Question

The revenue for a school performance can be modeled using the equation R=−30t²+200t, where t is the price per ticket in dollars. The school's cost for the performance is modeled using the equation C=−100t+750. How much does the school need to charge per ticket in order to break even?

Answer 1

To find the price per ticket that will result in the school breaking even, we need to set the revenue equation equal to the cost equation and solve for t. The price per ticket that will result in the school breaking even is $5.

Step-by-step explanation:

To find the price per ticket that will result in the school breaking even, we need to find the point where the revenue equation equals the cost equation. Let's set R (revenue) equal to C (cost) and solve for t (the price per ticket):

R = C

-30t² + 200t = -100t + 750

Combining like terms:

-30t² + 300t - 750 = 0

Next, we can use the quadratic formula to solve for t:

t = (-b ± √(b² - 4ac)) / 2a

Using the coefficients from the quadratic equation, we have:

t = ( -300 ± √(300² - 4(-30)(-750))) / 2(-30)

Simplifying:

t = ( -300 ± √(90000 - 90000)) / -60

t = ( -300 ± √0) / -60

Since the discriminant (√0) is equal to zero, we only have one solution:

t = -300 / -60

t = 5

Therefore, the school needs to charge $5 per ticket in order to break even.