Question

Revenue from Marginal Revenue Suppose that a company's marginal revenue from the manufacture and sale of egg beaters is \(\frac{d r}{d x}=2-\frac{2}{(x+1)^{2}}\)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 . \quad \$

Short answer

The total revenue from a production run of 3,000 eggbeaters is estimated to be \((6 - 4ln2)*1000\) dollars.

Step-by-step solution

Write the given

The marginal revenue, \(\frac{d r}{d x}=2-\frac{2}{(x+1)^{2}}\), is given. This represents the amount of revenue the company earns for each additional unit of production.

Set up the integral

We use integral calculus to find total revenue. For this, we integrate the marginal revenue from the start of the production run (\(x=0\)) to the end (\(x=3\)). So, we need to evaluate the integral: \(\int_0^3(2-\frac{2}{(x+1)^2})dx\).

Solve the Integral

Break it up into two separate integrals, use basic integration techniques to evaluate each integral. So it will be: \(\int_0^3 2 dx - \int_0^3 \frac{2}{(x+1)^2} dx \). The integral of 2 with respect to x is 2x. The second integral is a basic formula of derivative of \(ln |x|\), i.e. \(\frac{1}{x}\). Therefore, the integral simplifies to: \(= [2x]_0^3 - [2ln|x+1|]_0^3\).

Evaluate the Integral

Next, we substitute the limits of the integral (0 and 3) to get the final result. Remember that when computing the definite integral, we subtract the function evaluated at the lower limit from that at the upper limit: \(= 2*3 - 2*ln(3+1) - (2*0 - 2*ln(0+1)) = 6 - 2ln4 = 6 - 2*ln2^2 = 6 - 4ln2.\)

Interpret the Result

The result of \(6 - 4ln2\) is in thousands of dollars. So, to get the revenue in dollars, we have to multiply by 1,000. So the total revenue from a production run of 3,000 eggbeaters is \((6 - 4ln2)*1000\) dollars.