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.
### 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.
