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Chapter 4 | Applications of Derivatives
n ( p ) =1000−5 p . If they charge $50 per day or less, they will rent all their cars. If they charge $200 per day or more, they will not rent any cars. Assuming the owners plan to charge customers between $50 per day and $200 per day to rent a car, how much should they charge to maximize their revenue? Solution Step 1: Let p be the price charged per car per day and let n be the number of cars rented per day. Let R be the revenue per day. Step 2: The problem is to maximize R . Step 3: The revenue (per day) is equal to the number of cars rented per day times the price charged per car per day—that is, R = n × p . Step 4: Since the number of cars rented per day is modeled by the linear function n ( p ) =1000−5 p , the revenue R can be represented by the function R ( p ) = n × p = ⎛ ⎝ 1000−5 p ⎞ ⎠ p =−5 p 2 +1000 p . Step 5: Since the owners plan to charge between $50 per car per day and $200 per car per day, the problem is to find the maximum revenue R ( p ) for p in the closed interval ⎡ ⎣ 50, 200 ⎤ ⎦ . Step 6: Since R is a continuous function over the closed, bounded interval ⎡ ⎣ 50, 200 ⎤ ⎦ , it has an absolute maximum (and an absolute minimum) in that interval. To find the maximum value, look for critical points. The derivative is R ′( p ) =−10 p +1000. Therefore, the critical point is p =100 When p =100, R (100) = $50,000. When p =50, R ( p ) = $37,500. When p =200, R ( p ) =$0. Therefore, the absolute maximum occurs at p =$100. The car rental company should charge $100 per day per car to maximize revenue as shown in the following figure.
Figure 4.67 To maximize revenue, a car rental company has to balance the price of a rental against the number of cars people will rent at that price.
4.34 A car rental company charges its customers p dollars per day, where 60≤ p ≤150. It has found that the number of cars rented per day can be modeled by the linear function n ( p ) =750−5 p . How much should the company charge each customer to maximize revenue?
Example 4.36
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