Problem 74 - University Calculus: Early Transcendentals

Question

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 \(x=0\) to \(x=3\).

Step-by-Step Solution

Answer

Total expected revenue is 7,500 dollars.

Step 1: Set Up the Definite Integral

To find the total revenue, integrate the marginal revenue function from 0 to 3. The marginal revenue function is given by \( \frac{dr}{dx} = 2 - \frac{2}{(x+1)^2} \). Thus, we want to calculate \[ \int_{0}^{3} \left( 2 - \frac{2}{(x+1)^2} \right) \, dx. \]

Step 2: Integrate the Expression

We need to evaluate \[ \int \left( 2 - \frac{2}{(x+1)^2} \right) \, dx. \]This splits into two separate integrals: \[ \int 2 \, dx - \int \frac{2}{(x+1)^2} \, dx. \]The first integral is straightforward: \[ \int 2 \, dx = 2x. \]For the second integral, let \( u = x+1 \), then \( du = dx \). This gives \[ \int \frac{2}{u^2} \, du = -\frac{2}{u}. \]Re-substituting \( u = x+1 \), we obtain \[ -\frac{2}{x+1}. \]

Step 3: Evaluate the Definite Integral

Substituting back, the indefinite integral of the given function is:\[ 2x - \frac{2}{x+1}. \]We now evaluate this from 0 to 3:\[ \left[ 2x - \frac{2}{x+1} \right]_0^3 = \left(2(3) - \frac{2}{3+1}\right) - \left(2(0) - \frac{2}{0+1}\right). \]This simplifies to:\[ (6 - \frac{2}{4}) - (0 - 2) = 6 - 0.5 + 2 = 7.5. \]

Step 4: Calculate Total Revenue

The result of the integral is 7.5, which represents thousands of dollars, as the original function \( \frac{dr}{dx} \) was in thousands of dollars. Therefore, from a production run of 3,000 eggbeaters, the company should expect a total revenue of 7,500 dollars.

Key Concepts

Marginal Revenue

Understanding Marginal Revenue is essential in determining how revenue changes with each additional unit sold. In economics, it refers to the additional income generated by selling one more unit of a good or service. You can think of it as the derivative of total revenue concerning the quantity sold. This concept is vital for making pricing and production decisions.

• If the marginal revenue is high, it might be beneficial to increase production.
• Conversely, if it's low or negative, reducing production might be wise.

In calculus, marginal revenue is often given as a function, like the one provided in our exercise: \( \frac{d r}{d x}=2-\frac{2}{(x+1)^{2}} \). This function tells us how much revenue changes with a slight change in the number of products sold, specifically in thousands of units. To find the total revenue over a range of production, like from 0 to 3, we integrate over that interval.

Definite Integral

A Definite Integral represents the total accumulation of a quantity over an interval. In our context, it helps us calculate the total revenue generated by manufacturing and selling a certain number of units. If the integrand is the marginal revenue, then the definite integral of that function gives us the total revenue.

• The limits of the integral specify the interval over which we accumulate the quantity.
• The process involves integrating the marginal revenue function between the lower and upper limits, such as 0 and 3 in our case.

The definite integral that we set up from 0 to 3 thousand units is: \[ \int_{0}^{3} \left( 2 - \frac{2}{(x+1)^2} \right) \, dx. \]By evaluating this, we obtain the accumulated value of total revenue possible in thousands of dollars from these units.

Substitution Method

The Substitution Method is a technique used in integration to simplify complex problems. Often known as 'u-substitution,' this method helps in transforming difficult integrals into more manageable ones. The basic idea is to choose a substitution that simplifies the integrand.

• To apply this, identify a part of the equation to substitute for a new variable, usually denoted by \( u \).
• This new substitution should turn the integral into a simpler form that is easier to solve.

In our exercise, we used substitution for the integral \( \int \frac{2}{(x+1)^2} \, dx \) by letting \( u = x+1 \), thus \( du = dx \). This gave us a simpler expression to integrate: \[ \int \frac{2}{u^2} \, du = -\frac{2}{u}. \]After integration, the substitution was reversed to express the solution in terms of the original variable, completing our integration step.