Chapter 4, Problem 11P

Logistic Population Growth The table and scatter plot give the population of black flies in a closed laboratory container over an 18-day period.

1. (a) Use the Logistic command on your calculator to find a logistic model for these data.
2. (b) Use the model to estimate the time when there were 400 flies in the container.

Time (days)	Number of flies
0	10
2	25
4	66
6	144
8	262
10	374
12	446
16	492
18	498

To determine

To find: The population logistic model for the data of the population of the black flies in a closed laboratory container over an 18-day period.

Answer to Problem 11P

The population logistic model that gives the best fits of the data point is

$y=\frac{500.8558}{1+e^{-0.4981\left(x-7.8176\right)}}$, where the variable $y$ represents the number of flies and $x$ represents the time in days.

Explanation of Solution

Given:

The data is listed below in table by using the logistics command graphing calculator.

Times (days)	Number of flies
0	10
2	25
4	66
6	144
8	262
10	374
12	446
16	492
18	498

Formula used:

A logistic growth model is a function of the form $f\left(t\right)=\frac{c}{1+a{e}^{-bt}}$, where $a,b$ and $c$ are positive constants.

Calculation:

Let us consider the variable $x$ represents the time in days and the variable $y$ represents the number of flies.

Use online graphing calculator and draw the logistic growth function of the form $f\left(t\right)=\frac{c}{1+a{e}^{-bt}}$ that gives the best fits of the data point as shown below in Figure 1.

From Figure 1, it is noticed that the logistic growth model $y=\frac{500.8558}{1+e^{-0.4981\left(x-7.8176\right)}}$ gives the best fit of the data point with the scatter plot.

Therefore, The population logistic model that gives the best fits of the data point is $y=\frac{500.8558}{1+e^{-0.4981\left(x-7.8176\right)}}$, where the variable $y$ represents the number of flies and $x$ represents the time in days.

To determine

To estimate: The value of time when there are 400 flies in the container.

Answer to Problem 11P

The time when the number of flies in the container are 400 at $x=10.584\mathrm{days}$.

Explanation of Solution

From part (a) the number of flies $y$ at the time $x$ given by the logistic growth model $y=\frac{500.8558}{1+e^{-0.4981\left(x-7.8176\right)}}$.

Here need to calculate the time, when the numbers of flies in the container are 400.

Substitute $y=400$ in $y$ and the time $x$ calculated as follows,

$\begin{array}{c}400=\frac{500.8558}{1+e^{-0.4981\left(x-7.8176\right)}}\\ 1+e^{-0.4981\left(x-7.8176\right)}=\frac{500.8558}{400}\\ e^{-0.4981\left(x-7.8176\right)}=1.2521395-1\\ e^{-0.4981\left(x-7.8176\right)}=0.2521395\end{array}$

Take $\ln$ both sides and solve for $x$.

$\begin{array}{l}\ln e^{-0.4981\left(x-7.8176\right)}=\ln \left(0.2521395\right)\\ -0.4981\left(x-7.8176\right)=-1.37777277\\ 0.4981x=1.37777277+3.89394656\\ x=\frac{5.27171933}{0.4981}\end{array}$

Thus, value of time when the number of flies in the container is $x=10.584\mathrm{days}$.

Therefore, the time when the number of flies in the container are 400 at $x=10.584\mathrm{days}$