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Chapter 5: Problem 64

Suppose that a company's marginal revenue from the manufacture and sale of eggbeaters is $$\frac{d r}{d x}=2-2 /(x+1)^{2}.$$ Where \(r\) is measured in thousands of dollars and \(x\) in thousands of units. How much money should the company expect from a production run of \(x=3\) thousand eggbeaters? To find out, integrate the marginal revenue from \(\bar{x}=0\) to \(x=3.\)

Short Answer

Expert verified
The company should expect 4.5 thousand dollars from producing 3 thousand eggbeaters.

Step by step solution

01

Identify the marginal revenue function

The marginal revenue function given in the problem is \( \frac{dr}{dx} = 2 - \frac{2}{(x+1)^2} \). This function represents the rate of change of revenue with respect to the number of eggbeaters produced and sold.
02

Set up the definite integral

To find the total revenue from producing 3 thousand eggbeaters, set up the definite integral from \(x = 0\) to \(x = 3\) with the function \( \frac{dr}{dx} \). This is expressed as: \[ r = \int_{0}^{3} \left( 2 - \frac{2}{(x+1)^2} \right) \, dx \]
03

Integrate the function

Break down the integral into two separate integrals:\[ \int_{0}^{3} 2 \, dx - \int_{0}^{3} \frac{2}{(x+1)^2} \, dx \]First, integrate \( 2 \, dx \):\[ \int_{0}^{3} 2 \, dx = 2x \Bigg|_0^3 = 2(3) - 2(0) = 6 \]
04

Integrate the second part

Next, integrate \( \frac{2}{(x+1)^2} \, dx \). Use substitution: let \( u = x + 1 \), then \( du = dx \). The limits change from \( x = 0 \) to \( u = 1 \), and from \( x = 3 \) to \( u = 4 \). Now the integral becomes:\[ \int_{1}^{4} \frac{2}{u^2} \, du = 2 \left( -\frac{1}{u} \right) \Bigg|_1^4 \]\[ = 2 \left( -\frac{1}{4} + \frac{1}{1} \right) = 2 \left( \frac{1 - 0.25}{1} \right) = 2 \times 0.75 = 1.5 \]
05

Combine the results

Subtract the result of the second integral from the first:\[ 6 - 1.5 = 4.5 \]Therefore, the total revenue from producing 3 thousand eggbeaters is 4.5 thousand dollars.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Marginal Revenue
Marginal revenue is a crucial concept in economics that relates to the additional revenue generated from selling one more unit of a product. It's essentially the derivative of the revenue function with respect to the quantity of goods sold. In business, knowing the marginal revenue helps in understanding how changes in sales volume affect revenue.
  • Marginal revenue is often used to decide pricing strategies.
  • It plays a vital role in determining the optimal production quantity.
  • In this exercise, the marginal revenue function is given as \( \frac{dr}{dx} = 2 - \frac{2}{(x+1)^2} \).
Understanding this function helps in calculating the total revenue generated from a specific number of sales, providing insights into the company's financial performance.
Definite Integrals
Definite integrals are used to calculate the total accumulation of a quantity, which, in this context, is the revenue over a specific range of production levels. In calculus, a definite integral gives the net area under a curve within a specified interval. To calculate the total revenue from producing a certain number of units, we set up a definite integral of the marginal revenue function.
The expression for definite integrals generally looks like this:
\[ \int_{a}^{b} f(x) \, dx \]
  • \(a\) and \(b\) are the limits of integration, representing the range over which revenue is calculated.
  • In our exercise, \(a = 0\) and \(b = 3\), reflecting the production of 3,000 eggbeaters.
  • The integral is evaluated over these limits to find the accumulated revenue.
Substitution Method
The substitution method is a technique to simplify calculating integrals, especially when direct integration is challenging. This approach involves substituting a part of the integral with a new variable to make it easier to handle.
In the given exercise, to integrate \( \frac{2}{(x+1)^2} \), we use substitution:
  • Let \( u = x + 1 \), then \( du = dx \).
  • The integration limits change accordingly, from \( x = 0 \) to \( u = 1 \) and from \( x = 3 \) to \( u = 4 \).
  • This simplifies the integral to \( \int \frac{2}{u^2} \ du \), making it easier to evaluate.
This method often streamlines the integration process and allows for more straightforward calculations.
Revenue Calculation
Revenue calculation using calculus involves integrating the marginal revenue function over a specific interval to find the total revenue. By using the steps of setting up a definite integral and applying necessary integration techniques, we accurately determine the revenue accrued over the sales of a given quantity.
In this exercise:
  • First, calculate the definite integral of \( 2 \, dx \) from 0 to 3, resulting in a revenue component of 6.
  • Next, integrate \( \frac{2}{(x+1)^2} \, dx \) using substitution, yielding a component of 1.5.
  • Subtract 1.5 from 6 to get the total revenue, which is 4.5 thousand dollars.
Calculating revenue this way provides a comprehensive view of financial expectations.
Mathematical Economics
Mathematical economics integrates mathematical principles with economic theories to systematically solve economic problems. It uses mathematical constructs to form models, which help in making predictions and analyzing data. In the context of this exercise, the principles of calculus assist in understanding the relationship between production and revenue.
  • Mathematical economics helps in making data-driven business decisions.
  • It solidifies abstract economic concepts using precise mathematical language.
  • Through integration and derivation, complex economic relationships become quantifiable.
This exercise exemplifies how mathematical tools can enhance economic analysis, aiding businesses in maximizing their productivity and profitability.

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