To solve this problem, let's calculate the total profit function $P(x)$ as the difference between the revenue function $R(x)$ and the cost function $C(x)$:

Given:

• $C(x) = 20x^2 + 250$
• $R(x) = -0.5x^2 + 500x - 100$

The profit function is: $P(x) = R(x) - C(x)$

Substituting the values of $R(x)$ and $C(x)$: $P(x) = (-0.5x^2 + 500x - 100) - (20x^2 + 250)$

Simplify the expression: $P(x) = -0.5x^2 + 500x - 100 - 20x^2 - 250$ $P(x) = (-0.5x^2 - 20x^2) + 500x + (-100 - 250)$ $P(x) = -20.5x^2 + 500x - 350$

Thus, the total profit function is: $P(x) = -20.5x^2 + 500x - 350$

This matches option (3) in the provided choices.

Would you like further clarification or steps? Here are 5 related questions for deeper understanding:

1. What happens to the profit function $P(x)$ as $x$ increases significantly?
2. How would you find the number of units $x$ that maximizes profit?
3. What does the coefficient of $x^2$ in $P(x)$ tell you about the behavior of the profit function?
4. How can you determine the breakeven points where $P(x) = 0$?
5. How would changes in the cost or revenue functions affect the profit function?

Tip: When working with functions like profit, always check for constraints such as non-negative values of $x$ since it often represents units or quantities.