Find the break-even point for the firm whose cost function C and revenue function R are given. (Round your answers to the nearest whole number.)
C(x) = 180x + 15,000; R(x) = 270x

To find the break-even point, we set the cost function \( C(x) \) equal to the revenue function \( R(x) \) and solve for \( x \). The break-even point is where the firm's cost equals its revenue, meaning no profit or loss is made.

Given:
- Cost function: \( C(x) = 180x + 15,000 \)
- Revenue function: \( R(x) = 270x \)

We equate the two functions:

\[
C(x) = R(x)
\]

\[
180x + 15,000 = 270x
\]

### Step-by-step solution:
1. Subtract \( 180x \) from both sides:

\[
15,000 = 270x - 180x
\]

\[
15,000 = 90x
\]

2. Divide both sides by 90:

\[
x = \frac{15,000}{90} = 166.67
\]

Rounding to the nearest whole number:

\[
x = 167
\]

### Conclusion:
The break-even point occurs when the firm sells **167 units**.

Would you like further details, or do you have any questions?

Here are 5 related questions to explore:
1. What happens if the cost per unit increases to 200?
2. How would the break-even point change if the fixed costs rise to 20,000?
3. What is the profit when 200 units are sold?
4. What if the revenue per unit decreases to 250?
5. How does the break-even point shift if the fixed cost is reduced to 10,000?

**Tip:** Break-even analysis is crucial for understanding at what sales level a business starts to cover its costs!

Find the break-even point for the firm whose cost function C and revenue function R are given. (Round your answers to the nearest whole number.) C(x) = 180x + 15,000; R(x) = 270x

Find the break-even point where the cost function C(x) = 180x + 15,000 equals the revenue function R(x) = 270x. This problem involves algebraic concepts of linear equations and is suitable for students in Grades 9-11.

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