Can anyone help me please!

Question

A hockey arena seats 1800 people when full. when the price of ticket to a hockey game is priced at $10, the game will be sold out. The manager of the arena is trying to maximuze revenue bu increasing the ticket price, but she find that for every dollar she increase the price, 50 less seats are filled.

a) write the quadratic equation in form of y= ax^2 + bx + c

b) what is the maximum revenue the company can expect? how many seats will be filled when this occur. show how to complete the square to find the two answers.

c) show numerically that your answer from part b must be correct.

 

Thank you so much! :)

Answer 1

Total hockey arena seats are 1800 .

The price of each ticket is $10 .

For every $1 increase in price of the ticket the number of seats decreases by 50 .

If x times the price $10 increases by $1 then x times of 50 seats will decreases .

Mathematically the  revenue function R = Price of each ticket * number of seats filled .

                                                     R = (10 + 1*x )( 1800 - 50*x)

 

(a)

The  revenue function R= (10 + 1*x )( 1800 - 50*x)

 R= (10 + x )( 1800 - 50x)

 R= 18000 -500x +1800x - 50x²

R= - 50x² +1300x +18000

So the quadratic form of revenue function is  R= - 50x² +1300x +18000 .

Answer 2

 

(b)

 The revenue function is R= - 50x² +1300x +18000

To find the maximum revenue , make the first derivative of revenue function to zero .

R' = -100 x +1300 = 0

-100 x +1300 = 0

100 x  = 1300

x = 13 

Now put x = 13 in  revenue function , to get maximum revenue .

R = - 50 (13)² +1300 (13) +18000 

R = - 8450 +16900 +18000 

R = 26450 .

So the maximum revenue is 26450 .

To find out the number of seats filled put x = 13 in ( 1800 - 50*x)

= 1800 - 50*13

= 1800 - 650 

= 1150

So total numbers of seats filled is 1150 .

 

 

Answer 3

(c)

To prove the maximum revenue that we have calculated in part (b) , consider two different values of x .

let x = 12 

The  revenue is 

R= - 50 (12)² +1300 (12) +18000

R= - 7200 +15600 +18000

R= 26400 .

The revenue when x = 12 is 26400 .

let x = 13 

The revenue is 

R= - 50 (13)² +1300 (13) +18000

R = - 8450 +16900 +18000 

R = 26450 .

The revenue when x = 13 is 26450 .

let x = 14

The revenue is 

R= - 50 (14)² +1300 (14) +18000

R= - 9800 +18200 +18000

R= 26400 .

The revenue when x = 14 is 26400 .

X- value	12	13	14
revenue	26400	26450	26400

Hence maximum revenue will be earned when x = 13 .

In other words maximum revenue will be earned when the cost of each ticket is $23  and  1150 seats are filled .