Chapter 3, Problem 94RE

Solution

(a)

To calculate : The initial population of deer.

The initial population of deer is $12$ .

Given information :

The population $P$ of deer after $t$ years in Briggs State Park is modeled by the function:

  $P\left(t\right)=\frac{1200}{1+99e^{-0.4t}}$ .

Calculation :

Consider the function $P\left(t\right)=\frac{1200}{1+99e^{-0.4t}}$ .

To find the initial population of deer, substitute $t=0$ in the given function.

  $\begin{array}{c}P\left(0\right)=\frac{1200}{1+99e^{-0.4\left(0\right)}}\\ =\frac{1200}{1+99e^0}\\ =\frac{1200}{1+99\left(1\right)}\left[\text{by }{e}^0=1\right]\\ =\frac{1200}{100}\\ =12\end{array}$

Therefore, the initial population of deer is $12$ .

(b)

To calculate : The number of years when there will be $1000$ deer.

The population of deer will be $1000$ after $15.5$ years.

Given information :

The population $P$ of deer after $t$ years in Briggs State Park is modeled by the function:

  $P\left(t\right)=\frac{1200}{1+99e^{-0.4t}}$ .

Calculation :

Consider the function $P\left(t\right)=\frac{1200}{1+99e^{-0.4t}}$ .

To find the number of years when there will be $1000$ deer, substitute $P=1000$ in the given function and solve for $t$ .

  $\begin{array}{c}1000=\frac{1200}{1+99e^{-0.4\left(t\right)}}\\ 1+99e^{-0.4\left(t\right)}=\frac{1200}{1000}\\ 99e^{-0.4\left(t\right)}=1.2-1\\ e^{-0.4\left(t\right)}=\frac{0.2}{99}\\ -0.4t=\ln \left(\frac{0.2}{99}\right)\\ t\approx 15.5\end{array}$

Therefore, after $15.5$ years the population of deer will be $1000$ .

(c)

To calculate : The maximum number of deer planned for the park.

The maximum number of deer planned for the park is $1200$ .

Given information :

The population $P$ of deer after $t$ years in Briggs State Park is modeled by the function:

  $P\left(t\right)=\frac{1200}{1+99e^{-0.4t}}$ .

Calculation :

Consider the function $P\left(t\right)=\frac{1200}{1+99e^{-0.4t}}$ .

The maximum number of deer planned for the park is:

  $\begin{array}{c}\underset{P\to \infty }{\lim }P\left(t\right)=\underset{P\to \infty }{\lim }\frac{1200}{1+99e^{-0.4t}}\\ =\frac{1200}{1+99e^{-0.4\left(\infty \right)}}\\ =\frac{1200}{1+99\left(0\right)}\\ =1200\end{array}$

Therefore, the maximum number of deer planned for the park is $1200$ .