Question

Western music concert is organised every year in the stadium that can hold 36000 spectators. With ticket price of 10, the average attendance has been 24000. Some financial expert estimated that price of a ticket should be determined by the function 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 x can be represented as x r (a) 15x (b) 15- (c) 15x-- 3000 3000 30 (ii) The range of x is (a) [24000, 36000] (b) (0, 24000) (c) (0, 36000 (iii) The value of x for which revenue is maximum, is (a) 20000 (b) 21000 (c) 22500​

Answer 1

Step-by-step explanation:

Based on the above information, answer the following questions.

(i) The revenue, R as a function of xcan be represented as

(a) 15x−x23000 (b) 15−x23000 (c) 15x−130000 (d) 15x−x3000(ii) The range of x is

(a) [24000,36000] (b) [0,24000] (c) [0,36000] (d) none of these(iii) The value of xfor which revenue is maximum, is

(a) 20000 (b) 21000 (c) 22500 (d) 25000(iv) When the revenue is maximum, the price of the ticket in

(a) Rs.5 (b) Rs.5.5 (c) Rs.7 (d) Rs. 7.5(v) How any spectators should be present to maximize the revenue?

(a) 21500 (b) 21000 (c) 22000 (d) 225005

Answer 2

Note: the given question is incomplete and incorrect and the correct question is attached down.

Given:

Total capacity of the stadium$=36000$

Price of one ticket$=Rs. 10$

Average attendance of the concert$=24000$

Function with which the price of the ticket should be calculate is,

$\[p\left( x \right) = 15 - \frac{x}{{3000}}\]$

Solution:

Part (i):

Assume that p is the price per ticket and x is the number of the tickets sold.

Find the revenue function.

Revenue function,

$\[\begin{array}{l}R\left( x \right) = p \times x\\ \Rightarrow R\left( x \right) = \left( {15 - \frac{x}{{3000}}} \right)x\\ \Rightarrow R\left( x \right) = \left( {15x - \frac{{{x^2}}}{{3000}}} \right)\end{array}\]$

Therefore, the revenue function can be represented as $\[R\left( x \right) = \left( {15x - \frac{{{x^2}}}{{3000}}} \right)\]$

Hence, the correct answer is option (a). i.e.,  $\[R\left( x \right) = \left( {15x - \frac{{{x^2}}}{{3000}}} \right)\]$.

Part (ii):

FInd the range of x, i.e.,  the number of the tickets sold.

Know that more 36000 tickets cannot be sold.

Therefore, the range of x is $[0,36000]$.

Hence, the correct answer is option (c). i.e., $[0,36000]$.

Part (iii):

Find the value of x for which revenue is maximum.

Know that,

Revenue function, $\[R\left( x \right) = \left( {15x - \frac{{{x^2}}}{{3000}}} \right)\]$.

To find maxima, put $\[R'\left( x \right) = 0\]$.

Differentiate $R(x)$ with respect to x.

$\[\left( {15x - \frac{{{x^2}}}{{3000}}} \right) = 0\]$

$\[ \Rightarrow 15 - \frac{{2x}}{{3000}} = 0\]$

$\[ \Rightarrow 15 - \frac{x}{{1500}} = 0\]$

$\[\begin{array}{l} \Rightarrow x = 15 \times 1500\\ \Rightarrow x = 22500\end{array}\]$

Hence, the correct answer is option (c). i.e., 22500.

Part (iv)

Find the price of one ticket when revenue is maximum, i.e., $x=22500$.

Price of a ticket$\[ = 15 - \frac{x}{{3000}}\]$

                          $\[\begin{array}{l} = 15 - \frac{{22500}}{{3000}}\\ = 15 - \frac{{15}}{2}\\ = \frac{{15}}{2}\\ = Rs.7.5\end{array}\]$

Hence, the correct answer is option (d). i.e., Rs. 7.5.

Part (v)

Find the number of spectators that should be present to maximize the revenue.

Know that, number of spectators $=$ maximum tickets sold

Therefore, the number of spectators that should be present to maximize the revenue $=22500$

Hence, the correct answer is option (d). i.e., 22500.