Question t1 of 18 Ins questen: 1 porites) posstre Submil test Suppose that the manufacturer of a gas clothes dryer has found that, when the unit price is \( p \) dollars, the revenue \( R \) (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 \( \$ \) \( \square \) (Simplify your answer.) The maximum revenue is \( \$ \) \( \square \) (Simplify your answer.)

To find the unit price that maximizes revenue, you can use the revenue function \( R(p) = -7p^2 + 14,000p \). To find the price, first, identify the vertex of the parabola represented by this equation. The formula for the vertex \( p \) in a quadratic equation \( ax^2 + bx + c \) is given by \( p = -\frac{b}{2a} \). Here, \( a = -7 \) and \( b = 14,000 \). Calculating that gives: \[ p = -\frac{14000}{2 \times -7} = \frac{14000}{14} = 1000 \] So, the unit price that should be established to maximize revenue is \( \$1000 \). Next, we find the maximum revenue by substituting \( p = 1000 \) back into the revenue function: \[ R(1000) = -7(1000)^2 + 14,000(1000) \] Calculating this yields: \[ R(1000) = -7(1,000,000) + 14,000,000 = -7,000,000 + 14,000,000 = 7,000,000 \] Thus, the maximum revenue is \( \$7,000,000 \). To summarize: The unit price that should be established to maximize revenue is \( \$1000 \). The maximum revenue is \( \$7,000,000 \).