Chapter 3.2, Problem 46E

Solution

a.

To find: The initial population of deer.

The initial population of deer is 11.

Given:

The function $P\left(t\right)=\frac{1001}{1+90e^{-0.2t}}$ gives the population of deer after $t$ years.

Calculation:

To find the initial population of deer, put $t=0$ in the given function and simplify. That is,

  $\begin{array}{c}P\left(0\right)=\frac{1001}{1+90e^{-0.2\left(0\right)}}\\ =\frac{1001}{1+90e^0}\\ =\frac{1001}{1+90\left(1\right)}\\ =\frac{1001}{91}\\ =11\end{array}$

Hence, the initial population of deer is 11.

b.

To find: The time taken to reach the deer population of 600.

24 or 25 years.

Given:

The function $P\left(t\right)=\frac{1001}{1+90e^{-0.2t}}$ gives the population of deer after $t$ years.

Calculation:

Put $P\left(t\right)=600$ in the given function and solve it for $t$ . That is,

  $\begin{array}{c}600=\frac{1001}{1+90e^{-0.2t}}\\ 1+90e^{-0.2t}=\frac{1001}{600}\\ 90e^{-0.2t}=\frac{401}{600}\\ e^{-0.2t}=\frac{401}{54000}\\ -0.2t=\ln \left(\frac{401}{54000}\right)\\ t=-\frac{\ln \left(\frac{401}{54000}\right)}{0.2}\\ t=\text{24}\text{.5138}\\ t\approx 24\text{or}25\end{array}$

Hence, the deer population will reach 600 in about 24 or 25 years.

c.

To find: The maximum number of deer possible.

The maximum possible number of deer is 1001.

Given:

The function $P\left(t\right)=\frac{1001}{1+90e^{-0.2t}}$ gives the population of deer after $t$ years.

Calculation:

As $t\to \infty$ , it follows that

  $\begin{array}{c}\underset{t\to \infty }{\lim }P\left(t\right)=\underset{t\to \infty }{\lim }\left(\frac{1001}{1+90e^{-0.2t}}\right)\\ =\frac{1001}{1+90\left(0\right)}\\ =1001\end{array}$

Hence, the maximum possible number of deer is 1001.