The price $p$ (in dollars) and the quantity $x$ sold of a certain product satisfy the demand equation $x=-5p+200$. Answer parts (a) through $(\mathrm{g})$.
(a) Find a model that expresses the revenue $R$ as a function of $p$. (Remember, $R=xp$.)
$$R(p)=200p-5p^2$$
(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 $\{p\mid 0\le p\le 40\}$.
(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 $x=-5p+200$. The revenue $R$ is given by the product of the price and the quantity sold, i.e., $R=xp$.

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

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 $\{p\mid 0\le p\le 40\}$.

Step 4 ：The revenue function is a quadratic function of the form $R(p)=-5p^2+200p$. The maximum value of a quadratic function $f(x)=a{x}^2+bx+c$ 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 $\boxed{{\displaystyle 20}}$.