Question

Maximizing Revenue A lawn mower manufacturer has found that the revenue, in dollars, from sales of zero-turn mowers is a function of the unit price \(p,\) in dollars, that it charges. If the revenue \(R\) is $$ R(p)=-\frac{1}{2} p^{2}+2900 p $$ what unit price \(p\) should be charged to maximize revenue? What is the maximum revenue?

Short answer

The unit price that maximizes revenue is \(2900, and the maximum revenue is \)4205000.

Step-by-step solution

- Identify the given function

The revenue function is given as follows: \[ R(p) = -\frac{1}{2} p^2 + 2900p \]

- Find the derivative of the revenue function

To maximize the revenue, find the critical points by first finding the first derivative of the revenue function with respect to the unit price, \( p \). \[ R'(p) = \frac{d}{dp} \left( -\frac{1}{2} p^2 + 2900p \right) R'(p) = -p + 2900 \]

- Find the critical points

Set the first derivative equal to zero and solve for \( p \) to find the critical points. \[-p + 2900 = 0 p = 2900 \]

- Determine maximum revenue

To determine if this critical point is a maximum, we can use the second derivative test. Find the second derivative and evaluate it at \( p = 2900 \). \[ R''(p) = \frac{d}{dp} (-p + 2900) R''(p) = -1 \]Since \( R''(p) = -1 \) is negative, the function is concave down at \( p = 2900 \), indicating a maximum.

- Calculate the maximum revenue

Substitute \( p = 2900 \) back into the original revenue function to find the maximum revenue: \[ R(2900) = -\frac{1}{2} (2900)^2 + 2900(2900) R(2900) = -\frac{1}{2} (8410000) + 8410000 R(2900) = -4205000 + 8410000 R(2900) = 4205000 \]

Key concepts

Revenue Function

The revenue function is an equation that helps businesses understand their income from sales based on the price they charge for a product. In this exercise, the revenue function is a quadratic equation, \[ R(p) = -\frac{1}{2} p^2 + 2900p \]. This means revenue, denoted as \( R(p) \), depends on the unit price \( p \). Understanding the revenue function is the first step in maximizing your revenue. You plug different values of \( p \) into this equation to see how changing the price affects your earnings.

Calculus Optimization

Calculus optimization involves using calculus tools to find the optimal point where a function reaches its maximum or minimum value. In this context, it means using derivatives to find the price \( p \) that maximizes the revenue function. The first step is to find the first derivative of the revenue function. The first derivative \( R'(p) \) helps us locate the critical points where the function could have a maximum or minimum value. Once you have \( R'(p) = -p + 2900 \), you set it to zero. Solving \( -p + 2900 = 0 \) gives us the critical point \( p = 2900 \).

Critical Points

Critical points are values in the domain of the function where the first derivative is zero or undefined. These points are potential candidates for local maxima or minima. To find the critical points in our revenue function, we first found \[ R'(p) = -p + 2900 \]. Setting this derivative equal to zero gives us \( p = 2900 \). This means that when the price is 2900 dollars, we might have a maximum or minimum revenue. Identifying critical points is crucial because these are where the function's behavior changes.

Second Derivative Test

The second derivative test helps us determine whether a critical point is a maximum, a minimum, or a point of inflection. To use it, we take the second derivative of the revenue function. For our revenue function, the second derivative is \[ R''(p) = -1 \]. Since this is a constant negative value, the revenue function is concave down at \( p = 2900 \), indicating that \( p = 2900 \) is indeed a maximum. Once you confirm the maximum, you can substitute this value back into the original function to find the maximum revenue. For \( p = 2900 \), the maximum revenue is 4,205,000 dollars.