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The price-demand function is given as:
\[
p(x) = 80 - 3x \quad \text{where} \quad 1 \leq x \leq 20
\]

### Part (A): What is the revenue function?
The revenue function $ R(x) $ is found by multiplying the price $ p(x) $ by the number of units sold, $ x $. Thus, the revenue function is:
\[
R(x) = x \cdot p(x)
\]
Substituting $ p(x) = 80 - 3x $:
\[
R(x) = x(80 - 3x) = 80x - 3x^2
\]
So, the revenue function is:
\[
R(x) = 80x - 3x^2
\]

### Domain of the Revenue Function:
The domain of the revenue function $ R(x) = 80x - 3x^2 $ is the same as the domain of the price-demand function, which is $ 1 \leq x \leq 20 $. Therefore, the domain is $ [1, 20] $.

### Part (B): Complete the Table
We are given a table with values for $ x $ and are asked to compute $ R(x) $ using the revenue function $ R(x) = 80x - 3x^2 $. Let's calculate the missing values.

#### Calculation:
For $ x = 4, 8, 12, 16, 20 $:
1. $ R(4) = 80(4) - 3(4^2) = 320 - 48 = 272 $
2. $ R(8) = 80(8) - 3(8^2) = 640 - 192 = 448 $
3. $ R(12) = 80(12) - 3(12^2) = 960 - 432 = 528 $
4. $ R(16) = 80(16) - 3(16^2) = 1280 - 768 = 512 $
5. $ R(20) = 80(20) - 3(20^2) = 1600 - 1200 = 400 $

So, the completed table looks like this:

| $ x $ (Millions) | $ R(x) $ (Millions $) |
|--------------------|--------------------------|
| 1                  | 77                       |
| 4                  | 272                      |
| 8                  | 448                      |
| 12                 | 528                      |
| 16                 | 512                      |
| 20                 | 400                      |

---

Would you like more details on any part of the solution or have further questions?

### Related Questions:
1. How do you derive the revenue function from a price-demand function?
2. What are the possible interpretations of the revenue function's domain?
3. How do the values of $ R(x) $ change as $ x $ increases, and why?
4. How would you find the maximum revenue from this revenue function?
5. What does it mean if the revenue function is quadratic?

### Tip:
The revenue function $ R(x) = 80x - 3x^2 $ is quadratic, and its graph is a parabola. The vertex of this parabola can help determine the maximum revenue.

Use the price-demand function p(x) = 80 - 3x (where 1 <= x <= 20) to answer parts (A) and (B): What is the revenue function, and what is the domain of the revenue function? Complete the revenue table for given values of x.

This problem involves finding the revenue function based on a given price-demand equation p(x) = 80 - 3x. We calculate the quadratic revenue function R(x) = 80x - 3x^2 and complete the revenue table for values of x ranging from 1 to 20.

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