Suppose that the manufacturer of a gas clothes dryer has found that, when the unit price is \( p \) dollars, the \( \left(\right. \) evenue \( R \) (in dollars) is \( R(p)=-7 p^{2}+14,000 p \). What unit price should be established for the dryer to maximize revenue? What is the maximum revenue? The unit price that should be established to maximize revenue is s. (Simplify your answer.) The maximum revenue is \( \$ \square \) (Simplify your answer.)

To maximize revenue, we first need to find the vertex of the quadratic revenue function \( R(p) = -7p^2 + 14,000p \). The unit price \( p \) that maximizes revenue can be found using the formula \( p = -\frac{b}{2a} \). Here, \( a = -7 \) and \( b = 14,000 \). Calculating this gives: \[ p = -\frac{14,000}{2 \times -7} = \frac{14,000}{14} = 1,000 \] So, the unit price that should be established to maximize revenue is **$1,000**. Next, we can find the maximum revenue by substituting \( p = 1,000 \) back into the revenue function: \[ R(1,000) = -7(1,000)^2 + 14,000(1,000) = -7,000,000 + 14,000,000 = 7,000,000 \] Thus, the maximum revenue is **$7,000,000**. In summary, the unit price should be established at **$1,000**, and the maximum revenue is **$7,000,000**.