Question

In a population undergoing logistic growth, the rate at which new individuals are added to the population is highest at Select one: O a population densities that are close to 0 • b. population densities that are close to, but just below, K c. population densities that are greater than K d. population densities around one-half of K Clear my choice

Answer 1

Logistic population growth model is given by the equation:

dN di =rnik - N

where, dN/dt is the rate of change of population, r is the intrinsic growth rate, N is the current population size and K is the carrying capacity for the population.

Growth rate is maximum when population size (N) reaches half its carrying capacity (K), i.e. when N=K/2.

This is because when N is very small than K, then population size is very small (fewer mates) and fewer individuals can be added per unit time. When population size approaches K, there are very resources left and growth rate ceases.

However, when N=K/2, population growth is maximum as there are ample number of mates and resources are also available.

When N=K/2, the equation above will become,

dN/dt= (1/2) rN

Hence, the correct choice should be:

(d) population densities around one-half of K.

An example to understand the question and how to solve mathematically:

We can understand the problem using a hypothetical scenario. Consider a population, with following values : r = 0.1, K = 1000.

Case1: when population density is close to zero, N=0

Using the equation,

dN di =rnik - N

Putting the values of r, N, K we will get,

dN df = (0.1) (0) 1000 - 0 1000

which will be equal to zero.

Thus, when N~0, rate of addition of new individual is zero.

Case 2: Population density close to K, but below K

Considering N =900, we get,

dN (0,1)(900) 1000 – 900. 1000

dN/dt = 9

Thus 9 individuals will be added in this case.

Case 3: when population densities are greater than K

Considering N=1100, we get,

dN 1000 – 1100 -= (0.1) (1100) 1000

dN/dt = - 11

Here, the negative sign indicates a negative growth rate. Population size will reduce in this case.

Case 4: when population is one-half of K, N=K/2

Here, N = 500, we will get,

dN dt = (0.1) (500x1000 - 500 1000

dN/dt = 25

Thus, 25 individuals will be added in this case.

Hence, it is clear that population growth is maximum when N=K/2.

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