The price $p$ (in dollars) and the quantity $x$ sold of a certain product satisfy the demand equation $x=-5 p+200$. Answer parts (a) through $(\mathrm{g})$. (a) Find a model that expresses the revenue $R$ as a function of $p$. (Remember, $R=x p$.) \[ R(p)=200 p-5 p^{2} \] (Simplify your answer. Use integers or decimals for any numbers in the expression.) (b) What is the domain of $R$ ? Assume that $R$ is nonnegative. A. The domain is $\{p \mid 0 \leq p \leq 40\}$. (Simplify your answers. Type integers or decimals.) B. The domain is the set of all real numbers. (c) What price $p$ maximizes revenue?

Step 1 ：The price $p$ (in dollars) and the quantity $x$ sold of a certain product satisfy the demand equation $x=-5 p+200$. The revenue $R$ is given by the product of the price and the quantity sold, i.e., $R=x p$.

Step 2 ：Substituting the demand equation into the revenue equation, we get a model that expresses the revenue $R$ as a function of $p$: $R(p)=200 p-5 p^{2}$.

Step 3 ：The domain of $R$ is the set of all prices $p$ for which the revenue $R$ is nonnegative. Since the quantity $x$ sold cannot be negative, the price $p$ cannot be less than 0 or greater than 40. Therefore, the domain of $R$ is $\{p \mid 0 \leq p \leq 40\}$.

Step 4 ：The revenue function is a quadratic function of the form $R(p) = -5p^2 + 200p$. The maximum value of a quadratic function $f(x) = ax^2 + bx + c$ where $a < 0$ is obtained at $x = -b/2a$. In this case, $a = -5$ and $b = 200$.

Step 5 ：Substituting $a = -5$ and $b = 200$ into the formula $x = -b/2a$, we find that the price $p$ that maximizes revenue is $p = 20.0$.

Step 6 ：Final Answer: The price $p$ that maximizes revenue is \(\boxed{20}\).