Question

each year, 40% of a salmon population is extracted from a farming pond. at the beginning of the next year, the pond is stocked with an additional fixed amount of n caught wild salmon. let pn denote the amount of fish at the beginning of the n-th year assume that the initial salmon population on the pond is p0=5000
a. Write a recursion to describe Pn. b. Determine the value of N so the amount of fish remains constant at the beginning of each year.

Answer 1

(a)     $p_{n} =p_{n-1}-0.4p_{0} +n$

(b)       $n = {\frac{p_{n-1} -p_{n} }{0.4p_{0}} }$

Given,

         Initial salmon population $p_{0}=5000$

         Extraction rate = 40% per year

(a) Formula for recursion,

         For the first year,

                $p_{1} =p_{0} -0.4p_{0}+n$

         For the second year,

                $p_{2} =p_{1} -0.4p_{0} +n$

         For the n'th year,

               $p_{n} =p_{n-1}-0.4p_{0} +n$

(b) The value of n,

        From the above equation,

                $p_{n} =p_{n-1}-0.4p_{0} +n$

                  $n = {\frac{p_{n-1} -p_{n} }{0.4p_{0}} }$

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Answer 2

Each year, 40% of the salmon population is extracted from a farming pond.

a. The recursion to describe Pn is Pn = 0.6 Pn-1 + N.

b. The value of N is 3000.

What is recursion?

According to the recursion equation, the quantity of fish at the start of the nth year. Pn, is equal to 0.6 times the quantity at the start of the year before, Pn-1, plus the constant quantity of wild salmon that is restocked every year, N.

b. We can set the recursion equation equal to the initial population of 5000 fish in order to get the value of N that ensures the quantity of fish stays constant at the start of each year. Now we have the equation.

0.6Pn-1 + N = 5000. Solving for N gives us N = 3000.

Therefore, a. Pn = 0.6 Pn-1 + N. b.  3000.

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