The question asks for solving various parts related to the demand equation, which is given as:

$x = -5p + 200$

Where:

• $p$ is the price in dollars.
• $x$ is the quantity sold of a certain product.

Let's analyze each part of the problem:

(a) Find a model that expresses the revenue $R$ as a function of $p$.

We know that revenue $R$ is given by the formula:

$R = p \times x$

Substitute $x = -5p + 200$ into this equation:

$R(p) = p \times (-5p + 200)$

Simplify this:

$R(p) = -5p^2 + 200p$

This is the revenue function in terms of $p$.

(b) What is the domain of $R$, assuming that $R \geq 0$?

To find the domain, we ensure the revenue is nonnegative, i.e., $R(p) \geq 0$.

We already have:

$R(p) = -5p^2 + 200p$

This is a quadratic equation, and the revenue will be non-negative between the roots of the equation:

$-5p^2 + 200p = 0$

Factor this equation:

$p(-5p + 200) = 0$

So, $p = 0$ or $p = 40$.

Therefore, the domain is $0 \leq p \leq 40$.

(c) What price $p$ maximizes revenue?

Since the revenue function is quadratic and opens downward (as the coefficient of $p^2$ is negative), the price that maximizes revenue is at the vertex of the parabola. The formula for the vertex is:

$p = -\frac{b}{2a}$

For $R(p) = -5p^2 + 200p$, $a = -5$ and $b = 200$. Substituting into the vertex formula:

$p = -\frac{200}{2(-5)} = 20$

Thus, the price that maximizes revenue is $p = 20$.

(d) What is the maximum revenue?

To find the maximum revenue, substitute $p = 20$ into the revenue function:

$R(20) = -5(20)^2 + 200(20)$ $R(20) = -5(400) + 4000 = -2000 + 4000 = 2000$

The maximum revenue is $R = 2000$.

(e) How many units are sold at this price?

We use the demand equation $x = -5p + 200$ and substitute $p = 20$:

$x = -5(20) + 200 = -100 + 200 = 100$

So, 100 units are sold at this price.

(f) Graph $R$ and choose the correct graph.

The revenue function $R(p) = -5p^2 + 200p$ is a downward-opening parabola. Based on the calculated maximum at $p = 20$ and $R = 2000$, the correct graph is D.

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

Here are five related questions:

1. How do you calculate the roots of a quadratic function?
2. What happens to the revenue if the price increases beyond the maximum?
3. How can you find the vertex of a parabola algebraically?
4. How do we interpret the negative coefficient of $p^2$ in the revenue function?
5. What other factors can affect the domain of a revenue function?

Tip: When solving quadratic functions, always check if the parabola opens upwards or downwards, as this determines whether the vertex is a maximum or minimum.