sales level. For example, if we expect sales grow to 3 thousand dollars in the long run, then 3-s is the unattained sales. The model states that the growth rate G is proportional to the product of the sales level s, and the unattained sales. Assume that the constant of proportionality is 0.3 and that the sales grow to 2 thousand dollars in the long run. a.Find the formula for unattained sales. b.Write an equation that shows the proportionality relation for G. c.On the basis of the equation from the part b, make a graph of G as a function of s. d.At what sales level is the growth rate as large as possible? e.What is the largest possible growth rate?
Decay of Litter Litter such as leaves falls to the forest floor, where the action of insects and bacteria initiates the decay process. Let A be the amount of litter present, in grams per square meter, as a function of time t in years. If the litter falls at a constant rate of L grams per square meter per year, and if it decays at a constant proportional rate of k per year, then the limiting value of A is R=L/k. For this exercise and the next, we suppose that at time t=0, the forest floor is clear of litter. a. If D is the difference between the limiting value and A, so that D=RA, then D is an exponential function of time. Find the initial value of D in terms of R. b. The yearly decay factor for D is ek. Find a formula for D in term of R and k. Reminder:(ab)c=abc. c. Explain why A=RRekt.

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