Explore Westonci.ca, the leading Q&A site where experts provide accurate and helpful answers to all your questions. Get immediate and reliable answers to your questions from a community of experienced experts on our platform. Explore comprehensive solutions to your questions from a wide range of professionals on our user-friendly platform.
Sagot :
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.
Thanks for using our service. We aim to provide the most accurate answers for all your queries. Visit us again for more insights. Thanks for using our platform. We aim to provide accurate and up-to-date answers to all your queries. Come back soon. Get the answers you need at Westonci.ca. Stay informed with our latest expert advice.
