Questions: A car rental agency rents 180 cars per day at a rate of 39 dollars per day. For each 1 dollar increase in the daily rate, 4 fewer cars are rented. (a) If x is the number of cars rented in a day, and p is the daily rental price per car, what is the price-demand equation (assuming it is linear)? p= (b) What is the daily revenue function for the agency? (Hint: Revenue is the product of the price per item and the number of items.) R(x)= (c) How many cars should be rented to maximize daily revenue? cars (d) What is the maximum daily revenue for the agency? (e) What daily rental price should the agency charge to maximize daily revenue?

A car rental agency rents 180 cars per day at a rate of 39 dollars per day. For each 1 dollar increase in the daily rate, 4 fewer cars are rented.
(a) If x is the number of cars rented in a day, and p is the daily rental price per car, what is the price-demand equation (assuming it is linear)?
p=
(b) What is the daily revenue function for the agency? (Hint: Revenue is the product of the price per item and the number of items.)
R(x)=

(c) How many cars should be rented to maximize daily revenue?
 cars
(d) What is the maximum daily revenue for the agency?
 
(e) What daily rental price should the agency charge to maximize daily revenue?
Transcript text: A car rental agency rents 180 cars per day at a rate of 39 dollars per day. For each 1 dollar increase in the daily rate, 4 fewer cars are rented. (a) If $x$ is the number of cars rented in a day, and $\rho$ is the daily rental price per car, what is the price-demand equation (assuming it is linear)? \[ p=\square \] (b) What is the daily revenue function for the agency? (Hint: Revenue is the product of the price per item and the number of items.) \[ R(x)= \] $\square$ (c) How many cars should be rented to maximize daily revenue? $\qquad$ cars (d) What is the maximum daily revenue for the agency? \$ $\qquad$ (e) What daily rental price should the agency charge to maximize daily revenue? \$ $\square$
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Solution

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Solution Steps

Solution Approach

(a) To find the price-demand equation, we need to express the price \( p \) as a function of the number of cars rented \( x \). We know that at 180 cars, the price is $39. For each $1 increase in price, 4 fewer cars are rented. This gives us a linear relationship.

(b) The daily revenue function \( R(x) \) is the product of the price per car and the number of cars rented. Using the price-demand equation from part (a), we can express \( R(x) \) in terms of \( x \).

(c) To maximize the daily revenue, we need to find the value of \( x \) that maximizes \( R(x) \). This can be done by finding the vertex of the quadratic revenue function.

Step 1: Price-Demand Equation

To derive the price-demand equation, we start with the initial conditions: when \( x = 180 \), \( p = 39 \). For each $1 increase in price, 4 fewer cars are rented. Thus, the price-demand equation can be expressed as: \[ p = 84 - 0.25x \]

Step 2: Daily Revenue Function

The daily revenue function \( R(x) \) is the product of the price per car and the number of cars rented: \[ R(x) = x \cdot p = x \cdot (84 - 0.25x) = 84x - 0.25x^2 \]

Step 3: Maximizing Daily Revenue

To find the number of cars that maximizes daily revenue, we take the derivative of the revenue function and set it to zero: \[ R'(x) = 84 - 0.5x = 0 \] Solving for \( x \) gives: \[ x = 168 \]

Step 4: Maximum Daily Revenue

Substituting \( x = 168 \) back into the revenue function to find the maximum revenue: \[ R(168) = 84(168) - 0.25(168^2) = 7056 \]

Final Answer

The price-demand equation is \( p = 84 - 0.25x \), the number of cars that should be rented to maximize daily revenue is \( \boxed{168} \), and the maximum daily revenue for the agency is \( \boxed{7056} \).

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