Suppose that a population develops according to the logistic equation $\frac { d P } { d t } = 0.02 P - 0.0002 P ^ { 2 }$ where $t$ is measured in weeks. What is the carrying capacity?
A) $K = 100$
B) $K = 0.02$
C) $K = 0.0002$
D) $K = 300$
E) $K = 200$

Question

Quizplus · Accepted Answer

The carrying capacity, $K$, in the logistic growth model $\frac{dP}{dt} = rP\left(1 - \frac{P}{K}\right)$ is the value of $P$ at which the population growth stops increasing, which occurs when $\frac{dP}{dt} = 0$. In the given equation, $\frac{dP}{dt} = 0.02P - 0.0002P^2$, comparing it to the standard form, it's clear that $0.0002P^2$ corresponds to $\frac{rP^2}{K}$. Solving for $K$ when the population growth rate equals zero gives $K = 100$, as it is the value of $P$ that makes the equation balance out at a point where population growth ceases to increase.

Suppose That a Population Develops According to the Logistic Equation dPdt=0.02P−0.0002P2\frac { d P } { d t } = 0.02 P - 0.0002 P ^ { 2 }dtdP​=0.02P−0.0002P2

Suppose That a Population Develops According to the Logistic Equation $\frac { d P } { d t } = 0.02 P - 0.0002 P ^ { 2 }$