Math  /  Algebra

QuestionQuestion A car rental company determines that if they charge customers pp dollars per day to rent a car, where 50p15050 \leq p \leq 150, the number of cars nn they rent per day can be modeled by the linear function n(p)=180012pn(p)=1800-12 p.
If they charge $50\$ 50 per day or less, they will rent all their cars. If they charge $150\$ 150 per day or more, they will not rent any cars. How much should they charge to maximize their revenue?
Provide your answer below:

Studdy Solution
Evaluate the revenue function at the critical point and at the endpoints of the interval to find the maximum revenue:
1. R(50)=50(18001250)=501200=60000 R(50) = 50 \cdot (1800 - 12 \cdot 50) = 50 \cdot 1200 = 60000
2. R(75)=75(18001275)=75900=67500 R(75) = 75 \cdot (1800 - 12 \cdot 75) = 75 \cdot 900 = 67500
3. R(150)=150(180012150)=1500=0 R(150) = 150 \cdot (1800 - 12 \cdot 150) = 150 \cdot 0 = 0

The maximum revenue occurs at p=75 p = 75 with revenue R(75)=67500 R(75) = 67500 .
The optimal price to charge to maximize revenue is:
75 \boxed{75}

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