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. $$\frac{d P}{d t}=0.006 P(200-P)$$

Short answer

The carrying capacity of the population is 200. The size of the population when it is growing the fastest is 100 and the rate at which the population is growing when it is growing the fastest is 60 units per year.

Step-by-step solution

Identify the Carrying Capacity

From the given logistic growth equation \(\frac{dP}{dt}=0.006P(200-P)\), we can identify the carrying capacity, \(K\), as the value which makes the expression in the bracket equal to zero when population \(P\) is positive. Thus \(K = 200\). Hence the carrying capacity of the population is 200.

Find the Population Size at its Fastest Growth

The population \(P\) grows fastest when the rate \(\frac{dP}{dt}\) is at a maximum. This happens when \(P= \frac{K}{2}\). Thus, when population is growing fastest, \(P = \frac{200}{2} = 100\). Hence, the size of the population at its fastest growth is 100.

Find the Growth Rate at its Fastest

The rate of growth at the fastest growing point can be obtained by substituting \(P = 100\) into the expression for \(\frac{dP}{dt}\). Thus, \(\frac{dP}{dt}=0.006(100)(200-100) = 0.006*100*100 = 60\). The population is therefore growing at a rate of 60 units per year at its fastest.