Question

The current price of amateur theater tickets is $20, and the venue typically sells 500 tickets. A survey found that for each $1 increase in ticket price, 10 fewer tickets are sold.1. What price per ticket will maximize the revenue? 2. What is the maximum revenue?

Answer 1

Let the price for one $1 increase be

`=x`

Let the revenue be represented as

`=R`

The number of tickets sold will be represented below as

`=500-10x`

The price per ticket will be represented below as

`=20+x`

The revenue is calculated using the formula below

`R=Nu\text{mber of tickets}* price\text{ per tickets}`

By substituting the values above, we will have

`\begin{gathered} R=Nu\text{mber of tickets}* price\text{ per tickets} \\ R=(500-10x)(20+x) \\ \end{gathered}`

Expanding the brackets, we will have

`\begin{gathered} R=(500-10x)(20+x) \\ R=500(20+x)-10x(20+x) \\ R=10000+500x-200x-10x^2 \\ R=10000+300x-10x^2 \end{gathered}`

Step 1;

We would calculate the value of x that will give maximum revenue

We will use the expression below

`x=-(b)/(2a)`

The general form of a quadratic equation is given below as

`\begin{gathered} R=-10x^2+300x+10000 \\ y=ax^2+bx+c \\ By\text{ comparing coefficents, we will have} \\ a=-10,b=300,c=10000 \end{gathered}`

By substituting the values, we will have

`\begin{gathered} x=-(b)/(2a) \\ x=-(300)/((2*-10)) \\ x=(-300)/(-20) \\ x=15 \end{gathered}`

Therefore,

The price per ticket that will maximize profit will be

`\begin{gathered} =20+x \\ =20+15 \\ =35 \end{gathered}`

Hence,

The price per ticket that will maximize profit will be = $35

Step 2:

To calculate the maximum revenue, we will substitute the value of x= 15 in the equation for the revenue given below

`\begin{gathered} R=-10x^2+300x+10000 \\ R=-10(15)^2+300(15)+10000 \\ R=-2250+4500+10000 \\ R=12,250 \end{gathered}`

Hence,

The maximum revenue is = $12,250