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

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 Submit answer Next item
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Solution

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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

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