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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

La función demanda de un determinado producto, donde el precio es en dólares para una demanda unidades de dicho producto, es: q = 10,000e^{-0.02p}. Calcule el precio para el cual se obtiene el ingreso máximo por la venta del producto.

This problem explores how to calculate the price that maximizes revenue using a demand function q = 10,000e^{-0.02p}. It covers key calculus concepts such as differentiation and revenue maximization. Suitable for undergraduate calculus students.

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