The **main objective** of this question is to find the **maximum revenue** for the given **conditions**.

This question **uses** the concept of **revenue**. **Revenue** is the **sum of the average** selling **price** multiplied by a **number** of units sold, which is the a**mount of money** generated by a **business’s typical operations**.

## Expert Answer

**First,** we have to find the **demand function**.

Let $p(x) $ be the **demand function, **so:

\[ \space p(27000) \space = \space 10 \]

\[ \space p(33000) \space = \space 8 \]

**Now**:

\[ \space (x_1, \space y_1) \space = \space (27000, \space 10) \]

\[ \space (x_2, \space y_2) \space = \space (33000, \space 8) \]

This r**epresents** the two** points** on the **straight line,**Â so:

\[ \space \frac{y_1 \space – \space y_2}{x_1 \space – \space x_2} \space = \space \frac{10 \space – \space 8}{27000 \space – \space 33000} \]

**Now** **simplifying** the above **equation** results in:

\[ \space – \frac{1}{3000} \]

Now the straight line equation is:

\[ \space y \space = \space 19 \space – \space \frac{1}{3000}xÂ \]

**Now** we have to find the **maximum** revenue. We **know** that:

\[ \space p(x) \space = \space -\frac{1}{3000}x \space + \space 19 \]

\[ \space R(x) \space = \space x . \space p(x) \]

By **putting values**, we get:

\[ \space = \space 19 x \space – \space \frac{1}{3000}x^2 \]

**Now**:

\[ \space R” \space = \space 0 \space = \space – \frac{2}{3000}x \space + \space x \]

By **simplifying**, we get:

\[ \space x \space = \space 28500 \]

**Thus**:

\[ \space p(28500) \space = \space – \frac{1}{3000}(28500) \space + \space 19Â \]

\[ \space = \space 9.50 \]

## Numerical Answer

The **ticket price** should be **set** to $ 9.50 dollars $ in **order** to get the** maximum** **revenue**.

## Example

In the above question, if the average attendance is reduced to 25,000 with a ticket price of 10, find the ticket price which should give maximum revenue.

**First,** we have to find the **demand function**.

Let $p(x) $ be the **demand function,**Â so:

\[ \space p(27000) \space = \space 10 \]

\[ \space p(33000) \space = \space 8 \]

**Now**:

\[ \space (x_1, \space y_1) \space = \space (25000, \space 10) \]

\[ \space (x_2, \space y_2) \space = \space (33000, \space 8) \]

This r**epresents** the two** points** on the **straight line,Â **so:

\[ \space \frac{y_1 \space – \space y_2}{x_1 \space – \space x_2} \space = \space \frac{10 \space – \space 8}{25000 \space – \space 33000} \]

**Now** **simplifying** the above **equation** results in:

\[ \space – \frac{1}{4000} \]

Now the straight line equation is:

\[ \space y \space = \space 19 \space – \space \frac{1}{4000}xÂ \]

**Now** we have to find the **maximum** revenue. We **know** that:

\[ \space p(x) \space = \space -\frac{1}{4000}x \space + \space 19 \]

\[ \space R(x) \space = \space x . \space p(x) \]

By **putting values**, we get:

\[ \space = \space 19 x \space – \space \frac{1}{4000}x^2 \]

**Now**:

\[ \space R” \space = \space 0 \space = \space – \frac{2}{4000}x \space + \space x \]

By **simplifying**, we get:

\[ \space x \space = \space 38000 \]

**Thus**:

\[ \space p(38000) \space = \space – \frac{1}{4000}(38000) \space + \space 19Â \]

\[ \space = \space 11.875 \]

Thus, the **ticket price** **should** be** set** to $ 11.875 $ to get the **maximum revenue**.