The demand for tickets to an amusement park is given by p = 70 - 0.2q, where p is the price of a ticket in dollars and q is the number of people attending at that price. a. What price generates an attendance of 3,000 people? What is the total revenue at that price? What is the total revenue if the price is $20?
b. Write the revenue function as a function of attendance, q, at the amusement park.
c. What attendance maximizes revenue?
d. What price should be charged to maximize revenue?
e. What is the maximum revenue? Can we determine the corresponding profit?

Step-by-step explanation:

From the information given:

p = 70 - 0.02q

when:

q = 3000

p = 70 - 0.02 (3000)

p = 70 - 60

p = 10 dollars

The total revenue = price × number of people

= 10 × 3000

= 30000 dollars

If the price is $20

20 = 70 - 0.02q

0.02q = 70 - 20

0.02q = 50

q = 50/0.02

q = 2500

∴

If price is $20, The total revenue = 20  × 2500

= 50000 dollars

b)

The revenue function R = price × attendance

R = p × q

R = (70 - 0.02q)q

R = 70q - 0.02q²

c)

R(q) = 70q - 0.02q²

Taking the differential of the value on the right hand side.

R'(q) = 70 × (1) - 0.02 × (2q)

R'(q) = 70 - 0.04q

If we set R'(q) = 0

70 - 0.04q = 0

70 = 0.04q

q = 70/0.04

q = 1750   (critical point)

R"(q) = 0 - 0.04 × 1

R"(q) = -0.04

R"(1750) = -0.04 < 0

∴

R is maximum when q = 1750

d)

At q = 1750

p = 70 - 0.02(q)

p = 70 - 0.02(1750)

p = 70 - 35

p = 35 dollars

e)

R(q) = 70q - 0.02q²   because R is max at q = 1750

R(1750) =  70 × 1750 - 0.02  × (1750)²

R_(max) = 122500 - 61250

R_(max) = 61250 dollars

Given that the actual price is 20 dollars when the price is changed to 35 dollars, we have a maximum revenue of:

Suppose:

p = 20 and q = 2500  (from a)

Then;

R = p × q

R = 20 × 2500

R = 50000 dollars

If R = 35, then q = 1750

R_{max} = 35 × 1750

R_{max} = 61250 dollars

∴

Profit = R_{max}  - R

Profit = 61250 - 50000

Profit = 11250 dollars

$5.41

step-by-step explanation:

$20 - $14.51 = $5.49.