Question about the domain of the demand equation after expressing the Revenue as a function of x?

Question

Suppose the question is: the price $p$ and the quantity $x$ sold of a certain product obey the demand equation $x=-20p+500$ where $0<p \leq 25$. Once you find $R(x)$ when graphing $R(x)$ is the domain going to the the same as $0<p \leq 25$ where the point $(0,0)$ will remain open and $(25,0)$ closed? Or because $x$ is quantity demanded and you can technically have $0$ quantity demanded and $0$ revenue will the domain include the point $(0,0)$ closed?

Answer 1

For $x(p) = -20p + 500$, if the domain is $0 < p \leq 25$, then $0 > -20p \geq -500$ so that $500 > -20p + 500 \geq 0$. So the range of the demand $x$ is $0 \leq x < 500$.

As for the revenue function $R$, we need to know how it's actually defined. Presumably, revenue is price times quantity demanded: $$ R(p) = p \cdot x(p) = p \cdot (-20p + 500) $$ where the domain in terms of $p$ is $0 < p \leq 25$. If we insist on expressing $R$ as a function of $x$, then we can invert the given demand function: $$ x(p) = -20p + 500 \iff p(x) = \frac{x - 500}{-20} = 25 - \frac{1}{20}x $$ and substitute: $$ R(x) = p(x) \cdot x = \left(25 - \frac{1}{20}x \right) \cdot x $$ where the domain in terms of $x$ is $0 \leq x < 500$.