A manufacturer sells 80 boats per month at \( \$ 25000 \) per boat, and each month demand is increasing at a rate of 2 boats per month. What is the fastest the price could drop before the monthly revenue starts to drop? [Revenue \( = \) Price \( \times \) Quantity] per month

To determine the fastest rate at which the price of the boats can drop before the monthly revenue starts to decrease, let's analyze the given information step-by-step. ### Given: - **Initial Quantity (Q₀):** 80 boats per month - **Initial Price (P₀):** \$25,000 per boat - **Increase in Quantity (ΔQ):** 2 boats per month - **Revenue Formula:** \( R = P \times Q \) ### Goal: Find the maximum rate at which the price (\( v \)) can decrease so that the revenue \( R \) does not start to drop. ### Analysis: 1. **Define Variables Over Time:** - Let \( t \) represent the time in months. - **Quantity over time:** \( Q(t) = 80 + 2t \) - **Price over time:** \( P(t) = 25,000 - vt \) 2. **Revenue Over Time:** \[ R(t) = P(t) \times Q(t) = (25,000 - vt)(80 + 2t) \] 3. **Differentiate Revenue with Respect to Time:** To find when the revenue starts to drop, we need to find when the derivative of \( R(t) \) with respect to \( t \) becomes zero. \[ \frac{dR}{dt} = \frac{d}{dt}[(25,000 - vt)(80 + 2t)] \] Applying the product rule: \[ \frac{dR}{dt} = (-v)(80 + 2t) + (25,000 - vt)(2) \] 4. **Set the Derivative to Zero:** \[ -v(80 + 2t) + 2(25,000 - vt) = 0 \] Simplifying: \[ -80v - 2vt + 50,000 - 2vt = 0 \] At \( t = 0 \) (initial moment): \[ -80v + 50,000 = 0 \] Solving for \( v \): \[ 80v = 50,000 \\ v = \frac{50,000}{80} \\ v = 625 \] ### Conclusion: The price can **drop at most \$625 per month** before the monthly revenue begins to decline. **Answer:** The price may decrease by up to $625 per month before revenue begins to fall.