Pleas 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?

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.

step-by-step explanation:

y = | -2(-2) |     →     y = | 4 |     →     y = 4       ⇒   (-2, 4)

y = | -2(-1) |     →     y = | 2 |     →       y =   2       ⇒   (-1, 2)    

y = | -2(0) |     →     y = | 0 |     →       y = 0         ⇒   (0, 0)

y = | -2(1) |   c →     y = | -2 |     →     y =   2         ⇒   (1, 2)    

y = | -2(2) |     →     y = | - 4 |     →     y =   4         ⇒   (2, 4)