Question

Mapping the population of fish in a lake year and Pn+1 A quadratic model to predict the number of fish (in thousands) in a lake based on the number of fish that were in the lake the preceding year is given by the function: Pn+1 = (2-0.01pm)P, -16 where p., is the number of fish in the lake (in thousands) at the start of the is the number of fish in the lake at the start of next year. 1.) Suppose there are 60,000 fish in the lake at the start of this year, how many fish will be in the lake of the next year? Is this an increase or a decrease in the fish population? 2.) Starting with a population of 70,000 fish in the lake at the start of this year, compute the population of fish in the lake for the next 10 years. What do you notice? 3.) Starting with a population of 85,000 fish in the lake at the start of this year, compute the population of fish in the lake for the next 10 years. What do you notice? 4.) Based on the previous two problems, what population would remain unchanged year after year? This would be considered a stable population. 5.) Algebraically solve for the stable population by solving the equation p=(2-0.01p)p-16. You should have two solutions, what can you say about both solutions? 6.) In the model, the term 16 represents 16,000 fish that are harvested each year so we subtract 16 to figure out the number of next year's population. This would be considered a sustainable amount to harvest. Suppose the amount to harvest increased to 20,000 fish, what would be the model in this scenario? If the number of fish in the lake at the previous year is 85,000, how many fish will be present at the start of this year? 7.) What is the new stable population? Is this better than the previous model in terms of sustainability? 8.) What happens if you try to harvest 30,000 fish each year? Does this lead to a stable population? 9.) Based on this model, if you were in charge of managing the lake and its resources, how many fish would you permit to be harvested each year?

Answer 1

We will use the following function in sage (or python) to get the population after t years (in thousands):

sage: def f(n,t=0): 
....:     if t>0: ....:         return (2-f(n,t-1)*0.01)*f(n,t-1)-16 
....:     else: 
....:         return n 
....:

1) In this case, the code provides the following result:

sage: f(60,1)                                                                                                                                                                                                        
68.0000000000000

This is an increase in the population.

2) In this case, the code provides the following result:

sage: f(70,10)                                                                                                                                                                                                       
79.9983794848289

This implies that there have been an increase over the 10 years

3)

In this case, the code provides the following result:

sage: f(85,10)                                                                                                                                                                                                       
80.0004266459798

This implies that there have been an decrease over the 10 years.

4) Based on the last two parts' solution, we can see that a population of around 80 thousand will stay stable in a long run.

5) From the equation, we get that

p = (2 - 0.01p)p - 16 (1 – 0.01php 16 = 0 0.01p2 - p + 16 = 0 =p2 – 100p + 1600 = 0 p= 80, 20

From the previous problems, we can see that the first solution is a stable solution. And the second solution is a unstable one.