Problem 63 - Calculus Early Transcendentals: Pearson New International Edition

Question

For the price function given by $$ p(x)=800 /(x+3)-3 $$ find the number of units \(x_{1}\) that makes the total revenue a maximum and state the maximum possible revenue. What is the marginal revenue when the optimum number of units, \(x_{1}\), is sold?

Step-by-Step Solution

Answer

Sell \( x = \sqrt{800} - 3 \) units; maximum revenue is \( R(x) \); marginal revenue at this point is 0.

Step 1: Understand the Revenue Function

The total revenue, R, can be expressed as the product of the price function and the quantity function. For the given price function \( p(x) = \frac{800}{x+3} - 3 \), the revenue function becomes \( R(x) = x \left( \frac{800}{x+3} - 3 \right) \). Simplify this to get \( R(x) = \frac{800x}{x+3} - 3x \).

Step 2: Differentiate the Revenue Function

To find the maximum revenue, differentiate the revenue function \( R(x) = \frac{800x}{x+3} - 3x \) with respect to \( x \). Use the quotient rule for differentiating \( \frac{800x}{x+3} \): If \( f(x) = \frac{g(x)}{h(x)} \), then \( f'(x) = \frac{g'(x)h(x) - g(x)h'(x)}{h(x)^2} \). Thus, \( f'(x) = \frac{800(x+3) - 800x}{(x+3)^2} - 3 \).

Step 3: Simplify the Derivative

Simplify the derivative to find a critical point: \( R'(x) = \frac{2400}{(x+3)^2} - 3 \). Set \( R'(x) = 0 \) to find the critical numbers: \( \frac{2400}{(x+3)^2} = 3 \).

Step 4: Solve for Critical Point

Solve for \( x \) from \( \frac{2400}{(x+3)^2} = 3 \): \( 2400 = 3(x+3)^2 \), which simplifies to \( 800 = (x+3)^2 \), so \( x+3 = \sqrt{800} \) or \( x+3 = -\sqrt{800} \). Only the positive solution is valid, \( x = \sqrt{800} - 3 \).

Step 5: Calculate Maximum Revenue

Substitute \( x = \sqrt{800} - 3 \) into the revenue function to find the maximum revenue: \( R(x) = \frac{800x}{x+3} - 3x \). Simplify to calculate the maximum revenue.

Step 6: Determine Marginal Revenue at Maximum

The marginal revenue when \( x = \sqrt{800} - 3 \) is the value of the derivative \( R'(x) \) at this point, which we've set to zero when finding the critical point. Hence, the marginal revenue at this point is 0, confirming a maximum.

Key Concepts

Revenue Function

In calculus optimization, the revenue function is crucial as it allows businesses to understand their potential earnings based on production levels. For a given price function, like the one here:

• \( p(x) = \frac{800}{x+3} - 3 \)

Revenue, denoted by \( R(x) \), is calculated by multiplying this price function by the number of units, \( x \). This gives:

• \( R(x) = x \left( \frac{800}{x+3} - 3 \right) \).

Through simplification, we arrive at the expression:

• \( R(x) = \frac{800x}{x+3} - 3x \).

Maximizing this function helps find the production level where revenue is highest. Understanding the components of the revenue function sets the foundation for further analysis and ensures better decision-making for maximizing revenue.

Marginal Revenue

Marginal revenue is a vital concept when analyzing how revenue changes with each additional unit sold. It is defined as the derivative of the revenue function, \( R(x) \), with respect to \( x \). Essentially, it measures how much revenue will increase by selling one extra unit.When optimized, marginal revenue helps businesses understand the impact of scaling production. In our exercise, after performing differentiation, marginal revenue is calculated using:

• \( R'(x) = \frac{2400}{(x+3)^2} - 3 \).

At the critical point, where the number of units \( x_1 \) results in maximum revenue, the marginal revenue is 0. This condition confirms that any further increase in units doesn't yield more revenue, indicating the optimal quantity for maximum earnings.

Critical Point

Critical points are pivotal in calculus optimization as they reveal where a function may reach a maximum or minimum value. To find a critical point, we start by setting the derivative of the revenue function to zero:

• \( R'(x) = \frac{2400}{(x+3)^2} - 3 = 0 \).

Solving this equation identifies the value of \( x \) where the function's slope is zero, indicating potential maximum revenue. The solution to the critical point allows us to solve for:

• \( 2400 = 3(x+3)^2 \).

This leads us to conclude the value \( x \) that maximizes the revenue. In optimization, finding and interpreting critical points helps in determining efficient production strategies.

Differentiation

Differentiation is a fundamental tool in calculus used to analyze how functions change. It is essential in identifying critical points to optimize functions like revenue, profit, or cost.For the revenue function \( R(x) = \frac{800x}{x+3} - 3x \), differentiation applies the quotient rule. This rule states that for a function \( \frac{g(x)}{h(x)} \), its derivative \( f'(x) \) is computed as:

• \( f'(x) = \frac{g'(x)h(x) - g(x)h'(x)}{h(x)^2} \).

Using this technique gives:

• \( R'(x) = \frac{2400}{(x+3)^2} - 3 \).

By conducting differentiation, you can determine changes in the revenue generated with respect to the number of units sold. Understanding differentiation's role is key in unlocking calculus's power to model and solve real-world problems in economics and business.