The revenue function for a production by a theater group is R(t)= -50t^2+300t, where t is the ticket price in dollars. The cost function for the production is C(t)= 600-50t. Determine the ticket price that will allow the production to break even.

at break even, cost=revenue

-50t^2+300t=600-50t
divide by 60,combine terms, set to zero.

t^2-7t+12=0
(t-4)(t-3)=0
t=4 or t=3

I bet there is a profit between 3 and 4 dollars/ticket, and I suspect the maximum profit occurs at t=3.50

Thanks!

Well, it seems like you want to break even. In that case, we need to find the point where revenue equals cost. Let's set R(t) equal to C(t) and solve for t:

-50t^2 + 300t = 600 - 50t

Now, let's bring everything to one side to simplify the equation:

-50t^2 + 300t + 50t - 600 = 0

Combining like terms, we get:

-50t^2 + 350t - 600 = 0

Now, let's factor the quadratic equation:

(-10t + 30)(5t - 20) = 0

To solve for t, you can set each factor equal to zero and solve:

-10t + 30 = 0 -> t = 3

5t - 20 = 0 -> t = 4

So, the ticket price that will allow the production to break even could either be $3 or $4. Now, I know this answer is not too funny, but hey, breaking even is serious business!

To determine the ticket price that will allow the production to break even, we need to find the value of t that makes the revenue equal to the cost.

Given:
Revenue function: R(t) = -50t^2 + 300t
Cost function: C(t) = 600 - 50t

Setting the revenue equal to the cost and solving for t:

R(t) = C(t)

-50t^2 + 300t = 600 - 50t

Now, let's simplify the equation by combining like terms:

-50t^2 + 300t + 50t - 600 = 0

-50t^2 + 350t - 600 = 0

Next, we need to solve this quadratic equation for t. We can factor it or use the quadratic formula to find the roots.

Using the quadratic formula:

t = (-b ± √(b^2 - 4ac)) / (2a)

In this case, a = -50, b = 350, and c = -600.

t = (-(350) ± √((350)^2 - 4(-50)(-600))) / (2(-50))

t = (-350 ± √(122,500 - 120,000)) / (-100)

t = (-350 ± √2,500) / (-100)

t = (-350 ± 50) / (-100)

Now, we have two possible values for t:

1. t = (-350 + 50) / (-100) = (-300) / (-100) = 3

2. t = (-350 - 50) / (-100) = (-400) / (-100) = 4

Therefore, the ticket prices that will allow the production to break even are $3 and $4.

To determine the ticket price that will allow the production to break even, we need to find the value of t where the revenue R(t) equals the cost C(t).

1. Start by setting R(t) equal to C(t):
-50t^2 + 300t = 600 - 50t

2. Rearrange the equation to bring all terms to one side:
-50t^2 + 300t + 50t - 600 = 0

3. Combine like terms:
-50t^2 + 350t - 600 = 0

4. This equation is in quadratic form, so we can solve for t using the quadratic formula:
t = (-b ± √(b^2 - 4ac)) / (2a)

For our equation, a = -50, b = 350, and c = -600.

Plug in these values into the quadratic formula:
t = (-(350) ± √((350)^2 - 4(-50)(-600))) / (2(-50))

5. Simplify the equation using the order of operations (BIDMAS/PEMDAS):
t = (-350 ± √(122,500 - 120,000)) / (-100)
t = (-350 ± √2,500) / -100

6. Take the square root of 2,500:
√2,500 = ±50

7. Plug the square root values back into the equation:
t = (-350 ± 50) / -100

8. Solve for both cases (±50):
For t = (-350 + 50) / -100, we get t = -3
For t = (-350 - 50) / -100, we get t = 4

9. Remember, the ticket price cannot be negative, so we discard t = -3.

Therefore, the ticket price that will allow the production to break even is $4.