Chapter 6.5, Problem 23E

(a)

To determine

To find : The carrying capacity of the population.

Answer to Problem 23E

The answer is:

The carrying capacity of $200$ individuals

Explanation of Solution

Given information:

The differential equation is: $\frac{dP}{dt}=0.006P(200-P)$

Calculation:

The logistic differential equation is:

  $\frac{dP}{dt}=kP(M-P)$

Where $M$ is maximal carrying capacity.

  $\begin{array}{c}\frac{dP}{dt}=0.006P(200-P)\\ \Rightarrow M=200\end{array}$

Therefore the required carrying capacity of $200$ individuals.

(b)

To determine

To find: The size of the population when it is growing fastest.

Answer to Problem 23E

The answer is:

  $100$ individuals

Explanation of Solution

Given information:

The differential equation is: $\frac{dP}{dt}=0.006P(200-P)$

Calculation:

When the carrying capacity is cut in half, the growth is fastest (this is stated in question 3 under exploration 2 in  textbook examples). Since the carrying capacity from section a) is $200$ people, half of that number is $100$ people.

Therefore the required $100$ individuals of the population when it is growing fastest .

(c)

To determine

To find: The rate at that the population when it is growing fastest.

Answer to Problem 23E

The answer is:

  $60$ people per year.

Explanation of Solution

Given information:

The differential equation is: $\frac{dP}{dt}=0.006P(200-P)$

Calculation:

The logistic differential equation is:

  $\frac{dP}{dt}=kP(M-P)$ Where $M$ is maximal carrying capacity.

Comparing its to the given differential equation $\frac{dP}{dt}=0.006P(200-P)$ , then gives

  $M=200$

To determine the pace at which the population is growing while it is doing so at a rate that is the fastest, must discover the rate at which $P=100$

  $P=100$ is then substituted into the differential equation to produce:

  $\begin{array}{c}\frac{dP}{dt}=0.006(100)(200-100)\\ =0.6(100)\\ =60\end{array}$

Therefore the required rate at that the population when it is growing fastest is $60$ people per year.