Chapter 12, Problem 4P

a.

To determine

To find the recursive formula for the trout population if the trout population changes only because of reproduction.

Answer to Problem 4P

  $P_n=1.20P_{n-1}$ , $P_0=4000$ .

Explanation of Solution

Given:

The trout population increases by 20% through reproduction.

The population of trout at the end of nth year is $P_n$ .

Calculation:

Let the initial population be $P_0$ .

Then,

  $\begin{array}{l}{P}_1=P_0+0.2P_0=1.20P_0\\ P_2=P_1+0.2P_1=1.20P_1\\ P_3=P_2+0.2P_2=1.20P_2\\ \cdot \\ \cdot \\ \cdot \\ P_n=P_{n-1}+0.20P_{n-1}=1.20P_{n-1}\end{array}$

So, the recursive formula becomes $P_n=1.20P_{n-1}$ , $P_0=4000$ .

b.

To determine

To find the recursive formula for the trout population if each year 600 trout are harvested.

Answer to Problem 4P

  $P_n=1.20P_{n-1}+600$ , $P_0=4000$ .

Explanation of Solution

Given:

The trout population increases by 20% through reproduction and each year 600 trout are harvested.

The population of trout at the end of nth year is $P_n$ .

Calculation:

Let the initial population be $P_0$ .

Then,

  $\begin{array}{l}{P}_1=P_0+0.2P_0+600=1.20P_0+600\\ P_2=P_1+0.2P_1+600=1.20P_1+600\\ P_3=P_2+0.2P_2+600=1.20P_2+600\\ \cdot \\ \cdot \\ \cdot \\ P_n=P_{n-1}+0.20P_{n-1}+600=1.20P_{n-1}+600\end{array}$

So, the recursive formula becomes $P_n=1.20P_{n-1}+600$ , $P_0=4000$ .

c.

To determine

To find the recursive formula for the trout population if each year 250 additional trout are introduced.

Answer to Problem 4P

  $P_n=1.20P_{n-1}+250$ , $P_0=4000$ .

Explanation of Solution

Given:

The trout population increases by 20% through reproduction and each year 250 additional trout are introduced.

The population of trout at the end of nth year is $P_n$ .

Calculation:

Let the initial population be $P_0$ .

Then,

  $\begin{array}{l}{P}_1=P_0+0.2P_0+600=1.20P_0+250\\ P_2=P_1+0.2P_1+600=1.20P_1+250\\ P_3=P_2+0.2P_2+600=1.20P_2+250\\ \cdot \\ \cdot \\ \cdot \\ P_n=P_{n-1}+0.20P_{n-1}+600=1.20P_{n-1}+250\end{array}$

So, the recursive formula becomes $P_n=1.20P_{n-1}+250$ , $P_0=4000$ .

d.

To determine

To find the recursive formula for the trout population if each year 10% trout are harvested and 300 additional trout are introduced.

Answer to Problem 4P

  $P_n=\frac{10}{9}\left(1.20P_{n-1}+300\right)$ , $P_0=4000$ .

Explanation of Solution

Given:

The trout population increases by 20% through reproduction and each year 10% trout are harvested and 300 additional trout are introduced.

The population of trout at the end of nth year is $P_n$ .

Calculation:

Let the initial population be $P_0$ .

Then,

  $\begin{array}{c}{P}_1=P_0+0.2P_0+\left(0.10P_1+300\right)\\ 0.90P_1=P_0+0.2P_0+300\\ \frac{9}{10}{P}_1=1.20P_0+300\\ P_1=\frac{10}{9}\left(1.20P_0+300\right)\end{array}$

And,

  $\begin{array}{c}{P}_2=P_1+0.2P_1+\left(0.10P_2+300\right)\\ 0.90P_2=P_1+0.2P_1+300\\ \frac{9}{10}{P}_2=1.20P_1+300\\ P_2=\frac{10}{9}\left(1.20P_1+300\right)\end{array}$

So,

  $\begin{array}{c}{P}_n=P_{n-1}+0.2P_{n-1}+\left(0.10P_n+300\right)\\ 0.90P_n=P_{n-1}+0.2P_{n-1}+300\\ \frac{9}{10}{P}_n=1.20P_{n-1}+300\\ P_n=\frac{10}{9}\left(1.20P_{n-1}+300\right)\end{array}$

So, the recursive formula becomes $P_n=\frac{10}{9}\left(1.20P_{n-1}+300\right)$ , $P_0=4000$ .