A population \( P \) obeys the logistic model. It satisfies the equation \( \frac{d P}{d t}=\frac{5}{900} P(9-P) \) for \( P>0 \). (a) The population is increasing when 0 (b) The population is decreasing when \( P> \) (c) Assume that \( P(0)=4 \). Find \( P(84) \). \( P(84)= \)

To determine when the population \( P \) is increasing and decreasing, we analyze the given logistic equation \( \frac{d P}{d t}=\frac{5}{900} P(9-P) \). (a) The population is increasing when \( \frac{d P}{d t} > 0 \). For this to be true, both factors \( P \) and \( (9-P) \) must be positive. Since \( P > 0 \), the second factor \( (9-P) > 0 \) implies \( P < 9 \). Thus, the population is increasing when \( 0 < P < 9 \). (b) The population is decreasing when \( P > 9 \) because \( (9-P) \) becomes negative, making \( \frac{d P}{d t} < 0 \). (c) To find \( P(84) \), we use the logistic growth model formula. The general solution for the logistic model is given by: \[ P(t) = \frac{K}{1 + Ae^{-rt}}, \] where \( K \) is the carrying capacity, \( r \) is the growth rate, and \( A \) is a constant determined by the initial population. From the logistic equation, we identify: - \( K = 9 \) (carrying capacity), - \( r = \frac{5}{900} = \frac{1}{180} \). Now, since \( P(0) = 4 \): 1. Substitute into the logistic equation to find \( A \): \[ 4 = \frac{9}{1 + A} \implies 1 + A = \frac{9}{4} \implies A = \frac{9}{4} - 1 = \frac{5}{4}. \] 2. Now we can express \( P(t) \): \[ P(t) = \frac{9}{1 + \frac{5}{4} e^{-\frac{1}{180}t}}. \] 3. Now plug in \( t = 84 \): \[ P(84) = \frac{9}{1 + \frac{5}{4} e^{-\frac{84}{180}}}. \] First, calculate \( e^{-\frac{84}{180}} \) (approximately): - \( -\frac{84}{180} \approx -0.4667 \) - Therefore, \( e^{-0.4667} \approx 0.627 \). So we have: \[ P(84) = \frac{9}{1 + \frac{5}{4} \cdot 0.627} \approx \frac{9}{1 + 0.785} = \frac{9}{1.785} \approx 5.04. \] Finally, we can round it as: \[ P(84) \approx 5. \]