Respuesta :
To find the maximum revenue, we need to determine the ticket price that maximizes revenue and the corresponding number of tickets sold.
Let's denote:
- \( x \) as the number of $5 increases from the base price of $35.
- \( p \) as the increase in price from the base price of $35, so the price per ticket becomes \( 35 + 5x \).
- \( q \) as the decrease in the number of tickets sold for each $5 increase, so the number of tickets sold becomes \( 6000 - 400x \).
The revenue \( R \) is given by the product of the ticket price and the number of tickets sold:
\[ R = p \times q \]
Substituting the expressions for \( p \) and \( q \), we get:
\[ R = (35 + 5x) \times (6000 - 400x) \]
To find the maximum revenue, we can take the derivative of \( R \) with respect to \( x \), set it equal to zero, and solve for \( x \).
\[ \frac{dR}{dx} = 5(6000 - 400x) - 400(35 + 5x) \]
\[ \frac{dR}{dx} = 30000 - 2000x - 14000 - 2000x \]
\[ \frac{dR}{dx} = 16000 - 4000x \]
Setting \(\frac{dR}{dx} = 0\):
\[ 16000 - 4000x = 0 \]
\[ 4000x = 16000 \]
\[ x = 4 \]
So, \( x = 4 \) represents the number of $5 increases from the base price.
Now, to find the maximum revenue, substitute \( x = 4 \) back into the expression for \( R \):
\[ R = (35 + 5 \times 4) \times (6000 - 400 \times 4) \]
\[ R = (35 + 20) \times (6000 - 1600) \]
\[ R = 55 \times 4400 \]
\[ R = 242000 \]
Therefore, the maximum revenue is $242,000, and this is achieved by selling \( 6000 - 400 \times 4 = 4400 \) tickets.
Let's denote:
- \( x \) as the number of $5 increases from the base price of $35.
- \( p \) as the increase in price from the base price of $35, so the price per ticket becomes \( 35 + 5x \).
- \( q \) as the decrease in the number of tickets sold for each $5 increase, so the number of tickets sold becomes \( 6000 - 400x \).
The revenue \( R \) is given by the product of the ticket price and the number of tickets sold:
\[ R = p \times q \]
Substituting the expressions for \( p \) and \( q \), we get:
\[ R = (35 + 5x) \times (6000 - 400x) \]
To find the maximum revenue, we can take the derivative of \( R \) with respect to \( x \), set it equal to zero, and solve for \( x \).
\[ \frac{dR}{dx} = 5(6000 - 400x) - 400(35 + 5x) \]
\[ \frac{dR}{dx} = 30000 - 2000x - 14000 - 2000x \]
\[ \frac{dR}{dx} = 16000 - 4000x \]
Setting \(\frac{dR}{dx} = 0\):
\[ 16000 - 4000x = 0 \]
\[ 4000x = 16000 \]
\[ x = 4 \]
So, \( x = 4 \) represents the number of $5 increases from the base price.
Now, to find the maximum revenue, substitute \( x = 4 \) back into the expression for \( R \):
\[ R = (35 + 5 \times 4) \times (6000 - 400 \times 4) \]
\[ R = (35 + 20) \times (6000 - 1600) \]
\[ R = 55 \times 4400 \]
\[ R = 242000 \]
Therefore, the maximum revenue is $242,000, and this is achieved by selling \( 6000 - 400 \times 4 = 4400 \) tickets.
Answer: The maximum revenue is $242,000, and 4,400 tickets are sold to achieve this maximum.
Step-by-step explanation:
To find the maximum revenue and the number of tickets sold to achieve it, we need to follow these steps:
1. Calculate the number of tickets sold at $35 each:
- At $35 per ticket, all 6,000 seats are sold.
- Total revenue at $35 per ticket = 6,000 seats * $35 = $210,000.
2. Determine the impact of a $5 increase in ticket price:
- For each $5 increase, 400 fewer tickets are sold.
- Revenue after a $5 increase = (6,000 - 400) seats * ($35 + $5) = 5,600 seats * $40 = $224,000.
3. Continue increasing the ticket price by $5 until the revenue starts decreasing:
- Next increase: (5,200 seats * $45) = $234,000
- Further increase: (4,800 seats * $50) = $240,000
- Another increase: (4,400 seats * $55) = $242,000
4. Identify the maximum revenue and the corresponding number of tickets sold:
- The maximum revenue is $242,000.
- The number of tickets sold to achieve this maximum revenue is 4,400 seats.
Therefore, the maximum revenue is $242,000, and 4,400 tickets are sold to achieve this maximum.
