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 \( R(p)=-5 p^{2}+10,000 \mathrm{p} \). 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.)

We are given the revenue function R(p) = –5p² + 10,000p. Since this is a quadratic function that opens downward (because the coefficient of p² is negative), its maximum revenue occurs at the vertex of the parabola. The p-coordinate of the vertex is given by the formula:   p = –(b)/(2a) Here, a = –5 and b = 10,000, so:   p = –(10,000) / (2 × –5) = 10,000/10 = 1,000 Thus, to maximize revenue, the unit price should be set at $1,000. To find the maximum revenue, substitute p = 1,000 into the revenue function:   R(1,000) = –5(1,000)² + 10,000(1,000)       = –5(1,000,000) + 10,000,000       = –5,000,000 + 10,000,000       = 5,000,000 The maximum revenue is $5,000,000. So, the unit price that should be established is $1,000, and the maximum revenue is $5,000,000.