/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 13 The amount of vegetation eaten i... [FREE SOLUTION] | 91Ó°ÊÓ

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The amount of vegetation eaten in a day by a grazing animal is a function of the amount \(V\) of food available (measured as biomass, in units such as pounds per acre). \({ }^{21}\) 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 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 $$ G=2.5-4.8 e^{-0.004 V} $$ 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 $$ R=1.9-1.9 e^{-0.033 V} $$ 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?

Short Answer

Expert verified
The graph for both is concave down, western grey has a minimal biomass of about 189, and its satiation level is 2.5 pounds; red kangaroo is more efficient.

Step by step solution

01

Understanding the General Functional Response

The problem describes a functional response for the grazing animals, showing how their daily intake of food (G or R) depends on the vegetation biomass V. As V increases, the daily intake increases until it reaches a saturation point.
02

Graphing the Function for the Western Grey Kangaroo

Use the equation G=2.5-4.8e^{-0.004V} to plot the graph for values of V ranging from 0 to 2000. This will demonstrate how G increases as V increases, approaching a maximum level.
03

Analyzing Concavity of the Graph

Observe the graph of G plotted against V to determine its concavity. Concave down implies that the increase in daily intake slows down as V increases, which shows there is diminishing return on food intake as biomass increases.
04

Calculating Minimal Vegetation Biomass for Western Grey Kangaroo

Set G=0 and solve the equation G=2.5-4.8e^{-0.004V} for V. This will give you the minimal vegetation biomass threshold below which the kangaroo cannot eat.
05

Determining Satiation Level for Western Grey Kangaroo

Determine the maximum value of G, which occurs as V approaches infinity. This is the satiation level at G=2.5 pounds per day.
06

Graphing the Function for the Red Kangaroo

Use the equation R=1.9-1.9e^{-0.033V} to add this graph on top of the G(V) graph. This demonstrates how R against V behaves compared with G(V).
07

Comparing Grazing Efficiency

Find the minimal vegetation biomass level for the red kangaroo using its functional equation, similar to step 4. Compare these minimal vegetation levels to conclude which kangaroo is more efficient grazer.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Grazing Efficiency
Grazing efficiency is a measure of how effectively an animal can feed on available vegetation. It's particularly crucial for animals like kangaroos, whose habitat might offer varying amounts of grass and plants depending on the season. An efficient grazer is one who can find the necessary nutrients even when food is scarce.
For kangaroos, grazing efficiency can be assessed by examining the minimal vegetation biomass level. This is the lowest amount of vegetation biomass needed for the animal to start eating. Lower values here mean higher efficiency as it implies that even with scarce food, the animal can gather enough to survive.
To find this level for the western grey kangaroo, we solve the functional response equation, setting the daily intake \( G \) to zero. In =1, we found that the western grey kangaroo will eat when there is more than its minimal threshold of biomass available. Similarly, perform this calculation for the red kangaroo. By comparing the two thresholds, you can determine which kangaroo is more efficient at grazing. In this case, the red kangaroo has a lower threshold, indicating higher grazing efficiency.
Satiation Level
The satiation level is the maximum amount of food an animal will consume, regardless of available resources. Animals have a limit to how much they can eat in a given period, and this is crucial for managing their energy and nutritional needs.
For the western grey kangaroo, the satiation level is determined by analyzing the equation for its functional response. In mathematical terms, this is the limit of \( G = 2.5 - 4.8e^{-0.004V} \) as \( V \) approaches infinity. Essentially, as the biomass \( V \) increases beyond a certain point, the exponential term \( e^{-0.004V} \) becomes negligible.
This makes the functional response reach a steady value where \( G \) is equal to 2.5 pounds per day. This represents the maximum that the kangaroo will eat, even if there's a limitless supply of grass. Understanding satiation levels helps in wildlife management and conservation by ensuring that animals do not deplete their food sources unnecessarily.
Concavity Analysis
Concavity in graph analysis provides insights into how a function behaves as its variable changes. It's especially helpful in understanding the feeding patterns of kangaroos over various levels of available food.
For the western grey kangaroo, the graph of \( G \) against \( V \) is concave down. A concave down graph indicates a decreasing rate of growth. Practically, this signifies that as more biomass is available, each additional unit provides less of an increase in food intake.
This diminishing return mirrors how animals optimize their intake up until satiation. While initially increasing rapidly with biomass, the daily intake rate slows down as the kangaroo approaches its satiation level. Concavity analysis therefore helps in understanding the efficiency and eating patterns relative to varying food supplies.

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Most popular questions from this chapter

The manager of an employee health plan for a firm has studied the balance \(B\), in millions of dollars, in the plan account as a function of \(t\), the number of years since the plan was instituted. He has determined that the account balance is given by the formula \(B=60+7 t-50 e^{0.1 t}\). a. Make a graph of \(B\) versus \(t\) over the first 7 years of the plan. b. At what time is the account balance at its maximum? c. What is the smallest value of the account balance over the first 7 years of the plan?

If you borrow \(\$ 120,000\) at an APR of \(6 \%\) in order to buy a home, and if the lending institution compounds interest continuously, then your monthly payment \(M=M(Y)\), in dollars, depends on the number of years \(Y\) you take to pay off the loan. The relationship is given by $$ M=\frac{120000\left(e^{0.005}-1\right)}{1-e^{-0.06 Y}}. $$ a. Make a graph of \(M\) versus \(Y\). In choosing a graphing window, you should note that a home mortgage rarely extends beyond 30 years. b. Express in functional notation your monthly payment if you pay off the loan in 20 years, and then use the graph to find that value. c. Use the graph to find your monthly payment if you pay off the loan in 30 years. d. From part b to part \(\mathrm{c}\) of this problem, you increased the debt period by \(50 \%\). Did this decrease your monthly payment by \(50 \%\) ? e. Is the graph concave up or concave down? Explain your answer in practical terms. f. Calculate the average decrease per year in your monthly payment from a loan period of 25 to a loan period of 30 years.

Here is a model for the number of students enrolled in U.S. public high schools as a function of time since 1965 : $$ N=-0.02 t^{2}+0.44 t+11.65 . $$ In this formula \(N\) is the enrollment in millions of students, \(t\) is the time in years since 1965 , and the model is applicable from 1965 to \(1985 .\) a. Calculate \(N(7)\) and explain in practical terms what it means. b. In what year was the enrollment the largest? What was the largest enrollment? c. Find the average yearly rate of change in enrollment from 1965 to 1985 . Is the result misleading, considering your answer to part b?

Recall that the APR (the annual percentage rate) is the percentage rate on a loan that the Truth in Lending Act requires lending institutions to report on loan agreements. It does not tell directly what the interest rate really is. If you borrow money for 1 year and make no payments, then in order to calculate how much you owe at the end of the year, you must use another interest rate, the EAR (the effective annual rate), which is not normally reported on loan agreements. The calculation is made by adding the interest indicated by the EAR to the amount borrowed. The relationship between the APR and the EAR depends on how often interest is compounded. If you borrow money at an annual percentage rate APR (as a decimal), and if interest is compounded \(n\) times per year, then the effective annual rate EAR (as a decimal) is given by $$ \mathrm{EAR}=\left(1+\frac{\mathrm{APR}}{n}\right)^{n}-1 $$ For the remainder of this problem, we will assume an APR of \(10 \%\). Thus in the formula above, we would use \(0.1\) in place of APR. a. Would you expect a larger or a smaller EAR if interest is compounded more often? Explain your reasoning. b. Make a table that shows how the EAR depends on the number of compounding periods. Use your table to report the EAR if interest is compounded once each year, monthly, and daily. (Note: The formula will give the EAR as a decimal. You should report your answer as a percent with three decimal places.) c. If you borrow \(\$ 5000\) and make no payments for 1 year, how much will you owe at the end of a year if interest is compounded monthly? If interest is compounded daily? d. If interest is compounded as often as possiblethat is, continuously - then the relationship between APR and EAR is given by $$ \mathrm{EAR}=e^{\mathrm{APR}}-1 \text {. } $$ Again using an APR of 10\%, compare the EAR when the interest is compounded monthly with the EAR when the interest is compounded continuously.

The background for this exercise can be found in Exercises 11, 12, 13, and \(14 \mathrm{in} \mathrm{Sec}\) tion 1.4. A manufacturer of widgets has fixed costs of \(\$ 700\) per month, and the variable cost is \(\$ 65\) per thousand widgets (so it costs \(\$ 65\) to produce 1 thousand widgets). Let \(N\) be the number, in thousands, of widgets produced in a month. a. Find a formula for the manufacturer's total cost \(C\) as a function of \(N\). b. The highest price \(p\), in dollars per thousand widgets, at which \(N\) can be sold is given by the formula \(p=75-0.02 \mathrm{~N}\). Using this, find a formula for the total revenue \(R\) as a function of \(N\). c. Use your answers to parts a and \(b\) to find \(a\) formula for the profit \(P\) of this manufacturer as a function of \(N\). d. Use your formula from part c to determine the two break-even points for this manufacturer. Assume that the manufacturer can produce at most 500 thousand widgets in a month.

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