Questions: A manufacturer sells 70 boats per month at 23000 dollars per boat, and each month demand is increasing at a rate of 4 boats per month. What is the fastest the price could drop before the monthly revenue starts to drop?
dP/dt=
per month
Transcript text: A manufacturer sells 70 boats per month at 23000 dollars per boat, and each month demand is increasing at a rate of 4 boats per month. What is the fastest the price could drop before the monthly revenue starts to drop?
\[
\frac{d P}{d t}=\$
\]
$\square$ per month
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Solution
Solution Steps
Step 1: Define Variables
Let \( P \) be the price per boat and \( t \) be the time in months. The number of boats sold per month is given by \( 70 + 4t \).
Step 2: Revenue Function
The revenue \( R \) as a function of time is defined as:
\[
R = P \cdot (70 + 4t)
\]
Step 3: Differentiate Revenue
To find the rate of change of revenue with respect to time, we differentiate \( R \) with respect to \( t \):
\[
\frac{dR}{dt} = P \cdot \frac{d(70 + 4t)}{dt} + (70 + 4t) \cdot \frac{dP}{dt}
\]
This simplifies to:
\[
\frac{dR}{dt} = 4P + (70 + 4t) \cdot \frac{dP}{dt}
\]
Step 4: Set Revenue Derivative to Zero
To find the condition under which revenue does not decrease, we set \( \frac{dR}{dt} = 0 \):
\[
4P + (70 + 4t) \cdot \frac{dP}{dt} = 0
\]
Step 5: Solve for \( \frac{dP}{dt} \)
Rearranging the equation gives:
\[
(70 + 4t) \cdot \frac{dP}{dt} = -4P
\]
Thus, we can express \( \frac{dP}{dt} \) as:
\[
\frac{dP}{dt} = -\frac{4P}{70 + 4t}
\]
Step 6: Analyze the Result
The expression \( \frac{dP}{dt} = -\frac{4P}{70 + 4t} \) indicates the rate at which the price can drop before the revenue starts to decrease. The negative sign indicates a drop in price, and the fraction shows that the rate of price drop is dependent on the current price \( P \) and the number of boats sold, which increases over time.
Final Answer
\(\boxed{\frac{dP}{dt} = -\frac{4P}{70 + 4t}} \) dollars per month