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) Total revenue: \( R(x) = 20x + 4x^2 - \frac{x^3}{3} \), marginal revenue: \( R'(x) = 20 + 8x - x^2 \). (b) Increasing for \( x \geq 0 \). (c) Maximum at \( x = 4 \).

Step 1: Define the Total Revenue Function

The total revenue function, denoted as \( R(x) \), can be found by multiplying the price function \( p(x) \) by the quantity \( x \). Thus, \( R(x) = x \cdot p(x) = x(20+4x-\frac{x^2}{3}) = 20x + 4x^2 - \frac{x^3}{3} \).

Step 2: Find the Marginal Revenue Function

The marginal revenue function is the derivative of the total revenue function \( R(x) \). We differentiate \( R(x) = 20x + 4x^2 - \frac{x^3}{3} \) with respect to \( x \):\[ R'(x) = \frac{d}{dx}(20x) + \frac{d}{dx}(4x^2) - \frac{d}{dx}\left(\frac{x^3}{3}\right) = 20 + 8x - x^2. \]

Step 3: Determine Where Total Revenue is Increasing

The total revenue \( R(x) \) is increasing where its derivative \( R'(x) = 20 + 8x - x^2 \) is greater than zero. We solve the inequality \( 20 + 8x - x^2 > 0 \). First, find the roots of \( 20 + 8x - x^2 = 0 \) using the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \).Here, \( a = -1, b = 8, \text{ and } c = 20 \):\[ x = \frac{-8 \pm \sqrt{64 - 80}}{-2} = \frac{-8 \pm \sqrt{-16}}{-2}. \]Since the discriminant is negative, the equation has no real roots. Therefore, \( R(x) \) does not change sign, and because \( R'(x) \) starts positively, \( R(x) \) is always increasing for \( x \geq 0 \).

Step 4: Find x for Maximum Marginal Revenue

To find the maximum of the marginal revenue function \( R'(x) = 20 + 8x - x^2 \), we take its derivative to find critical points:\[ R''(x) = \frac{d}{dx}(20 + 8x - x^2) = 8 - 2x. \]Set \( R''(x) = 0 \) to find critical points:\[ 8 - 2x = 0 \quad \Rightarrow \quad x = 4. \]Check the second derivative to confirm a maximum: \( R''(x) < 0 \) when \( x > 4 \), indicating \( x = 4 \) is where the marginal revenue reaches its maximum.

Key Concepts

Total Revenue Function

The total revenue function is a crucial concept in understanding how businesses generate income from sales. It involves the price function, which in this case is given by the expression \( p(x) = 20 + 4x - \frac{x^2}{3} \). To find the total revenue function, denoted as \( R(x) \), we need to multiply this price function by the number of units sold, \( x \). This can be expressed as:

• \( R(x) = x \cdot p(x) \)
• Substituting the price function, we get: \( R(x) = x(20 + 4x - \frac{x^2}{3}) \)
• Simplifying gives: \( R(x) = 20x + 4x^2 - \frac{x^3}{3} \)

This function represents the total revenue, encapsulating how revenue changes with different quantities sold.

Marginal Revenue

Marginal revenue is the additional revenue generated from selling one more unit of a product. It's a critical factor in determining optimal pricing and production strategies. To calculate the marginal revenue, we need the derivative of the total revenue function, \( R(x) \). The derivative, \( R'(x) \), indicates the rate of change of total revenue:

• \( R'(x) = \frac{d}{dx}(20x + 4x^2 - \frac{x^3}{3}) \)
• This results in: \( R'(x) = 20 + 8x - x^2 \)

Understanding marginal revenue helps businesses decide whether an additional unit sold will increase overall profits.

Increasing Interval

The increasing interval of a revenue function indicates the range of units sold where total revenue continues to rise. To identify this, we examine where the derivative of the total revenue function is greater than zero. For \( R(x) \), this means checking when:

• \( R'(x) = 20 + 8x - x^2 > 0 \)

However, in this exercise, solving \( 20 + 8x - x^2 = 0 \) showed no real roots. This implies the derivative does not switch signs, always starting positive. Thus, the total revenue \( R(x) \) consistently increases for all non-negative \( x \). This behavior suggests a robust and steady revenue increase as more units are sold.

Derivative

The concept of a derivative is integral to both mathematical analysis and business strategies. It provides information about how a function changes at any given point. For the total revenue function \( R(x) = 20x + 4x^2 - \frac{x^3}{3} \), the derivative \( R'(x) = 20 + 8x - x^2 \) was used to derive insights into marginal revenue and increased intervals.
Taking the derivative is essentially finding the slope of the tangent to the curve at any point \( x \) and informs about the function's increase or decrease behavior. Moreover, finding the derivative's critical points helps determine maximum or minimum values of marginal cost or revenue, essential for strategic decision-making in business.