Question

In Exercises \(23-26,\) the logistic equation describes the growth of a population \(P,\) where \(t\) is measured in years. In each case, find (a) the carrying capacity of the population, (b) the size of the population when it is growing the fastest, and (c) the rate at which the population is growing when it is growing the fastest. $$\frac{d P}{d t}=10^{-5} P(5000-P)$$

Short answer

The carrying capacity of the population is 5000. The size of the population when it is growing the fastest is 2500. The rate at which the population is growing the fastest is 3.125 populations per year.

Step-by-step solution

Determine the Carrying Capacity

In the logistic differential equation, the carrying capacity is the value of \( P \) that makes \( \frac{dP}{dt} = 0 \). So it is the value of \( P \) when \( P(5000-P) = 0 \). This happens when \( P = 0 \) or when \( P = 5000 \). Since we are considering a growing population and the carrying capacity must be a positive value, the carrying capacity of this population is \( P = 5000 \). So the carrying capacity is 5000.

Determine the Size of the Population when it is Growing the Fastest

The population grows fastest at half the carrying capacity. Therefore, the size of the population is \( P = \frac{M}{2} = \frac{5000}{2} = 2500 \). So, the population size grows fastest when it reaches 2500.

Determine the Rate at which the Population is Growing the Fastest

To find the rate at which the population is growing its fastest, substitute \( P = 2500 \) into the population equation: \( \frac{dP}{dt} = 10^{-5} P(5000-P) = 10^{-5} * 2500 * (5000 - 2500) \). Simplifying this results in \( \frac{dP}{dt} = 3.125 \) per year. So the fastest rate of growth is 3.125 populations per year.