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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)=800−4pn(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

STEP 1

1. The revenue function R(p) R(p) is the product of the price per day p p and the number of cars rented per day n(p) n(p) .
2. The number of cars rented per day is given by the linear function n(p)=800−4p n(p) = 800 - 4p .
3. The price p p must be between 95and95 and 200.

STEP 2

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

STEP 3

Define the revenue function R(p) R(p) as the product of the price per day p p and the number of cars rented per day n(p) n(p) .
R(p)=p⋅n(p) R(p) = p \cdot n(p)

STEP 4

Substitute the expression for n(p) n(p) into the revenue function:
R(p)=p⋅(800−4p) R(p) = p \cdot (800 - 4p)
Simplify the expression:
R(p)=800p−4p2 R(p) = 800p - 4p^2

STEP 5

To find the critical points, take the derivative of R(p) R(p) with respect to p p and set it equal to zero:
R′(p)=800−8p R'(p) = 800 - 8p
Set the derivative equal to zero to find critical points:
800−8p=0 800 - 8p = 0
Solve for p p :
8p=800 8p = 800
p=100 p = 100

STEP 6

Check if the critical point p=100 p = 100 is within the given price range of 95to95 to 200.
Since 95≤100≤200 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⋅(800−4⋅95)=95⋅420=39900 R(95) = 95 \cdot (800 - 4 \cdot 95) = 95 \cdot 420 = 39900
R(100)=100⋅(800−4⋅100)=100⋅400=40000 R(100) = 100 \cdot (800 - 4 \cdot 100) = 100 \cdot 400 = 40000
R(200)=200⋅(800−4⋅200)=200⋅0=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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