In Problem 9, suppose we have the additional information that the population of splake in 2004 was estimated to be 5000. Use a logistic model to estimate the population of splake in the year 2020. What is the predicted limiting population? [Hint: Use the formulas in Problem 12.

Given, that in1990, thepopulation of splake in the lake was1000 and it was estimated to be 3000 in 1997 and 5000 in 2004. We have to find estimated population of splake in the year 2020 and the predicted limiting population.

Here, we have initial population,$p_0=1000$

$$\begin{gathered} p_a=3000 \\ {p}_b=5000 \end{gathered}$$

$t_a=7$(Because, 1997-1990=7)

$t_b=14$(Because, 2004-1990=14)

We will use the following formula to find theestimated population of splake in the year 2020,

$\mathit{p}\left(t\right)\mathbf{=}\frac{{\mathbf{p}}_{\mathbf{0}}{\mathbf{p}}_{\mathbf{1}}}{{\mathbf{p}}_{\mathbf{0}}\mathbf{+}\left(p_1-p_0\right){\mathbf{e}}^{\mathbf{-}\mathbf{A}{\mathbf{p}}_{\mathbf{1}}\mathbf{t}}}\mathbf{·}\mathbf{·}\mathbf{·}\mathbf{·}\mathbf{·}\mathbf{·}\left(1\right)$

To find the values of p₁and A, we will use the following formulas from problem 12,

$$\begin{gathered} {\mathit{p}}_{\mathbf{1}}\mathbf{=}\left[\frac{p_a{p}_b-2p_0{p}_b+p_0{p}_a}{p_a^2-p_0{p}_b}\right]{\mathit{p}}_{\mathbf{a}}\mathbf{,}\mathbf{·}\mathbf{·}\mathbf{·}\mathbf{·}\mathbf{·}\mathbf{·}\left(2\right) \\ \mathit{A}\mathbf{=}\frac{\mathbf{1}}{{\mathbf{p}}_{\mathbf{1}}{\mathbf{t}}_{\mathbf{a}}}\mathit{l}\mathit{n}\left[\frac{p_b\left(p_a-p_0\right)}{p_0\left(p_b-p_a\right)}\right]\mathbf{·}\mathbf{·}\mathbf{·}\mathbf{·}\mathbf{·}\mathbf{·}\left(3\right) \end{gathered}$$

One will find the values of and A, using the formulas from equation (2 and 3),

$$\begin{gathered} {\mathit{p}}_{\mathbf{1}}\mathbf{=}\left[\frac{\left(3000\right)\left(5000\right)-2\left(1000\right)\left(5000\right)+\left(1000\right)\left(3000\right)}{{\left(3000\right)}^2-\left(1000\right)\left(5000\right)}\right]\left(3000\right) \\ {\mathit{p}}_{\mathbf{1}}\mathbf{=}\mathbf{6000} \\ \mathit{A}\mathbf{=}\frac{\mathbf{1}}{\left(6000\right)\left(7\right)}\mathit{l}\mathit{n}\left[\frac{\left(5000\right)\left(3000-1000\right)}{\left(1000\right)\left(5000-3000\right)}\right] \\ \mathit{A}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{00003832} \end{gathered}$$

We will use these values of p₁ and A in equation (1) to find the estimated population of splake in the year 2020.

To find the estimated population of splake in the year 2020, we will substitute t=30 and other values from step1 and step3,

$$\begin{gathered} \mathit{p}\left(30\right)\mathbf{=}\frac{\left(1000\right)\left(6000\right)}{\left(1000\right)\mathbf{+}\left(6000-1000\right){\mathbf{e}}^{\mathbf{-}\left(0.00003832\right)\left(6000\right)\left(30\right)}} \\ \mathit{p}\left(30\right)\mathbf{=}\mathbf{5970} \end{gathered}$$

Hence, the estimated population of splake in the year 2020 is 5970.

Thus, thepredicted limiting population is 6000.