Question

A model for the population P(t) in a suburb of a large city is given by the initial-value problem dP/dt = P(10^-1 - 10^-7P), P(0) = 4000, where t is measured in months. What is the limiting value of the population? At what time will the population be equal to one-half of this limiting value? (Round your answer to one decimal place.) months

Answer 1

The limiting value of the population is approximately
`1 x 10^7`, and the population will be equal to one-half of this limiting value at around 3.5 months.

Step-by-step explanation:

The limiting value of the population can be found by setting the derivative of the population function equal to zero and solving for P. The given differential equation is
`\( (dP)/(dt) = P(10^(-1) - 10^(-7)P) \)`. Setting this equal to zero gives ( P = 0 ) or
`\( P = 10^6 \)`. Since the initial condition is ( P(0) = 4000 ), the limiting value is
`\( 10^6 \) (1 x 10^6)`.

To find the time at which the population is equal to one-half of the limiting value, we set
`\( P(t) = (1)/(2) * 10^6 \)`and solve for t. Substituting this into the differential equation and solving the resulting separable equation yields the time at which the population is one-half of the limiting value. The solution is
`\( t \approx 3.5 \)` months.

In summary, the limiting value of the population is
`\( 1 * 10^6 \)`, and the population will be equal to one-half of this limiting value at approximately 3.5 months. This conclusion is reached by solving the differential equation, applying the initial condition, and solving for the specific time at which the population reaches one-half of the limiting value.

Full Question:

What is the limiting value of the population? At what time will the population be equal to one-half of this limiting value? (Round your answer to one decimal place.) months