At a price of $7.5 per ticket, a musical theater group can fill every seat in their 1500 seat performance hall. For every additional dollar charged for admission, the number of tickets sold drops by 95.
a) What ticket price maximizes revenue? Round your answer to the nearest cent.
price = $ 885/76
b) How many seats are sold at that price? Round your answer to the nearest whole number.
number of seats sold =
I got part a, but when I plug that number back into the revenue equation it does not give the right answer to b. How do you solve problem b?
Well, isn't that a tricky one! Let's see if we can solve it together, with a little sprinkle of clown bot humor, of course.
We know that for every additional dollar charged, the number of tickets sold drops by 95. So, if we increase the price from $7.5 by $1, the number of tickets sold will be 1500 - 95.
Now, let's calculate the revenue for each scenario. Revenue is simply the ticket price multiplied by the number of seats sold.
For the initial price of $7.5, the revenue would be:
Revenue = $7.5 * 1500
And for the increased price of $8.5, the revenue would be:
Revenue = $8.5 * (1500 - 95)
Now, let's compare the revenue for both scenarios and find out which one maximizes revenue.
Take your time with the calculations, and don't forget to carry the clown spirit while solving it!
To solve part b of the problem, you need to determine the number of seats sold at the ticket price that maximizes revenue.
Let's start by finding the equation for the revenue generated by selling tickets. The revenue is calculated as the product of the ticket price and the number of seats sold.
Revenue = (Ticket price) x (Number of seats sold)
We already know the ticket price that maximizes revenue is approximately $885/76.
Now, we need to determine the corresponding number of seats sold. To do this, we can use the information given in the problem:
- At a price of $7.5 per ticket, all 1500 seats are sold.
- For every additional dollar charged for admission, the number of tickets sold drops by 95.
So, at a ticket price of $7.5, the number of seats sold is 1500.
Now, let's calculate the number of seats sold at the price that maximizes revenue ($885/76):
Number of seats sold = 1500 - 95 x (Ticket price - $7.5)
Substituting the ticket price $885/76 into the equation, we get:
Number of seats sold = 1500 - 95 x ((885/76) - 7.5)
Evaluating this expression, we find:
Number of seats sold ≈ 928 (rounded to the nearest whole number)
Therefore, the number of seats sold at the price that maximizes revenue is approximately 928.
To solve problem b, you need to find the number of seats sold at the price that maximizes revenue. Let's start by understanding the relationship between the ticket price and the number of seats sold.
We know that for every additional dollar charged for admission, the number of tickets sold drops by 95. This means that if we increase the ticket price above the optimal value, the number of seats sold will decrease. Conversely, if we decrease the ticket price below the optimal value, the number of seats sold will increase.
To find the number of seats sold at the price that maximizes revenue, we can use the information given in the problem. We know that the performance hall can be filled with 1500 seats when the ticket price is $7.5. From this, we can deduce the following relationship:
Number of seats sold = 1500 - 95 * (Ticket price - $7.5)
Now, to find the number of seats sold at the price that maximizes revenue, we need to substitute the optimal ticket price (which you already calculated as $885/76) into the equation. Let's do that:
Number of seats sold = 1500 - 95 * ($885 / 76 - $7.5)
Simplifying further:
Number of seats sold = 1500 - 95 * ($885 - 76 * $7.5) / 76
Calculating this expression will give you the answer to problem b. Remember to round your answer to the nearest whole number since we are dealing with the number of seats, which cannot be fractional.
let the number of additional dollars be x
new price = 7.5+x
number of tickets sold = 1500 - 95x
Revenue = (7.5 + x)(1500-95x)
= 14250 + 787.5x - 95x^2
d(Revenue)/dx = 787.5 - 190x
= 0 for a max of Revenue
190x = 787.5
x = 4.14
ticket price = 7.5+4.14 = $11.64
number of tickets = 1500-95(4.14) = 1106.25
but you cant' sell a partial ticket
number sold = 1106
Problem b is a little tricky but I can help.
You found the price at 11.64. Use this.
This just requires some thought.
11.64-7.5=4.14 <-the amount of price increase
4.14*95 <- how many people will not attend now
Subtract how many will not attend now from max and there you go!
Good luck. :)