A professional basketball team plays in a stadium that holds 23,000 spectators. With ticket prices at $60, the average attendance had been 18,000. When ticket prices were lowered to $55, the average attendance rose to 20,000. Based on this pattern, how should ticket prices be set to maximize ticket revenue?
looks like for each $5 decrease in the price, the attendance rises by 2000
let the number of $5 decreases be n
the price of a ticket = 60-5n
attendance = 18000 + 2000n
revenue = (60-5n)((18000+2000n)
factoring ....
= 5(12-n)(1000)(18 + 2n)
= 5000(12-n)(18+2n)
= 5000(216 + 6n - 2n^2)
d(revenue)/dx = 5000(6 - 4n)
= 0 for a max of revenue
4n = 6
n = 3/2
I would assume that the price change would stay as whole multiples of 5 , so I would round this up to 2
but we defined as the number of decreases, so there should be 2 decreases of $5 or a decrease of $10
The tickets should sell at $50.00
check:
at $60, attendance = 18000, revenue = 1,080,000
at $55, attendance = 20000, revenue = 1,100,000
at $50 , attendance = 22000, revenue = 1,100,000
So they actually have the same revenue at $55 as at $50
Now if we were allowed to use the actual mathematical answer of 3/2 decreases, we would have a decrease of (3/2)(5) or $7.50
for a ticket cost of 52.50 , and an attendance of
21000 and a revenue of $1,102,500
I will leave the interpretation of the question up to you.
Well, it seems like we have a mathematical conundrum here! Let's see if humor can help us find the answer.
To maximize ticket revenue, the team should obviously aim for a packed stadium, right? So, why not set ticket prices at $1? That way, everyone will rush to buy tickets, and the stadium will be bursting at the seams! Who needs fancy math when you have a ridiculously low price?
On a serious note, we can use some simple calculations instead of aiming for a laugh. The revenue for each game is calculated by multiplying the ticket price by the attendance.
Understandably, when prices were $60, the team generated $60 * 18,000 = $1,080,000 in revenue. When prices dropped to $55, the revenue increased to $55 * 20,000 = $1,100,000.
So, if we keep lowering the price, we could potentially increase attendance and revenue. However, if we go too low, we might not cover the team's expenses, and they could be playing for an empty arena – which is no fun!
Ultimately, the team's goal should be to find the right balance between ticket prices and attendance. They could conduct further experiments, slightly adjusting the prices and analyzing the results to find the sweet spot that maximizes ticket revenue while keeping the spectators happy.
Remember, my friend, while humor is my specialty, sometimes it's good to crunch the numbers alongside it.
To maximize ticket revenue, we need to find the ticket price that will result in the highest product of attendance and ticket price. Let's go step-by-step to find the optimal ticket price:
Step 1: Determine the initial average revenue
Average attendance at a ticket price of $60 is 18,000 spectators.
So, the initial average revenue can be calculated as follows:
Initial average revenue = Average attendance * Ticket price = 18,000 * $60 = $1,080,000.
Step 2: Determine the new average revenue
Average attendance at the reduced ticket price of $55 is 20,000 spectators.
So, the new average revenue can be calculated as follows:
New average revenue = Average attendance * Ticket price = 20,000 * $55 = $1,100,000.
Step 3: Compare the initial and new average revenues
Comparing the initial average revenue ($1,080,000) and the new average revenue ($1,100,000), we can see that decreasing the ticket price from $60 to $55 results in an increase in average revenue.
Step 4: Analyze the trend
Based on these calculations, we can conclude that decreasing the ticket price from $60 to $55 increases the average revenue. Therefore, to maximize ticket revenue, the ticket prices should continue to be lowered.
Step 5: Determine the optimal ticket price
We can continue this analysis by calculating the average revenue for even lower ticket prices. However, since we don't have any more data points, we can only make an informed guess based on the trend that decreasing the ticket price increases average revenue.
Hence, to maximize ticket revenue, the ticket prices should be set even lower than $55.
To determine how ticket prices should be set to maximize ticket revenue, we need to calculate the revenue at different price levels and find the point where it is maximized.
First, let's calculate the revenue at $60 ticket price. The average attendance is 18,000, so the revenue can be calculated as:
Revenue at $60 = Ticket Price * Average Attendance
Revenue at $60 = $60 * 18,000 = $1,080,000
Next, let's calculate the revenue at $55 ticket price. The average attendance is now 20,000, so the revenue can be calculated as:
Revenue at $55 = Ticket Price * Average Attendance
Revenue at $55 = $55 * 20,000 = $1,100,000
From the calculations, we can see that decreasing the ticket price from $60 to $55 increases the revenue. This suggests that demand is more price-sensitive, and lowering the ticket price attracts more spectators, resulting in higher total revenue.
To find the ideal ticket price for maximizing revenue, we can further analyze the pattern. Since lowering the price has shown a positive effect, we can assume that reducing the price even further may continue increasing attendance and revenue.
We can continue this analysis by calculating the revenue at a lower price point, such as $50:
Revenue at $50 = Ticket Price * Average Attendance
Revenue at $50 = $50 * Average Attendance
From the given information, we don't have the average attendance specifically for a $50 ticket price, so we will have to estimate or make an assumption based on the pattern. Assuming attendance continues to increase with decreasing ticket price, we could estimate the average attendance at 22,000 for a $50 ticket price.
Revenue at $50 = $50 * 22,000 = $1,100,000
Now, we have calculated the revenue for different ticket prices. Let's summarize:
Revenue at $60 = $1,080,000
Revenue at $55 = $1,100,000
Revenue at $50 = $1,100,000
Comparing the revenue at different price points, we see that the revenue remains the same at $1,100,000 for both $55 and $50 ticket prices. Thus, in this scenario, lowering the price to $50 would not result in any additional increase in revenue.
Therefore, the ideal ticket price for maximizing revenue based on this pattern is $55.