If the demand for a product is given by \( q=85-6 p^{2} \), find the price that will maximize revenue. (Round your answer to the nearest cent.) \[ \text { dollars } \]

To determine the price \( p \) that maximizes revenue for the product with the demand function \( q = 85 - 6p^2 \), follow these steps: 1. **Revenue Function:** Revenue \( R \) is given by: \[ R = p \times q = p(85 - 6p^2) = 85p - 6p^3 \] 2. **Maximize Revenue:** To find the price that maximizes revenue, take the derivative of \( R \) with respect to \( p \) and set it to zero: \[ \frac{dR}{dp} = 85 - 18p^2 = 0 \] Solving for \( p \): \[ 18p^2 = 85 \quad \Rightarrow \quad p^2 = \frac{85}{18} \quad \Rightarrow \quad p = \sqrt{\frac{85}{18}} \approx 2.173 \text{ dollars} \] 3. **Rounding to the Nearest Cent:** Rounding \( p \) to the nearest cent: \[ p \approx \$2.17 \] **Final Answer:** \[ \$2.17 \]