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.