The University of Oregon girls club soccer team is raffling off a bicycle to raise money for new equipment. If they charge 2 dollars per ticket, they will sell 600 tickets. For each 50 cent increase in ticket price, they will sell 30 fewer tickets. What ticket price should they charge to maximize their income?
The team should charge "" dollars to maximize their income.
Ticket price should be 6 dollars so that they sell 360 tickets and get income 2160 dollars.
current price of ticket = $2.00
number of sales = 600
let the number of 50 cent increases be n
price per ticket = 2 +.5n
number of sales = 600-30n
income = (2+.5n)(600-30n)
= 1200 - 60n + 300n - 15n^2
= 1200 + 240n - 15n^2
not sure if you know Calculus, so I will not use that method
short cut:
the n of the vertex is -b/(2a) = -240/-30 = 8
So there should be 8 increases of .50 cents, or a $4.00 increase, which makes the price of each ticket
$6.00
To determine the ticket price that maximizes the University of Oregon girls club soccer team's income, we need to find the price that generates the highest revenue.
Let's break down the problem step-by-step:
1. Start with the given information:
- Initial ticket price: $2
- Number of tickets sold at the initial price: 600
- Decrease in ticket sales for every 50 cent increase in price: 30
2. To solve this problem, we need to set up an equation for revenue.
Revenue = Ticket Price × Number of Tickets Sold
3. Calculate the revenue for the initial ticket price:
Initial revenue = $2 × 600 = $1200
4. Next, determine how the ticket price and ticket sales are related:
- For each 50 cent increase in the ticket price, 30 fewer tickets are sold.
- This means that for each 50 cent decrease in the ticket price, 30 more tickets are sold.
5. To maximize revenue, we need to find the price at which the total revenue is highest. We can do this by calculating the revenue for different ticket prices.
6. Let's create a table to summarize the information:
Ticket Price (in dollars) | Number of Tickets Sold | Revenue
--------------------------------------------
2 | 600 | $1200
2.5 | 570 | $1425
3 | 540 | $1620
3.5 | 510 | $1785
4 | 480 | $1920
7. As we can see from the table, the highest revenue is achieved when the ticket price is $4, resulting in a revenue of $1920.
Thus, the University of Oregon girls club soccer team should charge $4 per ticket to maximize their income.
To determine the ticket price that will maximize their income, we need to analyze the relationship between the ticket price and the number of tickets sold.
Let's break down the information provided:
1. The initial ticket price is $2, and they plan to sell 600 tickets.
2. For every 50 cent increase in the ticket price, they will sell 30 fewer tickets.
First, let's determine the effect of the price increase on the number of tickets sold. For each 50 cent increase in price, the sales decrease by 30 tickets. Thus, we can establish the relationship as follows:
If x is the number of 50 cent increases,
The number of tickets sold = 600 - (30 * x).
Now, let's calculate the ticket price after the increase. Each increase of 50 cents from the initial price of $2 means an increase of 0.5 * x dollars.
Thus, the ticket price after the increase = $2 + (0.5 * x).
Next, let's calculate the total income based on the number of tickets sold.
Total income = (Ticket price after the increase) * (Number of tickets sold).
Substituting the values, we get:
Total income = (2 + 0.5x) * (600 - 30x).
To maximize their income, we need to find the value of x that produces the highest total income. We can achieve this by finding the vertex of the quadratic equation represented by the total income.
The quadratic equation representing the total income is:
Income(x) = (2 + 0.5x)(600 - 30x).
To find the value of x that maximizes the income, we can take the derivative of the income function, set it equal to zero, and solve for x:
dIncome(x)/dx = 0.
Differentiating the income function:
dIncome(x)/dx = (0.5)(600 - 30x) - (2 + 0.5x)(-30).
Simplifying, we get:
0.5(600 - 30x) - 30(2 + 0.5x) = 0.
300 - 15x - 60 - 15x = 0.
-30x = -240.
x = 8.
Now that we have the value of x, we can find the ticket price that maximizes the income by substituting x back into the formula for the ticket price after the increase:
Ticket price after the increase = $2 + (0.5 * 8) = $6.
Therefore, the University of Oregon girls club soccer team should charge $6 per ticket to maximize their income.