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 find the price that maximizes revenue, we first express revenue \( R \) as a function of price \( p \). Revenue is given by the formula \( R = p \times q \). Substituting \( q \) from the demand equation, we have: \[ R = p(85 - 6p^{2}) = 85p - 6p^{3} \] Next, we need to take the derivative of \( R \) with respect to \( p \) and set it equal to zero to find the critical points: \[ \frac{dR}{dp} = 85 - 18p^{2} \] Setting the derivative equal to zero: \[ 85 - 18p^{2} = 0 \] Solving for \( p \): \[ 18p^{2} = 85 \] \[ p^{2} = \frac{85}{18} \] \[ p = \sqrt{\frac{85}{18}} \] Calculating this gives: \[ p \approx 2.08 \] Now, rounding to the nearest cent, the price that will maximize revenue is: **$2.08**