Concept explainers
OXYGEN CONTENT OF A POND The oxygen content t days after organic waste has been dumped into a pond is given by
percent of its normal level.
- a. Show that f(0) = 100 and f(10) = 75.
- b. Use the Intermediate Value Theorem to conclude that the oxygen content of the pond must have been at a level of 80% at some time.
- c. At what time(s) is the oxygen content at the 80% level?
Hint: Use the
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Chapter 2 Solutions
Applied Calculus for the Managerial, Life, and Social Sciences: A Brief Approach
- 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?arrow_forwardDecay 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.arrow_forwardWater Flea F. E Smith has reported on population growth of the water flea. In one experiment, he found that the time t, in days, required to reach a population of N is given by the relation e0.44t=NN0(228N0228N)4.46. Here N0 is the initial population size. If the initial population size is 50, how long is required for the population to grow to 125?arrow_forward
- 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?arrow_forwardMaximum Sales Growth This is a continuation of Exercise 10. In this exercise, we determine how the sales level that gives the maximum growth rate is related to the limit on sales. Assume, as above, that the constant of proportionality is 0.3, but now suppose that sales grow to a level of 4 thousand dollars in the limit. a. Write an equation that shows the proportionality relation for G. b. On the basis of the equation from part a, make a graph of G as a function of s. c. At what sales level is the growth rate as large as possible? d. Replace the limit of 4 thousand dollars with another number, and find at what sales level the growth rate is as large as possible. What is the relationship between the limit and the sales level that gives the largest growth rate? Does this relationship change if the proportionality constant is changed? e. Use your answers in part d to explain how to determine the limit if we are given sales data showing the sales up to a point where the growth rate begins to decrease.arrow_forward
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