1. **Problem a:** Find the book value of the tractor at the end of 2 years. Given the problem does not provide an explicit formula for the tractor's book value, we assume a depreciation or value function is needed. Since no formula is given, we cannot calculate the exact value without it. 2. **Problem b:** Find the relative rate of change of the book value of the tractor at the end of 2 years. Similarly, without a formula for the book value function $B(t)$, we cannot compute the relative rate of change, which is given by $$\frac{B'(t)}{B(t)}.$$ 3. **Problem 77:** Population growth model: Population at time $t$ months is given by $$P(t) = \frac{40 + 80e^{0.06t}}{20 + e^{0.06t}}$$ We want the relative rate of growth at 5 years, i.e., at $t = 60$ months. **Step 1:** Compute $P'(t)$ using the quotient rule: $$P'(t) = \frac{(80 \cdot 0.06 e^{0.06t})(20 + e^{0.06t}) - (40 + 80 e^{0.06t})(0.06 e^{0.06t})}{(20 + e^{0.06t})^2}$$ Simplify numerator: $$= \frac{4.8 e^{0.06t}(20 + e^{0.06t}) - 0.06 e^{0.06t}(40 + 80 e^{0.06t})}{(20 + e^{0.06t})^2}$$ Expand terms: $$= \frac{96 e^{0.06t} + 4.8 e^{0.12t} - 2.4 e^{0.06t} - 4.8 e^{0.12t}}{(20 + e^{0.06t})^2} = \frac{(96 - 2.4) e^{0.06t} + (4.8 - 4.8) e^{0.12t}}{(20 + e^{0.06t})^2} = \frac{93.6 e^{0.06t}}{(20 + e^{0.06t})^2}$$ **Step 2:** Calculate $P(60)$ and $P'(60)$: Calculate $e^{0.06 \times 60} = e^{3.6} \approx 36.5982$ $$P(60) = \frac{40 + 80 \times 36.5982}{20 + 36.5982} = \frac{40 + 2927.856}{56.5982} = \frac{2967.856}{56.5982} \approx 52.44$$ $$P'(60) = \frac{93.6 \times 36.5982}{(56.5982)^2} = \frac{3423.8}{3203.5} \approx 1.069$$ **Step 3:** Relative rate of growth at $t=60$: $$\frac{P'(60)}{P(60)} = \frac{1.069}{52.44} \approx 0.0204$$ So the relative rate of growth is approximately 0.0204 per month. 4. **Problem 78:** Marginal revenue from demand function: Demand function: $$p = 200 - 0.01 x \ln x$$ where $p$ is price per yacht (in hundreds), $x$ is number of yachts. **Step 1:** Revenue function: $$R(x) = x p = x \left(200 - 0.01 x \ln x\right) = 200x - 0.01 x^2 \ln x$$ **Step 2:** Marginal revenue is derivative $R'(x)$: Use product and chain rules: $$R'(x) = 200 - 0.01 \left(2x \ln x + x^2 \cdot \frac{1}{x}\right) = 200 - 0.01 (2x \ln x + x) = 200 - 0.01 x (2 \ln x + 1)$$ This gives the marginal revenue function. --- **Summary:** - Tractor book value problems cannot be solved without a formula. - Population relative growth at 5 years is approximately 0.0204 per month. - Marginal revenue function is $$R'(x) = 200 - 0.01 x (2 \ln x + 1).$$