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? \[ p=\$ 20 \] (Simplify your answer. Type an integer or a decimal.) (d) What is the maximum revenue? \[ \mathrm{R}=\$ \] (Simplify your answer. hype an integer or a decimal.)

Step 1 ：The revenue R is given by the product of the price p and the quantity x sold, which is given by the equation \(R = p(-5p + 200) = 200p - 5p^2\).

Step 2 ：We want to maximize this expression by completing the square. We can factor out a -5 to get \(-5(p^2 - 40p)\).

Step 3 ：To complete the square, we add \((40/2)^2 = 400\) inside the parenthesis and subtract \(-5 \cdot 400 = -2000\) outside. We are left with the expression \(-5(p^2 - 40p + 400) + 2000 = -5(p - 20)^2 + 2000\).

Step 4 ：Note that the \(-5(p - 20)^2\) term will always be nonpositive since the perfect square is always nonnegative. Thus, the revenue is maximized when \(-5(p - 20)^2\) equals 0, which is when \(p = 20\).

Step 5 ：So, the price that maximizes revenue is \(\boxed{20}\) dollars.

Step 6 ：To find the maximum revenue, we substitute \(p = 20\) into the revenue function \(R(p) = 200p - 5p^2\), we get \(R(20) = 200 \cdot 20 - 5 \cdot 20^2 = 4000 - 2000 = 2000\).

Step 7 ：So, the maximum revenue is \(\boxed{2000}\) dollars.