Math  /  Algebra

QuestionA charter flight charges a fare of $300\$ 300 per person plus $5\$ 5 per person for each unsold seat on the plane. The plane holds 100 passengers. Let xx represent the number of unsold seats. Complete parts (a) through (d). (a) Find an expression for the total revenue received for the flight R(x)R(x). (Hint: Multiply the number of people flying, 100x100-\mathrm{x}, by the price per ticket) R(x)=30000+200x5x2R(x)=30000+200 x-5 x^{2} (b) Choose the correct graph of the function, R(x)\mathrm{R}(\mathrm{x}), below. A. B. c. D. (c) The number of unsold seats that will produce the maximum revenue is 20. (Round to the nearest whole number as needed.) (d) The maximum revenue is $\$ \square (Round to the nearest whole number as needed.)

Studdy Solution

STEP 1

1. The plane has a capacity of 100 passengers.
2. The fare per person is 300.<br/>3.Anadditional300.<br />3. An additional 5 is charged per unsold seat.
4. x x represents the number of unsold seats.

STEP 2

1. Derive the expression for total revenue R(x) R(x) .
2. Analyze the graph of R(x) R(x) .
3. Determine the number of unsold seats for maximum revenue.
4. Calculate the maximum revenue.

STEP 3

To find the expression for total revenue R(x) R(x) , we start by calculating the number of passengers flying, which is 100x 100 - x .
The fare per person is 300+5x 300 + 5x because each unsold seat adds $5 to the fare of each sold seat.
The total revenue R(x) R(x) is the product of the number of passengers and the fare per person: R(x)=(100x)(300+5x) R(x) = (100 - x)(300 + 5x)

STEP 4

Expand the expression: R(x)=(100x)(300+5x)=100×300+100×5xx×300x×5x R(x) = (100 - x)(300 + 5x) = 100 \times 300 + 100 \times 5x - x \times 300 - x \times 5x
R(x)=30000+500x300x5x2 R(x) = 30000 + 500x - 300x - 5x^2
R(x)=30000+200x5x2 R(x) = 30000 + 200x - 5x^2
This matches the given expression for R(x) R(x) .

STEP 5

To choose the correct graph, note that R(x)=30000+200x5x2 R(x) = 30000 + 200x - 5x^2 is a quadratic function with a negative leading coefficient, indicating a downward-opening parabola.

STEP 6

To find the number of unsold seats that maximizes revenue, use the vertex formula for a parabola x=b2a x = -\frac{b}{2a} , where a=5 a = -5 and b=200 b = 200 .
x=2002×5=20 x = -\frac{200}{2 \times -5} = 20
This confirms that the number of unsold seats for maximum revenue is 20.

STEP 7

Substitute x=20 x = 20 into the revenue function to find the maximum revenue: R(20)=30000+200(20)5(20)2 R(20) = 30000 + 200(20) - 5(20)^2
R(20)=30000+40002000 R(20) = 30000 + 4000 - 2000
R(20)=32000 R(20) = 32000
The maximum revenue is $32000\$ 32000.
The maximum revenue is:
32000 \boxed{32000}

Was this helpful?

Studdy solves anything!

banner

Start learning now

Download Studdy AI Tutor now. Learn with ease and get all help you need to be successful at school.

ParentsInfluencer programContactPolicyTerms
TwitterInstagramFacebookTikTokDiscord