1. **Problem Statement:** We have two fan models, each with blades shaped as sectors of a circle. Each blade has a perimeter constraint, and the total area of the fan (all blades combined) must not exceed 1200 cm². Model A: 3 blades, each blade perimeter = 80 cm. Model B: 5 blades, each blade perimeter = 60 cm. We need to find the radius $r$ and sector angle $\theta$ (in radians) that maximize the area of one blade for each model, then find the total swept area, and advise on the suitable model. 2. **Formulas and Important Rules:** - The perimeter $P$ of a sector with radius $r$ and central angle $\theta$ is: $$P = 2r + r\theta$$ (two radii plus the arc length $r\theta$) - The area $A$ of a sector is: $$A = \frac{1}{2} r^2 \theta$$ - Total swept area for the fan is number of blades $n$ times area of one blade: $$A_{total} = n \times A = n \times \frac{1}{2} r^2 \theta$$ - The total fan area must not exceed 1200 cm²: $$A_{total} \leq 1200$$ 3. **Step 1: Express $\theta$ in terms of $r$ using the perimeter constraint:** For each blade: $$P = 2r + r\theta = r(2 + \theta)$$ So, $$\theta = \frac{P}{r} - 2$$ 4. **Step 2: Write the area of one blade as a function of $r$ only:** $$A = \frac{1}{2} r^2 \theta = \frac{1}{2} r^2 \left( \frac{P}{r} - 2 \right) = \frac{1}{2} r^2 \frac{P}{r} - \frac{1}{2} r^2 \times 2 = \frac{1}{2} P r - r^2$$ Simplify: $$A(r) = \frac{P}{2} r - r^2$$ 5. **Step 3: Maximize $A(r)$ with respect to $r$:** Take derivative: $$\frac{dA}{dr} = \frac{P}{2} - 2r$$ Set derivative to zero for maximum: $$\frac{P}{2} - 2r = 0 \implies r = \frac{P}{4}$$ 6. **Step 4: Calculate $\theta$ at $r = \frac{P}{4}$:** $$\theta = \frac{P}{r} - 2 = \frac{P}{P/4} - 2 = 4 - 2 = 2$$ So the optimal sector angle is $\theta = 2$ radians for both models. 7. **Step 5: Calculate the maximum area of one blade:** $$A = \frac{1}{2} r^2 \theta = \frac{1}{2} \left( \frac{P}{4} \right)^2 \times 2 = \frac{1}{2} \times \frac{P^2}{16} \times 2 = \frac{P^2}{16}$$ 8. **Step 6: Calculate total swept area for each model:** Model A ($n=3$, $P=80$): $$A = \frac{80^2}{16} = \frac{6400}{16} = 400 \text{ cm}^2$$ $$A_{total} = 3 \times 400 = 1200 \text{ cm}^2$$ Model B ($n=5$, $P=60$): $$A = \frac{60^2}{16} = \frac{3600}{16} = 225 \text{ cm}^2$$ $$A_{total} = 5 \times 225 = 1125 \text{ cm}^2$$ Both satisfy the total area constraint. 9. **Step 7: Advise on suitable model:** - Model A uses the full allowed area (1200 cm²), Model B uses less (1125 cm²). - Model A blades are larger, Model B has more blades but smaller area. - For better airflow and efficiency, Model A with larger blades and full area usage is recommended. --- **Part (b):** Given Model A's blade radius $r$ increases over time at rate: $$\frac{dr}{dt} = 0.05 \text{ cm/month}$$ Perimeter constraint must hold: $$P = 2r + r\theta = r(2 + \theta) = 80$$ Since $P$ is constant and $\theta = 2$ radians, $$80 = r(2 + 2) = 4r \implies r = 20 \text{ cm}$$ If $r$ increases, to keep $P=80$, $\theta$ must adjust: $$P = r(2 + \theta) = 80 \implies 2 + \theta = \frac{80}{r} \implies \theta = \frac{80}{r} - 2$$ Differentiate both sides w.r.t. $t$: $$\frac{d\theta}{dt} = -\frac{80}{r^2} \frac{dr}{dt}$$ At $r=20$ cm and $\frac{dr}{dt} = 0.05$ cm/month: $$\frac{d\theta}{dt} = -\frac{80}{20^2} \times 0.05 = -\frac{80}{400} \times 0.05 = -0.2 \times 0.05 = -0.01 \text{ radians/month}$$ So, as $r$ increases, $\theta$ decreases to maintain perimeter. **Summary:** - Radius $r$ increases at 0.05 cm/month. - Sector angle $\theta$ decreases at 0.01 radians/month to keep perimeter constant. This ensures the blade perimeter constraint is always maintained despite deformation.