A population is modeled by the differential equation.
$\frac { d P } { d t } = 1.4 P \left( 1 - \frac { P } { 4360 } \right)$
For what values of $P$ is the population increasing? Select the correct answer

Question

A population is modeled by the differential equation.

\frac { d P } { d t } = 1.4 P \left( 1 - \frac { P } { 4360 } \right)

For what values of

P

is the population increasing? Select the correct answer

Quizplus · Accepted Answer

The population is increasing where the derivative is positive. We need to solve for where the derivative is positive. \begin{align*} \frac { d P } { d t } &> 0 \ 1.4 P \left( 1 - \frac { P } { 4360 } \right)&>0\ \end{align*} For this inequality to hold, either $1.4 P > 0$ and $1 - \frac { P } { 4360 }>0$ (i.e. both terms are positive), or $1.4 P < 0 $ and $1 - \frac { P } { 4360 } < 0$ (i.e. both terms are negative). The second case gives no valid solutions since it implies that $P<0$. So we only need to solve the first case: \begin{align*} 1.4 P > 0 \quad &\Rightarrow \quad P>0\ 1 - \frac { P } { 4360 }>0 \quad &\Rightarrow \quad P<4360\ \end{align*} Therefore, the population is increasing when $0

A Population Is Modeled by the Differential Equation dPdt=1.4P(1−P4360)\frac { d P } { d t } = 1.4 P \left( 1 - \frac { P } { 4360 } \right)dtdP​=1.4P(1−4360P​)

A Population Is Modeled by the Differential Equation $\frac { d P } { d t } = 1.4 P \left( 1 - \frac { P } { 4360 } \right)$