Chapter MPS, Problem 11.7MPS

a.

To determine

Write a recursive formula.

Answer to Problem 11.7MPS

  $a_n=0.8a_{n-1}1000$

Explanation of Solution

Given information:

In a city fishing pond, the city starts with $4000$ trout. $20\%$ of the fish are caught each year, so the city adds $1000$ fish every spring.

Calculation:

Assume that $a_n$ is the number of fish in the pond after $n-1th$ month.

Initially, number of fish in the pond is $a_1=4000$ .

Each year $1000$ fish are added and $20\%$ of the fish are caught.

Thus,

  $\begin{array}{l}{a}_2=a_1-0.2a_1+1000\\ a_2=0.8a_1+1000\end{array}$

Then,

  $a_3=0.8a_2+1000$

Hence, the recursive formula is $a_n=0.8a_{n-1}1000$ .

b.

To determine

Find the number of fish in the pond after $5$ years.

Answer to Problem 11.7MPS

  $4672$

Explanation of Solution

Given information:

In a city fishing pond, the city starts with $4000$ trout. $20\%$ of the fish are caught each year, so the city adds $1000$ fish every spring.

Calculation:

The recursive formula for the number of fish after $\left(n-1\right)$ th month is,

  $a_n=0.8a_{n-1}1000$

Substitute $n=6$ ,

  $\begin{array}{l}{a}_n=0.8a_{n-1}1000\\ a_6=0.8a_5+1000\end{array}$

Find the value of $a_5$ as follows:

  $\begin{array}{l}{a}_2=0.8a_1+1000\\ a_2=0.8\times 4000+1000\\ a_2=4200\\ a_3=0.8a_2+1000\\ a_3=0.8\times 4200+1000\\ a_3=4360\\ a_4=0.8a_3+1000\\ a_4=0.8\times 4360+1000\\ a_4=4488\\ a_5=0.8a_4+1000\\ a_5=0.8\times 4488+1000\\ a_5=4590.4\\ a_5\approx 4590\end{array}$

Now, substitute $a_5=4590$ in formula $a_6=0.8a_5+1000$ .

  $\begin{array}{l}{a}_6=0.8a_5+1000\\ a_6=0.8\times 4590+1000\\ a_6=4672\end{array}$

Hence, the number of fish in the pond after $5$ years is $4672$ .

c.

To determine

Find the number of fish needed to maintain a constant population.

Answer to Problem 11.7MPS

  $5000$

Explanation of Solution

Given information:

In a city fishing pond, the city starts with $4000$ trout. $20\%$ of the fish are caught each year, so the city adds $1000$ fish every spring.

Calculation:

The recursive formula is $a_n=0.8a_{n-1}+1000$ .

For constant population, $a_n\&a_{n-1}$ will remain same.

  $\begin{array}{l}{a}_n=0.8a_{n-1}+1000\\ a=0.8a+1000,Use{a}_n=a,and,a_{n-1}=a\\ a-0.8a=1000\\ 0.2a=1000\\ a=\frac{1000}{0.2}\\ a=5000\end{array}$

Hence, the number of fish needed to maintain a constant population is $5000$ .