1. **Problem Statement:** We have a software development cost function given by $$C(h) = 5000 + 150h - 2h^2 + 0.01h^3$$ where $h$ is the number of developer hours. We need to: (a) Find the first differential $dC$ at $h=100$. (b) Find the second differential $d^2C$ at $h=100$. (c) Use the differentials to determine the hour allocation for maximum marginal productivity. 2. **Formulas and Rules:** - The first differential $dC$ approximates the change in cost for a small change in $h$ and is given by $dC = C'(h) dh$ where $C'(h)$ is the first derivative. - The second differential $d^2C$ relates to the curvature of the cost function and is given by $d^2C = C''(h) (dh)^2$ where $C''(h)$ is the second derivative. - Maximum marginal productivity occurs where the marginal cost rate changes from increasing to decreasing, i.e., where $C'(h)$ is maximized or $C''(h) = 0$. 3. **Calculate the first derivative $C'(h)$:** $$C'(h) = \frac{d}{dh} \left(5000 + 150h - 2h^2 + 0.01h^3\right) = 0 + 150 - 4h + 0.03h^2$$ 4. **Calculate the second derivative $C''(h)$:** $$C''(h) = \frac{d}{dh} C'(h) = \frac{d}{dh} \left(150 - 4h + 0.03h^2\right) = 0 - 4 + 0.06h = -4 + 0.06h$$ 5. **Evaluate $dC$ at $h=100$:** First, compute $C'(100)$: $$C'(100) = 150 - 4(100) + 0.03(100)^2 = 150 - 400 + 0.03 \times 10000 = 150 - 400 + 300 = 50$$ Assuming a small change $dh$, the differential is: $$dC = C'(100) dh = 50 dh$$ 6. **Evaluate $d^2C$ at $h=100$:** Compute $C''(100)$: $$C''(100) = -4 + 0.06 \times 100 = -4 + 6 = 2$$ Then, $$d^2C = C''(100) (dh)^2 = 2 (dh)^2$$ 7. **Determine hour allocation for maximum marginal productivity:** Set the second derivative to zero to find critical points: $$C''(h) = -4 + 0.06h = 0 \implies 0.06h = 4 \implies h = \frac{4}{0.06} = 66.67$$ Check the sign of $C''(h)$ around $h=66.67$: - For $h < 66.67$, $C''(h) < 0$ (marginal cost decreasing). - For $h > 66.67$, $C''(h) > 0$ (marginal cost increasing). Thus, $h = 66.67$ hours is the point of inflection where marginal productivity is maximized. **Final answers:** (a) $dC = 50 dh$ at $h=100$ (b) $d^2C = 2 (dh)^2$ at $h=100$ (c) Maximum marginal productivity occurs at approximately $h = 66.67$ hours.