Let's break down the questions step by step: ## Given Logistic Differential Equation \[ \frac{dy}{dt} = 0.05y - 0.0002y^2, \quad y(0) = 20 \] Where: - \( y(t) \) is the whale population at time \( t \). - \( t \) is time in years. ### a) What maximum value (carrying capacity \(P\)) will the whale population approach? The logistic equation is generally written as: \[ \frac{dy}{dt} = ry \left(1 - \frac{y}{K}\right) \] Comparing: \[ \frac{dy}{dt} = 0.05y - 0.0002y^2 \] \[ = y(0.05 - 0.0002y) \] So, \( r = 0.05 \) and \( \frac{r}{K} = 0.0002 \). Solve for \( K \): \[ \frac{r}{K} = 0.0002 \implies K = \frac{0.05}{0.0002} = 250 \] **Final Answer (a):** \[ \boxed{250} \] The maximum population (carrying capacity) is **250** whales. --- ### b) At what value of \( t \) will the whale population reach a value of \( \frac{P}{2} \)? We want \( y(t) = \frac{K}{2} = 125 \). The general solution for the logistic equation is: \[ y(t) = \frac{K}{1 + Ae^{-rt}} \] Where \( A = \frac{K - y_0}{y_0} \), \( y_0 = 20 \), \( K = 250 \), \( r = 0.05 \). Let's compute \( A \): \[ A = \frac{250 - 20}{20} = \frac{230}{20} = 11.5 \] Plugging into the solution: \[ y(t) = \frac{250}{1 + 11.5 e^{-0.05t}} \] Set \( y(t) = 125 \): \[ 125 = \frac{250}{1 + 11.5 e^{-0.05t}} \] \[ 1 + 11.5 e^{-0.05t} = \frac{250}{125} = 2 \] \[ 11.5 e^{-0.05t} = 2 - 1 = 1 \] \[ e^{-0.05t} = \frac{1}{11.5} \] \[ -0.05t = \ln\left(\frac{1}{11.5}\right) \] \[ -0.05t = -\ln(11.5) \] \[ t = \frac{\ln(11.5)}{0.05} \] Calculate \( \ln(11.5) \): \[ \ln(11.5) \approx 2.442 \] \[ t = \frac{2.442}{0.05} = 48.84 \] **Final Answer (b):** \[ \boxed{t \approx 48.8 \text{ years}} \] --- ## **Summary** **a)** The whale population will eventually approach **250**. **b)** The population will reach half that value (\(125\)) at **approximately 48.8 years**.