Chapter 12.1, Problem 82AYU

Trout Population A pond currently contains $2000$ trout. A fish hatchery decides to add $20$ trout each month. It is also known that the trout population is growing at a rate of $3\%$ per month. The size of the population after $n$ months is given by the recursively defined sequence

$p_0=\$3000p_n=1.03p_{n-1}+20$.

How many trout are in the pond after $2$ months? That is. what is $p_2$?

To determine

The number of trout in the pond after $2$ months, that is $P_2$ where, a pond currently contains $2000$ trout. A fish hatchery decides to add $20$ trout each month. It is also known that the trout population is growing at a rate of $3\%$ per month. The size of the populations after $n$ months is given by recursive sequence $P_0=2000P_n=1.03P_{n-1}+20$.

Answer to Problem 82AYU

Solution:

The number of trout in the pond after $2$ months, is $P_2=2162$.

Explanation of Solution

Given information:

A pond currently contains $2000$ trout. A fish hatchery decides to add $20$ trout each month. It is also known that the trout population is growing at a rate of $3\%$ per month. The size of the populations after $n$ months is given by recursive sequence $P_0=2000P_n=1.03P_{n-1}+20$.

Explanation:

Consider, the recursive relation, $P_0=2000P_n=1.03P_{n-1}+20$.

To find the number of trout in the pond after $2$ months, that is $P_2$, find the value of $P_1$ first.

Substitute $n=1$ in $P_n=1.03P_{n-1}+20$ .

$\Rightarrow P_1=1.03P_{1-1}+20$

$\Rightarrow P_1=1.03P_0+20$

Substitute $P_0=2000$ in above equation.

$\Rightarrow P_1=1.03\left(2000\right)+20=2080$.

Now to find $P_2$, substitute $n=2$ in $P_n=1.03P_{n-1}+20$ .

$\Rightarrow P_2=1.03P_{2-1}+20$

$\Rightarrow P_2=1.03P_1+20$

Substitute $P_1=2080$ in above equation.

$\Rightarrow P_2=1.03\left(2080\right)+20=2162.4$.

Therefore, the number of trout in the pond after $2$ months is approximately $P_2=2162$.