An amusement park has 24,000 visitors per day when it charges $30 per person. For each $1 increase in price, the park loses about 500 visitors. How much should the park charge to maximize daily revenue? What is the
maximum daily revenue?

R(x)=-100x^2+1200x+16000

(-8,0) and (20,0)

$14, cost that gives maximum revenue.

1400 visitors that give max rev.

x: price increase

When price = 8, visitors = 2000

For each increase(x) of 1 in price, visitors decrease by 100

price = 8 + x

visitors = 2000 - 100x

1. Revenue = price * visitors

R(x) = (8+x) (2000-100x)

R(x) = 16,000 - 800x + 2,000x - 100x²

R(x) = -100x² + 1,200x + 16,000

2. Coordinates of maximum point

R(x) is parabola curving down. Maximum point is at vertex

Vertex has x-coordinate = -b/2a

x = -1200/-200 = 6

R(6) = -100*36 + 1,200*6 + 16,000 = 19,600

Maximum point (6, 19600)

3. Admission cost / maximum revenue

Maximum revenue occurs when price increase = 6

Cost = 8 + 6 = 14

Admission cost of $14 gives maximum revenue

4. Number of visitors:

Visitors = 2000 - 100x = 2000 - 600 = 1400

answer: there are x+184 pages in olivia's book.

step-by-step explanation:

let the total number of pages in the book which olive has already read be x.

then if olive has 184 pages left to read, then the total number of pages are in olive's book= pages she has read+pages she has left to read

therefore, the total number of pages in olivia's book=

hence, there are x+184 pages in olivia's book.