If a venue sells 11,250 tickets to a basketball game at [tex]$145 each, it is found that increasing the price by $[/tex]5 means that 3,500 fewer tickets will be sold.

What is the price that would maximize revenue?

Save any rounding for the final calculation. Show your answer to two decimal places.



Answer :

To solve this problem, we need to determine the price that maximizes the revenue for the venue. Let's break down the problem systematically:

### Step 1: Define the Variables

- Initial price per ticket (P₀): [tex]$145 - Initial number of tickets sold (T₀): 11,250 tickets - Decrease in tickets sold per $[/tex]5 increase in price: 3,500 tickets

### Step 2: Express Tickets Sold as a Function of Price

Let's say the price per ticket is [tex]\( P \)[/tex] dollars. As given, increasing the price by every $5 reduces the number of tickets sold by 3,500.

The number of tickets sold [tex]\( T \)[/tex] as a function of the price [tex]\( P \)[/tex] can be expressed as:

[tex]\[ T(P) = 11,250 - \frac{3,500}{5} \times (P - 145) \][/tex]

Reducing this, we get:

[tex]\[ T(P) = 11,250 - 700 \times (P - 145) \][/tex]

### Step 3: Calculate the Revenue Function

Revenue [tex]\( R \)[/tex] is given by the product of the price per ticket and the number of tickets sold:

[tex]\[ R(P) = P \times T(P) \][/tex]

Substituting the expression of [tex]\( T(P) \)[/tex]:

[tex]\[ R(P) = P \times \left(11,250 - 700 \times (P - 145)\right) \][/tex]

### Step 4: Simplify the Revenue Function

Substitute and simplify inside the parentheses first:

[tex]\[ R(P) = P \times \left(11,250 - 700P + 101,500\right) \][/tex]
[tex]\[ R(P) = P \times (112,750 - 700P) \][/tex]
[tex]\[ R(P) = 112,750P - 700P^2 \][/tex]

### Step 5: Optimize the Revenue Function

To find the price [tex]\( P \)[/tex] that maximizes revenue, we need to find the critical points by taking the first derivative of [tex]\( R(P) \)[/tex] and setting it to zero:

[tex]\[ \frac{dR}{dP} = 112,750 - 1,400P \][/tex]

Set the derivative equal to zero to find the critical point:

[tex]\[ 112,750 - 1,400P = 0 \][/tex]
[tex]\[ 112,750 = 1,400P \][/tex]
[tex]\[ P = \frac{112,750}{1,400} \][/tex]
[tex]\[ P = 80.54 \][/tex]

### Step 6: Calculate the Maximum Revenue

Now, substitute [tex]\( P = 80.54 \)[/tex] back into the revenue function to get the maximum revenue:

[tex]\[ R(80.54) = 112,750(80.54) - 700(80.54)^2 \][/tex]

After performing the arithmetic (which is done internally to arrive at the final value):

[tex]\[ R(80.54) = 4,540,200.89 \][/tex]

### Step 7: Final Answer

The price that maximizes revenue is [tex]\( \boxed{80.54} \)[/tex] dollars per ticket, and the maximum revenue is [tex]\( \boxed{4,540,200.89} \)[/tex] dollars. Both answers are rounded to two decimal places, as required.