A manufacturer sells 80 boats per month at 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 Quantity]
per month

Ask by Hill Burns. in the United States

Jan 25,2025

Question

A manufacturer sells 80 boats per month at 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 Quantity]
per month

Ask by Hill Burns. in the United States

Jan 25,2025

UpStudy AI · Accepted Answer

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.
The price can decrease by up to \$625 per month before revenue starts to drop.

A manufacturer sells 80 boats per month at 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 Quantity]
per month

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A manufacturer sells 80 boats per month at 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 Quantity] per month

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preguntas relacionadas

Latest Calculus Questions

A manufacturer sells 80 boats per month at 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 Quantity]
per month