Math  /  Algebra

QuestionQuestion Owners of a car rental company that charges customers between $95\$ 95 and $200\$ 200 per day have determined that the number of cars rented per day can be modeled by the linear function n(p)=8004pn(p)=800-4 p, where pp is the daily rental charge. How much should the company charge each customer per day to maximize revenue? Do not include units or a dollar sign in your answer.
Provide your answer below:

Studdy Solution
Check if the critical point p=100 p = 100 is within the given price range of 95to95 to 200.
Since 95100200 95 \leq 100 \leq 200 , the critical point is within the range.
Evaluate the revenue function at the endpoints and the critical point to determine the maximum revenue:
R(95)=95(800495)=95420=39900 R(95) = 95 \cdot (800 - 4 \cdot 95) = 95 \cdot 420 = 39900
R(100)=100(8004100)=100400=40000 R(100) = 100 \cdot (800 - 4 \cdot 100) = 100 \cdot 400 = 40000
R(200)=200(8004200)=2000=0 R(200) = 200 \cdot (800 - 4 \cdot 200) = 200 \cdot 0 = 0
The maximum revenue occurs at p=100 p = 100 .
The company should charge each customer:
100 \boxed{100}

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