A population is modeled by the differential equation.
$\frac { d P } { d t } = 1.4 P \left( 1 - \frac { P } { 4560 } \right)$
For what values of $P$ is the population increasing?

Question

A population is modeled by the differential equation. $$\frac { d P } { d t } = 1.4 P \left( 1 - \frac { P } { 4560 } \right)$$ For what values of $P$ is the population increasing? A) $P > 1.4$ B) $P > 4560$ C) $P < 4580$ D) $0 < P < 4560$ E) $0 < P < 1.4$

Quizplus · Accepted Answer

The population is increasing when $\frac{dP}{dt}>0$. Simplifying the given differential equation, we get: $$\frac { d P } { d t } = 1.4 P - \frac { P^2 } { 4560 }$$ Setting this equal to zero and solving for $P$, we get the critical points: $$P=0 \quad \text{and} \quad P=4560$$ These divide the real line into three intervals: $$(-\infty,0), \quad (0,4560), \quad \text{and} \quad (4560,\infty)$$ To determine the sign of $\frac{dP}{dt}$ in each interval, we can test a value in that interval or use a sign chart. Using a test value in each interval (e.g. -1, 1000, and 5000), we can conclude that: $$\frac{dP}{dt} < 0 \quad \text{for} \quad P < 0 \quad \text{and} \quad P > 4560$$ $$\frac{dP}{dt} > 0 \quad \text{for} \quad 0 < P < 4560$$ Therefore, the population is increasing for $0

A Population Is Modeled by the Differential Equation dPdt=1.4P(1−P4560)\frac { d P } { d t } = 1.4 P \left( 1 - \frac { P } { 4560 } \right)dtdP​=1.4P(1−4560P​)

A Population Is Modeled by the Differential Equation $\frac { d P } { d t } = 1.4 P \left( 1 - \frac { P } { 4560 } \right)$