Answer:
The profit function is:

The maximum value is 406, 300 occurring when x = 640.
Step-by-step explanation:
The revenue function is:

And the cost function is:

Then the total profit function will be:

This is a quadratic function.
Therefore, the maximum value of the total profit will occur at its vertex point.
The vertex of a quadratic is given by:

In this case, a = -1, b = 1280, and c = -3300.
Then the point at which the maximum profit occurs is at:

And the maximum profit will be:
