To solve this problem, we are given the revenue function for the manufacturer:

$R(x) = 80x - 0.4x^2$

where $x$ represents the number of units sold, and $R(x)$ is the revenue in dollars. We need to find the maximum revenue and the number of units $x$ that maximizes this revenue.

Solution Steps:

1. Find the derivative of $R(x)$: To find the maximum, we need to take the derivative of $R(x)$ with respect to $x$ and set it equal to zero. $R'(x) = 80 - 0.8x$

2. Set the derivative to zero and solve for $x$: $80 - 0.8x = 0$ $0.8x = 80$ $x = \frac{80}{0.8} = 100$

   So, $x = 100$ units should be manufactured to achieve maximum revenue.

3. Calculate the maximum revenue by substituting $x = 100$ back into $R(x)$: $R(100) = 80(100) - 0.4(100)^2$ $R(100) = 8000 - 0.4 \times 10000$ $R(100) = 8000 - 4000 = 4000$

Answer:

• Maximum Revenue: $4000
• Number of Units for Maximum Revenue: 100 units

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

Here are some related questions to expand on this topic:

1. What is the significance of the second derivative in verifying maximum revenue?
2. How would the maximum revenue change if the function had a different quadratic term?
3. What are other methods to find maximum points of a quadratic function besides differentiation?
4. How does the concept of maximum revenue relate to the real-world pricing strategy?
5. How would we interpret the result if the derivative did not equal zero?

Tip: For optimization problems, always check the sign of the second derivative to confirm if the critical point is a maximum or a minimum.