Find the marginal profit for producing x units. (The profit is measured in dollars.)
P = −0.65x2 + 5,000x − 1,750,000

To find the marginal profit, you need to calculate the derivative of the profit function \( P(x) \) with respect to \( x \). The marginal profit represents the rate of change of profit with respect to the number of units produced.

Given the profit function:

\[
P(x) = -0.65x^2 + 5000x - 1,750,000
\]

Differentiate \( P(x) \) with respect to \( x \) to find the marginal profit \( P'(x) \):

\[
P'(x) = \frac{d}{dx} \left( -0.65x^2 + 5000x - 1,750,000 \right)
\]

The derivative of each term is as follows:

- The derivative of \( -0.65x^2 \) is \( -1.3x \) (since \( \frac{d}{dx} \left( ax^2 \right) = 2ax \))
- The derivative of \( 5000x \) is \( 5000 \)
- The derivative of \( -1,750,000 \) is \( 0 \) (since it's a constant)

So, the marginal profit function is:

\[
P'(x) = -1.3x + 5000
\]

### Therefore, the marginal profit for producing \( x \) units is \( P'(x) = -1.3x + 5000 \).

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

Here are some related questions you might consider:
1. What is the marginal profit at \( x = 1000 \) units?
2. How many units should be produced to maximize profit?
3. What is the maximum profit achievable?
4. How does the profit change as more units are produced?
5. What does a negative marginal profit indicate about the production?

**Tip:** Marginal profit is crucial for determining the optimal level of production, where producing one more unit neither increases nor decreases the profit significantly.

Find the marginal profit for producing x units. (The profit is measured in dollars.) P = −0.65x^2 + 5,000x − 1,750,000

Learn how to calculate the marginal profit for producing x units using differentiation. This problem involves finding the derivative of the profit function P(x) = -0.65x^2 + 5000x - 1,750,000. Ideal for students in Grades 11-12 or college-level calculus.

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