To find the maximum revenue for the revenue function $R(x) = 392x - 0.9x^2$, we need to use calculus to determine the critical points. Specifically, we'll take the derivative of the function and set it equal to zero to find the critical points, then check for the maximum.

Step 1: Find the derivative of the revenue function

The derivative of $R(x)$ with respect to $x$ is: $R'(x) = 392 - 1.8x$

Step 2: Set the derivative equal to zero to find the critical points

$392 - 1.8x = 0$ Solving for $x$: $1.8x = 392$ $x = \frac{392}{1.8} = 217.78$

Step 3: Verify that it's a maximum

To confirm that this critical point represents a maximum, we can check the second derivative of the function, $R''(x)$.

The second derivative of $R(x)$ is: $R''(x) = -1.8$ Since $R''(x) < 0$, the function is concave down, meaning $x = 217.78$ is indeed a maximum.

Step 4: Find the maximum revenue

Now that we know the value of $x$ that maximizes revenue, we substitute it back into the original revenue function to find the maximum revenue:

$R(217.78) = 392(217.78) - 0.9(217.78)^2$

Let me calculate that.

The maximum revenue is approximately $42,684.44$.

Would you like more details on any part of this solution, or do you have further questions?

Here are some related questions for deeper understanding:

1. How does the derivative help in finding the maximum or minimum of a function?
2. Why does the second derivative test confirm that $x = 217.78$ is a maximum?
3. What is the economic interpretation of the function $R(x) = 392x - 0.9x^2$?
4. How would you find the maximum if there were restrictions on the values of $x$?
5. How does this result change if the coefficients in the revenue function were different?

Tip: In optimization problems, always use the second derivative test to confirm whether you have a maximum or a minimum.