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Math  /  Algebra

QuestionQuestion A car rental company determines that if they charge customers pp dollars per day to rent a car, where 50≤p≤15050 \leq p \leq 150, the number of cars nn they rent per day can be modeled by the linear function n(p)=1800−12pn(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

STEP 1

1. The price per day p p is within the range 50≤p≤150 50 \leq p \leq 150 .
2. The number of cars rented per day n n is given by the function n(p)=1800−12p n(p) = 1800 - 12p .
3. Revenue R R is calculated as the product of the price per day and the number of cars rented: R(p)=p⋅n(p) R(p) = p \cdot n(p) .

STEP 2

1. Write the revenue function in terms of p p .
2. Find the critical points of the revenue function.
3. Determine the maximum revenue within the given price range.

STEP 3

Write the revenue function R(p) R(p) as a function of p p :
R(p)=p⋅n(p)=p⋅(1800−12p) R(p) = p \cdot n(p) = p \cdot (1800 - 12p)
Simplify the expression:
R(p)=1800p−12p2 R(p) = 1800p - 12p^2

STEP 4

To find the critical points, take the derivative of the revenue function R(p) R(p) with respect to p p :
R′(p)=ddp(1800p−12p2)=1800−24p R'(p) = \frac{d}{dp}(1800p - 12p^2) = 1800 - 24p
Set the derivative equal to zero to find critical points:
1800−24p=0 1800 - 24p = 0
Solve for p p :
24p=1800 24p = 1800
p=180024 p = \frac{1800}{24}
p=75 p = 75

STEP 5

Evaluate the revenue function at the critical point and at the endpoints of the interval to find the maximum revenue:
1. R(50)=50⋅(1800−12⋅50)=50⋅1200=60000 R(50) = 50 \cdot (1800 - 12 \cdot 50) = 50 \cdot 1200 = 60000
2. R(75)=75⋅(1800−12⋅75)=75⋅900=67500 R(75) = 75 \cdot (1800 - 12 \cdot 75) = 75 \cdot 900 = 67500
3. R(150)=150⋅(1800−12⋅150)=150⋅0=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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