Problem 61 - Calculus Early Transcendentals: Pearson New International Edition

Question

A price function, \(p\), is defined by $$ p(x)=20+4 x-\frac{x^{2}}{3} $$ where \(x \geq 0\) is the number of units. (a) Find the total revenue function and the marginal revenue function. (b) On what interval is the total revenue increasing? (c) For what number \(x\) is the marginal revenue a maximum?

Step-by-Step Solution

Answer

(a) Revenue: \( R(x) = 20x + 4x^2 - \frac{x^3}{3} \); Marginal Revenue: \( R'(x) = 20 + 8x - x^2 \). (b) Increasing on \( 0 \leq x < 10 \). (c) Marginal revenue maximized at \( x = 4 \).

Step 1: Determine the Revenue Function

The revenue function, denoted by \( R(x) \), is found by multiplying the price function \( p(x) \) by the number of units \( x \). Therefore, we have:\[ R(x) = x \cdot p(x) = x \cdot \left(20 + 4x - \frac{x^2}{3}\right) \].This simplifies to:\[ R(x) = 20x + 4x^2 - \frac{x^3}{3} \]. This is the total revenue function.

Step 2: Find the Marginal Revenue Function

Marginal Revenue is the derivative of the total revenue function \( R(x) \). Let’s calculate:\[ R'(x) = \frac{d}{dx} \left( 20x + 4x^2 - \frac{x^3}{3} \right) \].Taking the derivative term by term, we get:\[ R'(x) = 20 + 8x - x^2 \].This is the marginal revenue function.

Step 3: Find the Interval Where Total Revenue is Increasing

The total revenue function is increasing where its derivative, i.e., the marginal revenue function, is greater than zero.Thus, solve the inequality:\[ 20 + 8x - x^2 > 0 \].This can be rewritten as:\[ -x^2 + 8x + 20 > 0 \].Solve the quadratic equation \( -x^2 + 8x + 20 = 0 \) using the quadratic formula:\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]Here, \( a = -1, b = 8, c = 20 \):\[ x = \frac{-8 \pm \sqrt{8^2 - 4(-1)(20)}}{2(-1)} = \frac{-8 \pm \sqrt{144}}{-2} \].This yields \( x = -2 \) and \( x = 10 \).However, since \( x \geq 0 \), we consider the interval \( 0 \leq x < 10 \). The total revenue is increasing on this interval.

Step 4: Determine When Marginal Revenue is Maximized

To find the \( x \) value that maximizes marginal revenue, we take the derivative of the marginal revenue function, \( R'(x) = 20 + 8x - x^2 \), and set it to zero. This gives:\[ R''(x) = \frac{d}{dx}(20 + 8x - x^2) = 8 - 2x \].Set \( R''(x) = 0 \) to find critical points:\[ 8 - 2x = 0 \Rightarrow x = 4 \].Check the second derivative at \( x = 4 \):\( R''(4) = 8 - 2(4) = 0 \), indicating a possible inflection point rather than a local maximum.However, since the marginal revenue derived is a downward opening parabola (\( R'(x) = -x^2 + 8x + 20 \)), it attains a maximum value at \( x = 4 \).

Key Concepts

Price Function

The price function, often denoted as \( p(x) \), relates the price of goods or services to the quantity or number of units sold, represented by \( x \). In this exercise, the price function is given by the equation:
\[ p(x) = 20 + 4x - \frac{x^2}{3} \]
This function suggests that as more units \( x \) are produced or sold, the price per unit changes according to this quadratic function. A price function can have linear or non-linear terms, and it may increase, decrease, or follow complex curves based on the specific coefficients in the equation. The understanding of a price function is essential because it determines how prices adjust when quantity changes, impacting potential revenue or profit.

Marginal Revenue

Marginal revenue represents the additional revenue earned from selling one extra unit of a product or service. It is essentially the derivative of the total revenue function, which shows how changes in quantity sold affect total revenue. In mathematical terms, a marginal revenue (MR) function can be expressed as:
\[ MR(x) = \frac{d}{dx} R(x) \]
Where \( R(x) \) is the total revenue function. For this problem, the marginal revenue function, derived from the total revenue function \( R(x) = 20x + 4x^2 - \frac{x^3}{3} \), is:
\[ R'(x) = 20 + 8x - x^2 \]
The marginal revenue is crucial for determining the profit-maximizing level of output, as it shows precisely where additional production increases or decreases total revenue. Understanding its behavior, such as intervals where it is positive, helps businesses make strategic pricing and production decisions.

Total Revenue

Total revenue is the total income a company generates from selling a certain amount of goods or services. It is calculated as the product of the number of units sold, \( x \), and the price function \( p(x) \). Hence, the total revenue function \( R(x) \) is given by:
\[ R(x) = x \cdot p(x) \]
In this exercise, the total revenue function simplifies to:
\[ R(x) = 20x + 4x^2 - \frac{x^3}{3} \]
This function helps businesses to understand how their revenues change with respect to the quantity of goods sold. Recognizing the behavior of the total revenue function, such as determining intervals where it is increasing, can guide decisions regarding optimal production levels.

Derivative

The concept of a derivative is central in calculus and holds significant importance in business mathematics. A derivative measures how a function changes as its input changes, essentially describing the function's rate of change at any given point. Mathematically, it is represented by:
\[ \frac{d}{dx} \]
In revenue functions, the derivative is used to determine marginal revenue, which is the rate at which revenue changes with an additional unit sold. For example, given the total revenue function \( R(x) \), its derivative \( R'(x) \) provides insights into how each additional sale affects overall revenue.
Calculating derivatives allows businesses to find critical points and optimizes their pricing and production strategies based on these rates of changes, maximizing profit and efficiency.