1. **Problem 1: Magazine Sales Growth** Given the sales function: $$S = 200,000(1 - e^{-0.05t})$$ where $t$ is time in weeks. **a) Calculate sales after 1 week:** Use the formula with $t=1$: $$S = 200,000(1 - e^{-0.05 \times 1}) = 200,000(1 - e^{-0.05})$$ Calculate $e^{-0.05} \approx 0.9512$: $$S = 200,000(1 - 0.9512) = 200,000 \times 0.0488 = 9,760$$ **b) Plot sales over 52 weeks:** The function $S(t) = 200,000(1 - e^{-0.05t})$ shows sales increasing over time, approaching a maximum of 200,000 as $t$ grows large. **c) Maximum growth of sales:** The maximum sales is the limit as $t \to \infty$: $$\lim_{t \to \infty} 200,000(1 - e^{-0.05t}) = 200,000(1 - 0) = 200,000$$ --- 2. **Problem 2: Influenza Spread in Community** Given logistic growth: $$N(t) = \frac{1000}{1 + 999 e^{-0.3t}}$$ **a) Number infected after 10 and 20 days:** For $t=10$: $$N(10) = \frac{1000}{1 + 999 e^{-0.3 \times 10}} = \frac{1000}{1 + 999 e^{-3}}$$ Calculate $e^{-3} \approx 0.0498$: $$N(10) = \frac{1000}{1 + 999 \times 0.0498} = \frac{1000}{1 + 49.75} = \frac{1000}{50.75} \approx 19.7 \approx 20$$ For $t=20$: $$N(20) = \frac{1000}{1 + 999 e^{-0.3 \times 20}} = \frac{1000}{1 + 999 e^{-6}}$$ Calculate $e^{-6} \approx 0.00248$: $$N(20) = \frac{1000}{1 + 999 \times 0.00248} = \frac{1000}{1 + 2.48} = \frac{1000}{3.48} \approx 287.4 \approx 287$$ **b) Time for half the community infected ($N=500$):** Set $N(t) = 500$: $$500 = \frac{1000}{1 + 999 e^{-0.3t}}$$ Invert: $$1 + 999 e^{-0.3t} = \frac{1000}{500} = 2$$ $$999 e^{-0.3t} = 1$$ $$e^{-0.3t} = \frac{1}{999}$$ Take natural log: $$-0.3t = \ln\left(\frac{1}{999}\right) = -\ln(999)$$ $$t = \frac{\ln(999)}{0.3} \approx \frac{6.9078}{0.3} = 23.03 \approx 23 \text{ days}$$ **c) Limiting value of $N(t)$ as $t \to \infty$:** As $t$ increases, $e^{-0.3t} \to 0$, so: $$\lim_{t \to \infty} N(t) = \frac{1000}{1 + 0} = 1000$$ This means eventually the entire community contracts influenza, which is the carrying capacity or maximum population.