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

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

We need to find the values of P where the population is increasing. This means we need to find where the derivative of P with respect to time (dP/dt) is positive.

$\frac { d P } { d t } = 1.4 P \left( 1 - \frac { P } { 4560 } \right)$

We can break this into two parts: 1.4P and (1 - P/4560).

When P is less than 4560 but greater than 0, the second factor (1 - P/4560) is positive. Therefore, the value of dP/dt depends solely on the value of 1.4P.

When P is less than 4560/1.4 ≈ 3257.14, 1.4P is less than 4560, so the derivative is positive.

When P is greater than 4560/1.4 ≈ 3257.14, 1.4P is greater than 4560, so the derivative is negative.

Therefore, the population is increasing when 0 < P < 4560/1.4 ≈ 3257.14, or 0 < P < 3260 (rounded to the nearest integer). So the answer is D.

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 $\frac { d P } { d t } = 1.4 P \left( 1 - \frac { P } { 4560 } \right)$