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do the same for this but with x = —	8p +800

Based on your request, let's adapt the problem to the new demand equation $ x = -8p + 800 $.

### Given Demand Equation:
\[ x = -8p + 800 \]

### Part (a): Model for Revenue $ R $
Revenue $ R $ is given by $ R = xp $. Substitute $ x $ from the demand equation:
\[ R = (-8p + 800)p \]
\[ R = -8p^2 + 800p \]

Therefore, the model expressing revenue $ R $ as a function of $ p $ is:
\[ R = -8p^2 + 800p \]

### Part (b): Domain of $ R $
Assume $ R $ is nonnegative. Since $ x $ represents the quantity sold, $ x \geq 0 $:
\[ -8p + 800 \geq 0 \]
\[ 800 \geq 8p \]
\[ 100 \geq p \]
Also, since $ p \geq 0 $:
\[ 0 \leq p \leq 100 \]

So, the domain of $ R $ is $ 0 \leq p \leq 100 $.

### Part (c): Price $ p $ that Maximizes Revenue
The revenue function $ R $ is a quadratic function:
\[ R = -8p^2 + 800p \]
To find the maximum revenue, we find the vertex of the parabola. The vertex formula for $ y = ax^2 + bx + c $ is given by $ p = -\frac{b}{2a} $:
\[ a = -8, \, b = 800 \]
\[ p = -\frac{800}{2(-8)} \]
\[ p = \frac{800}{16} \]
\[ p = 50 \]

Therefore, the price $ p $ that maximizes revenue is $50.

### Part (d): Maximum Revenue
Substitute $ p = 50 $ into the revenue function:
\[ R = -8(50)^2 + 800(50) \]
\[ R = -8(2500) + 40000 \]
\[ R = -20000 + 40000 \]
\[ R = 20000 \]

Therefore, the maximum revenue is $20,000.

### Part (e): Number of Units Sold at Price $ p = 50 $
Substitute $ p = 50 $ into the demand equation:
\[ x = -8(50) + 800 \]
\[ x = -400 + 800 \]
\[ x = 400 \]

Therefore, 400 units are sold at the price of $50.

### Part (f): Graph $ R $
To graph $ R $, plot the intercepts and the vertex:
- Intercepts: $ (0, 0) $ and $ (100, 0) $
- Vertex: $ (50, 20000) $

### Part (g): Price for at least $13,650 in Revenue
Find the intersection of $ R $ and $ 13650 $:
\[ 13650 = -8p^2 + 800p \]
\[ -8p^2 + 800p - 13650 = 0 \]

Solve the quadratic equation:
\[ 8p^2 - 800p + 13650 = 0 \]
\[ p^2 - 100p + 1706.25 = 0 \]

Use the quadratic formula $ p = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} $:
\[ a = 1, \, b = -100, \, c = 1706.25 \]
\[ p = \frac{100 \pm \sqrt{10000 - 6825}}{2} \]
\[ p = \frac{100 \pm \sqrt{3175}}{2} \]
\[ p \approx 35 \text{ or } p \approx 65 \]

Therefore, to earn at least $13,650 in revenue, the company should charge a price between $35 and $65.

Would you like more details on any specific part of this solution?

Here are 5 questions you might ask:
1. Can you explain how to derive the quadratic formula?
2. What are the steps to graph a quadratic function?
3. How do you determine the intercepts of a quadratic function?
4. Can you show more examples of solving quadratic equations?
5. How does changing the coefficients affect the shape of the parabola?

**Tip:** When solving quadratic equations, always check for possible simplifications before applying the quadratic formula.

Explore how to maximize revenue using the demand equation x = -8p + 800. Learn to derive revenue function, find optimal price for maximum revenue, and determine units sold. Suitable for advanced high school students.

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