#### Answer

(a) The carrying capacity is $50$
$k = 0.02$
(b) $P(t) = \frac{50}{1+0.25e^{-0.02t}}$
(c) 41.5

#### 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:
$\frac{dP}{dt} = 0.02P-0.0004P^2$
$\frac{dP}{dt} = 0.02P(1-\frac{P}{50})$
The carrying capacity is $50$
$k = 0.02$
(b) The solution has this form:
$P(t) = \frac{M}{1+Ae^{-kt}}~~$ where $~~A = \frac{M-P_0}{P_0}$
We can find $A$:
$A = \frac{M-P_0}{P_0} = \frac{50-40}{40} = 0.25$
We can write the solution:
$P(t) = \frac{50}{1+0.25e^{-0.02t}}$
(c) We can find the population after 10 weeks:
$P(t) = \frac{50}{1+0.25e^{(-0.02)(10)}} = 41.5$