The problem involves finding the price $p$ at which the revenue is maximized. We are given the demand function for a product:

$q = 10,000 e^{-0.02p}$ where $q$ represents the demand and $p$ is the price in dollars.

Step-by-step approach to solve for maximum revenue:

1. Revenue Function: The revenue $R(p)$ is the product of the price $p$ and the quantity demanded $q$: $R(p) = p \cdot q = p \cdot 10,000 e^{-0.02p}$ So, the revenue function becomes: $R(p) = 10,000 p e^{-0.02p}$

2. **Max