Answer
(a) The carrying capacity is $1200$
$k = 0.04$
(b) $P(t) = \frac{1200}{1+19e^{-0.04t}}$
(c) 87.4
Work Step by Step
(a) A logistic equation has this form:
$\frac{dP}{dt} = kP(1-\frac{P}{M})$
We can consider the given logistic equation:
$\frac{dP}{dt} = 0.04P(1-\frac{P}{1200})$
The carrying capacity is $1200$
$k = 0.04$
(b) The solution has this form:
$P(t) = \frac{M}{1+Ae^{-kt}}~~$ where $~~A = \frac{M-P_0}{P_0}$
We can find $A$:
$A = \frac{M-P_0}{P_0} = \frac{1200-60}{60} = 19$
We can write the solution:
$P(t) = \frac{1200}{1+19e^{-0.04t}}$
(c) We can find the population after 10 weeks:
$P(t) = \frac{1200}{1+19e^{(-0.04)(10)}} = 87.4$
