Chapter 3.2, Problem 25E

Solution

To find: the logistic function that satisfies the given conditions.

  $P\left(t\right)\approx \frac{128}{1+7{\left(0.844\right)}^t}$

Given information:

Initial population = 16, maximum sustainable population eq 128, passing through (5, 32).

Definition Used:

Logistic Growth Function:

Let a , b , c , and k be positive constants, with $b<1$ . A logistic growth function in x is a function that can be written in the form

  $f\left(x\right)=\frac{c}{1+a\cdot b^x}\text{ or }f\left(x\right)=\frac{c}{1+a\cdot e^{-kx}}$

Where the constant c is the limit to growth.

Explanation:

Now, the required function will be of the form:

  $P\left(t\right)=\frac{c}{1+a\cdot b^t}$

The maximum sustainable population is the numerator of this function, c, so

  $c=128$

Now, let’s find the initial population by substituting $t=0$ ,

  $\begin{array}{c}P\left(0\right)=\frac{c}{1+a\cdot b^{\left(0\right)}}\\ =\frac{c}{1+a}\\ =\frac{128}{1+a}\end{array}$

Now, initial population is given to be 16, so,

  $\begin{array}{c}16=\frac{128}{1+a}\\ 1+a=\frac{128}{16}\\ 1+a=8\\ a=7\end{array}$

Now, to find the value of the constant b , substitute the given point $\left(5,32\right)$ ,

  $\begin{array}{c}P\left(5\right)=\frac{c}{1+a\cdot b^{\left(5\right)}}\\ 32=\frac{128}{1+7b^5}\\ 1+7b^5=\frac{128}{32}\end{array}$

  $\begin{array}{c}1+7b^5=4\\ 7b^5=3\\ b=\sqrt[5]{\frac{3}{7}}\\ b\approx 0.844\end{array}$

Hence, the required function is:

  $P\left(t\right)\approx \frac{128}{1+7{\left(0.844\right)}^t}$