You have two functions, a cost function \( C(x) = 3x^2 - 10x + 200 \) and a revenue function \( R(x) = -2x^2 + 100x + 50 \). Here are a few things you might want to analyze:

### 1. **Profit Function:**
The profit function \( P(x) \) is calculated as:
\[
P(x) = R(x) - C(x)
\]
Substitute the given equations:
\[
P(x) = (-2x^2 + 100x + 50) - (3x^2 - 10x + 200)
\]
Simplifying:
\[
P(x) = -2x^2 + 100x + 50 - 3x^2 + 10x - 200
\]
\[
P(x) = -5x^2 + 110x - 150
\]
Thus, the profit function is:
\[
P(x) = -5x^2 + 110x - 150
\]

### 2. **Break-Even Point:**
The break-even point occurs when profit is zero:
\[
P(x) = 0 \quad \text{or} \quad R(x) = C(x)
\]
This gives the equation:
\[
-2x^2 + 100x + 50 = 3x^2 - 10x + 200
\]
Simplifying:
\[
-5x^2 + 110x - 150 = 0
\]
We can solve this quadratic equation to find the break-even points.

Would you like me to solve for the break-even points, or explore other aspects of the problem?

### 5 Related Questions:
1. What is the maximum profit, and at what value of \(x\) does it occur?
2. How do the cost and revenue behave as \(x\) increases?
3. What are the marginal cost and marginal revenue functions?
4. What is the minimum cost and at what production level does it occur?
5. Can we determine the production levels for loss-making?

#### Tip:
Quadratic functions, like these, can give insights into the maximum or minimum values of profit, cost, and revenue by analyzing their critical points.

Explore how to analyze cost and revenue functions using quadratic equations. Learn how to derive the profit function and find the break-even point for given functions. Suitable for advanced high school students and beyond.

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