Question

Pleaseeee help ASAP it’s due today :(

An amusement park has 24,000 visitors per day when it charges $40 per person. For each $I increase in price,
the park loses about 300 visitors. How much should the park charge to maximize daily revenue? What is the
maximum daily revenue?

Answer 1

Let, price increased by x times.

New price = 40 + ( 1 × x ) = 40 + x

It is given that for each $1 increase park loses 300 visitors.

Number of visitors = ( 24000 - 300x )

So, revenue is given by :

R = ( 24000 - 300x )( 40 + x ) ....1)

To finding critical point :

R'(x) = 0

-300( 40 + x ) + ( 24000 - 300x ) = 0 .....By product law

-12000 - 300x + 24000 -300x = 0

600x = 12000

x = 20

So, revenue is maximum at x = 20 .

Putting x = 20 in equation 1) , we get :

R = ( 24000 - 300x )( 40 + x )

R = [ 24000 - 300(20)][40 + 20 ]

R = $1080000

Therefore, park should charge $( 40 + 20 ) = $60 for maximising the revenue and maximum revenue is $10,80,000 .

Hence, this is the required solution.