Question

II.
CASE STUDY:
17
Western music concert is organized every year in the stadium that can hold 36000
spectators. With ticket price of Rs. 10, the average attendance has been 24000. Some
financial expert estimated that price of a ticket should be determined by the function
1
P(x) = 15
where x is the number of tickets sold.
3000
Based on the above information, answer the following questions.
i) The revenue, R as a function of can be represented as
2
1
a) 15.6
b) 15 -
3000
c) 150-
X
d) 15x -
3000
3000
30000
ii) The range of x is
a) (24000, 36000] b) (0,24000]
c) (0,36000]
d) none of these
iii) The value of x for which revenue is maximum is
a) 20000
b) 21000
c) 22500
d) 25000
iv. When the revenue is maximum, the price of the ticket is (v) how many spectators should be present to maximize the revenue? ​

Answer 1

Answer:

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Step-by-step explanation:

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Answer 2

Answer:

(i) Revenue function R(x) = 15x - $\frac{x^{2}}{3000}$

(ii)  Range of x = (0, 36000]

(iii) Maximum revenue will be at x = 22500

(iv) When the revenue is maximum, the price of the ticket must be Rs.   7.50

(v) 22500 spectators must be present to maximize the revenue.

Step-by-step explanation:

Maximum capacity = 36000

Ticket price = Rs. 10

Average attendees = 24000

Price of the ticket = 15 - $\frac{x}{3000}$  x is the no. of tickets sold

(i) Revenue function R(x) = Selling price * Quantity

                                         =  (15 - $\frac{x}{3000}$) * x

                                         = 15x - $\frac{x^{2}}{3000}$

Therefore, R(x) = 15x - $\frac{x^{2}}{3000}$

(ii) Range of x =

Since maximum capacity is 36000, the upper limit is 36000

As 24000 is the average attendance,

The minimum value of x can be 0, the lower limit is 0.

Therefore, range = (0, 36000]

(iii) R(x) = 15x - $\frac{x^{2}}{3000}$

=> R'(x) = 15 - $\frac{1}{3000}$ * 2x

putting R'(x) = 0

15 - $\frac{1}{3000}$ * 2x = 0

=> 15 =   $\frac{2x}{3000}$

=> x = 22500

R''(x) = 0 - $\frac{1}{3000}$ * 2 < 0 Therefore, maximum at x = 22500

Therefore, maximum revenue will be at x = 22500

(iv) We know that Price of the ticket = 15 - $\frac{x}{3000}$  x is the no. of tickets sold

if x = 22500, the price of the ticket will be 15 - $\frac{22500}{3000}$

                                                                          = 15 - 7.5

                                                                          = 7.5

Therefore, when the revenue is maximum, the price of the ticket must be Rs. 7.50

(v) For maximum revenue, we have already proved that x = 22500

where x is the no. of tickets sold = no. of spectators.

Therefore, 22500 spectators must be present to maximize the revenue.