Question

An average of 70,000 people visit Riverside Park each day in the summer. The park charges $12.00 for admission. Consultants predict that for each $1.00 increase in the entrance price, the park would lose an average of 2,500 customers per day. (a) Express the daily revenue from ticket sales, R as a function of the number of $1.00 price increases, x. R=f(x)= (70000-2500x)(12+x) equation editorEquation Editor (b) What ticket price maximizes the revenue from ticket sales?

Answer 1

Answer:

The price that maximizes the revenue is $20

Step-by-step explanation:

(a) Express the daily revenue from ticket sales, R as a function of the number of $1.00 price increases,

R(x) = (70000-2500x)(12+x)

(b) What ticket price maximizes the revenue from ticket sales?

The ticket price that maximizes the revenue is associated with the x of the vertex of the parabola.

R(x) = (70000-2500x)(12+x) = 840000 + 70000x - 30000x - 2500x² =

-2500x² + 40000x + 840000

x vertex = -b/2a = -40000/2.(-2500) = 8

12+8 = 20

Answer 2

The price that maximizes revenue is $20 and the revenue function is 840000 - 40000x - 2500x².

How to calculate the revenue

• Average people = 70000
• Tickets price = $12.00
• Decrease in customer = 2500
• The ticket price for an individual = (12 + x)

R = f(x)

= (70000 - 2500x)(12 + x)

= 840000 + 70000x - 30000x - 2500x

= 840000 - 40000x - 2500x²

The first derivative of the revenue will be:

= 40000 - 5000x

x = 40000/5000

x = 8

Therefore, the price that will maximize revenue will be:

= 12 + x

= 12 + 8.

= 20

In conclusion, the price that maximizes revenue is $20

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