assume there is a certain population of fish in a pond whose growth is described by the logistic equation. it is estimated that the carrying capacity for the pond is 1100 fish. absent constraints, the population would grow by 20% per year.
if the starting population is given by ( p_0 = 600 ), then after one breeding season the population of the pond is given by
( p_1=)
after two breeding seasons the population of the pond is given by
( p_2=)

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Accepted Answer

# Explanation:
## Step1: Recall the logistic equation formula
The logistic equation for population growth is $p_{n + 1}=p_{n}+rp_{n}(1-\frac{p_{n}}{K})$, where $r$ is the growth rate, $K$ is the carrying capacity. Given $K = 1100$ and $r=20\%$ (since the population would grow by $20\%$ per year in the absence of constraints, $r = 0.2$), $p_{0}=600$.
For $n = 0$:
$$
\begin{align*}
p_{1}&=p_{0}+rp_{0}(1-\frac{p_{0}}{K})\
&=600+0.2\times600\times(1 - \frac{600}{1100})\
&=600+120\times\frac{1100 - 600}{1100}\
&=600+120\times\frac{500}{1100}\
&=600+\frac{60000}{1100}\
&=600+\frac{600}{11}\
&=\frac{6600 + 600}{11}\
&=\frac{7200}{11}\approx654.55
\end{align*}
$$
## Step2: Calculate $p_{2}$
Now, use $p_{1}\approx654.55$ to calculate $p_{2}$.
$$
\begin{align*}
p_{2}&=p_{1}+rp_{1}(1-\frac{p_{1}}{K})\
&\approx654.55+0.2\times654.55\times(1-\frac{654.55}{1100})\
&=654.55+130.91\times\frac{1100 - 654.55}{1100}\
&=654.55+130.91\times\frac{445.45}{1100}\
&=654.55+\frac{130.91\times445.45}{1100}\
&\approx654.55 + 53.17\
&=707.72
\end{align*}
$$

# Answer:
$p_{1}\approx655$, $p_{2}\approx708$

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QUESTION IMAGE

Question

Explanation:

Step1: Recall the logistic equation formula

Step2: Calculate \(p_{2}\)

Answer: