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 \( R(p)=-4 p^{2}+12,000 p \). (a) At what prices \( p \) is revenue zero? (b) For what range of prices will revenue exceed \( \$ 800,000 \) ? (a) At what prices \( p \) is revenue zero? The revenue equals zero when p is \( \$ 0,3000 \). (Use a comma to separate answers, but do not use commas in any individual numbers.) (b) For what range of prices will revenue exceed \( \$ 800,000 \) ? (Type your answer in interval notation. Round to the nearest cent as needed.)

To solve the problem, we will follow the steps outlined: ### Part (a): Finding Prices \( p \) Where Revenue is Zero The revenue function is given by: \[ R(p) = -4p^2 + 12000p \] To find the prices \( p \) where revenue is zero, we set \( R(p) \) equal to zero: \[ -4p^2 + 12000p = 0 \] We can factor out \( p \): \[ p(-4p + 12000) = 0 \] This gives us two solutions: 1. \( p = 0 \) 2. \( -4p + 12000 = 0 \) Now, we solve for \( p \) in the second equation: \[ -4p + 12000 = 0 \implies 4p = 12000 \implies p = \frac{12000}{4} = 3000 \] Thus, the prices \( p \) where revenue is zero are: \[ p = 0, 3000 \] ### Part (b): Finding Range of Prices Where Revenue Exceeds \( 800,000 \) We need to find the range of prices \( p \) such that: \[ R(p) > 800000 \] Setting up the inequality: \[ -4p^2 + 12000p > 800000 \] Rearranging gives: \[ -4p^2 + 12000p - 800000 > 0 \] To simplify, we can divide the entire inequality by -4 (remember to reverse the inequality sign): \[ p^2 - 3000p + 200000 < 0 \] Next, we will find the roots of the quadratic equation \( p^2 - 3000p + 200000 = 0 \) using the quadratic formula: \[ p = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] Here, \( a = 1 \), \( b = -3000 \), and \( c = 200000 \): \[ p = \frac{3000 \pm \sqrt{(-3000)^2 - 4 \cdot 1 \cdot 200000}}{2 \cdot 1} \] Calculating the discriminant: \[ (-3000)^2 - 4 \cdot 1 \cdot 200000 = 9000000 - 800000 = 820000 \] Now substituting back into the quadratic formula: \[ p = \frac{3000 \pm \sqrt{820000}}{2} \] Calculating \( \sqrt{820000} \): \[ \sqrt{820000} \approx 905.88 \] Now substituting this value back: \[ p = \frac{3000 \pm 905.88}{2} \] Calculating the two roots: 1. \( p_1 = \frac{3000 + 905.88}{2} \approx \frac{3905.88}{2} \approx 1952.94 \) 2. \( p_2 = \frac{3000 - 905.88}{2} \approx \frac{2094.12}{2} \approx 1047.06 \) Thus, the roots are approximately \( p_1 \approx 1952.94 \) and \( p_2 \approx 1047.06 \). The quadratic \( p^2 - 3000p + 200000 < 0 \) is satisfied between the roots: \[ (1047.06, 1952.94) \] ### Final Answers (a) The prices \( p \) where revenue is zero are: \[ 0, 3000 \] (b) The range of prices where revenue exceeds \( 800,000 \) is: \[ (1047.06, 1952.94) \]