1. **Problem statement:** Given the rate function of drug entering the bloodstream: $$R(t) = \frac{80t}{t^2 + 4}, \quad t \geq 0$$ We need to find: (a) The rate of drug entering after a long time. (b) When the drug entry rate is maximum and what that maximum rate is. (c) The total amount of drug entering in the first 6 hours. 2. **Step 1: Rate after a long time** As $t \to \infty$, analyze the behavior of $R(t)$: $$R(t) = \frac{80t}{t^2 + 4} = \frac{80t}{t^2(1 + \frac{4}{t^2})} = \frac{80}{t(1 + \frac{4}{t^2})}$$ Since $\frac{4}{t^2} \to 0$ as $t \to \infty$, we get: $$\lim_{t \to \infty} R(t) = \lim_{t \to \infty} \frac{80}{t} = 0$$ **Answer:** The rate approaches 0 mg/hr after a long time. 3. **Step 2: Find when $R(t)$ is maximum and the maximum rate** To find maxima, differentiate $R(t)$ with respect to $t$: $$R(t) = \frac{80t}{t^2 + 4}$$ Using quotient rule: $$R'(t) = \frac{(80)(t^2 + 4) - 80t(2t)}{(t^2 + 4)^2} = \frac{80t^2 + 320 - 160t^2}{(t^2 + 4)^2} = \frac{-80t^2 + 320}{(t^2 + 4)^2}$$ Set numerator to zero for critical points: $$-80t^2 + 320 = 0 \implies 80t^2 = 320 \implies t^2 = 4 \implies t = 2 \text{ (since } t \geq 0)$$ Check the sign of $R'(t)$ around $t=2$: - For $t < 2$, numerator positive (e.g., $t=1$: $-80(1) + 320 = 240 > 0$) - For $t > 2$, numerator negative (e.g., $t=3$: $-80(9) + 320 = -400 < 0$) So $t=2$ is a maximum. Calculate $R(2)$: $$R(2) = \frac{80 \times 2}{2^2 + 4} = \frac{160}{4 + 4} = \frac{160}{8} = 20$$ **Answer:** Maximum rate is 20 mg/hr at $t=2$ hours. 4. **Step 3: Total amount of drug in first 6 hours** Total amount is the integral of rate from 0 to 6: $$A = \int_0^6 R(t) dt = \int_0^6 \frac{80t}{t^2 + 4} dt$$ Use substitution: let $u = t^2 + 4$, then $du = 2t dt \implies t dt = \frac{du}{2}$ Rewrite integral: $$A = 80 \int_0^6 \frac{t}{t^2 + 4} dt = 80 \int_{u=4}^{u=6^2+4=40} \frac{1}{u} \cdot \frac{du}{2} = 40 \int_4^{40} \frac{1}{u} du$$ Integrate: $$40 [\ln|u|]_4^{40} = 40 (\ln 40 - \ln 4) = 40 \ln \frac{40}{4} = 40 \ln 10$$ Numerical approximation: $$40 \ln 10 \approx 40 \times 2.302585 = 92.1034$$ **Answer:** Total drug amount in first 6 hours is approximately 92.1 mg. --- **Final answers:** (a) Rate after long time: $0$ mg/hr (b) Maximum rate: $20$ mg/hr at $t=2$ hours (c) Total amount in 6 hours: $40 \ln 10 \approx 92.1$ mg