Question

Revenue 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 $$ R(p)=-4 p^{2}+4000 p $$ (a) At what prices \(p\) is revenue zero? (b) For what range of prices will revenue exceed \(\$ 800,000 ?\)

Short answer

Revenue is zero at \(p = 0\) and \(p = 1000\). Revenue exceeds \(800,000\) when \(276.39 < p < 723.61\).

Step-by-step solution

Setting Revenue to Zero for Part (a)

Given the revenue function \( R(p) = -4p^2 + 4000p \). To find when the revenue is zero, set \( R(p) = 0 \): \( -4p^2 + 4000p = 0 \).

Solving the Equation for Part (a)

Factor the equation \( -4p(p - 1000) = 0 \). This gives two solutions: \( p = 0 \) and \( p = 1000 \).

Setting up for Part (b)

Given the revenue function \( R(p) = -4p^2 + 4000p \), we need to find when \( R(p) > 800,000 \). Set up the inequality: \( -4p^2 + 4000p > 800,000 \).

Solving the Inequality for Part (b)

Rearrange the inequality: \(-4p^2 + 4000p - 800,000 > 0 \). Now, solve the quadratic equation \( -4p^2 + 4000p - 800,000 = 0 \) using the quadratic formula: \( p = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) where \( a = -4 \), \( b = 4000 \), and \( c = -800,000 \).

Applying the Quadratic Formula

Compute the discriminant first: \( b^2 - 4ac = 4000^2 - 4(-4)(-800,000) = 16,000,000 - 12,800,000 = 3,200,000 \). The solutions are \( p = \frac{-4000 \pm \sqrt{3,200,000}}{2(-4)} \). This simplifies to \( p = \frac{4000 \pm 1788.85}{8} \).

Finding the Price Range for Part (b)

Finding the exact values: \( p = \frac{4000 + 1788.85}{8} \approx 723.61 \) and \( p = \frac{4000 - 1788.85}{8} \approx 276.39 \). Thus, the revenue exceeds \( 800,000 \) in the range \(276.39 < p < 723.61 \).

Summary

The prices \( p \) where revenue is zero are \( 0 \ \text{andsyyyy}\ 1000 \). For the revenue to exceed \( 800,000 \), the price \( p \) must be in the range \( 276.39 < p < 723.61 \).

Key concepts

Quadratic Equations

A quadratic equation is a second-degree polynomial in the form \( ax^2 + bx + c = 0 \). In this formula, \(a\), \(b\), and \(c\) are constants with \(a e 0\). The solutions to quadratic equations can be found using various methods such as factoring, completing the square, or the quadratic formula.

In our problem, the revenue function \( R(p) = -4p^2 + 4000p \) is a quadratic equation, where \(p\) represents the unit price. to find when the revenue is zero, we set \( R(p) = 0 \), giving us the equation \( -4p^2 + 4000p = 0 \).

- Factoring this equation, we have \( -4p(p - 1000) = 0 \), leading to the solutions \( p = 0 \) and \( p = 1000 \). These are the prices at which the revenue is zero.

Revenue Maximization

Revenue maximization involves finding the price \( p \) that yields the highest revenue. The given revenue function \( R(p) = -4p^2 + 4000p \) represents a downward-opening parabola because the coefficient of \( p^2 \) is negative. This means the vertex of the parabola will give us the maximum revenue.

To find the vertex of a quadratic function \( ax^2 + bx + c \), we use \( p = -\frac{b}{2a} \). For \( R(p) = -4p^2 + 4000p \), we have \( a = -4 \) and \( b = 4000 \). Thus, the price that maximizes revenue is \( p = -\frac{4000}{2(-4)} = 500 \).

Therefore, the unit price of \$500 maximizes the revenue.

Solving Inequalities

Solving inequalities involves finding the range of values that satisfy an inequality. In our exercise, we needed to find the range of prices \( p \) for which the revenue exceeds \$800,000. This means solving the inequality \( -4p^2 + 4000p > 800,000 \).

First, we rearrange it as a quadratic equation: \( -4p^2 + 4000p - 800,000 > 0 \). We solve the corresponding equality \( -4p^2 + 4000p - 800,000 = 0 \) using the quadratic formula \( p = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = -4 \), \( b = 4000 \), and \( c = -800,000 \).

The discriminant \( \Delta \) is calculated as: \( \Delta = 4000^2 - 4(-4)(-800,000) = 3,200,000 \). The solutions are \( p = \frac{4000 \pm \sqrt{3,200,000}}{8} \approx 723.61 \text{ and } 276.39 \).
Thus, the price range for which revenue exceeds \$800,000 is \( 276.39 < p < 723.61 \).