Problem 67 - Applied Calculus

Question

A company's marginal revenue function is \(M R(x)=700 x^{-1}\) and its marginal cost function is \(M C(x)=500 x^{-1}\) (both in thousands of dollars), where \(x\) is the number of units \((x>1)\). Find the total profit from \(x=100\) to \(x=200\)

Step-by-Step Solution

Answer

The total profit from \(x=100\) to \(x=200\) is approximately $138,600.

Step 1: Define Profit Function

The total profit is given by the integral of the marginal revenue minus the integral of the marginal cost over the interval from \(x = 100\) to \(x = 200\). We set up the profit function as \( P(x) = \int (MR(x) - MC(x)) \, dx \).

Step 2: Set Up the Integral

Substitute the marginal revenue and marginal cost functions into the profit function: \( P(x) = \int_{100}^{200} (700x^{-1} - 500x^{-1}) \, dx \).

Step 3: Simplify the Integral Expression

Simplify the integrand: \( 700x^{-1} - 500x^{-1} = 200x^{-1} \). Thus, the integral becomes \( P(x) = \int_{100}^{200} 200x^{-1} \, dx \).

Step 4: Integrate

Integrate the simplified function: \( \int 200x^{-1} \, dx = 200 \ln|x| + C \), where \(C\) is the integration constant. We are interested in the definite integral, so \(C\) will not affect our result.

Step 5: Evaluate the Definite Integral

Evaluate \(200 \ln|x|\) from \(x = 100\) to \(x = 200\): \[ 200 \ln|200| - 200 \ln|100| \]. Use the properties of logarithms to simplify:\[ 200 (\ln 200 - \ln 100) = 200 \ln \left( \frac{200}{100} \right) = 200 \ln 2 \].

Step 6: Calculate the Final Answer

Compute the numerical value of \(200 \ln 2\). Using a calculator, find \(\ln 2 \approx 0.693\). Thus, \(200 \ln 2 \approx 200 \times 0.693 = 138.6\). This result is in thousands of dollars.

Key Concepts

Marginal Cost

Marginal cost is an important concept in economics and refers to the cost of producing one additional unit of a good. In this exercise, the marginal cost function is given as \(MC(x) = 500x^{-1}\), indicating how the cost changes with the production of different quantities of units, \(x\), where \(x > 1\). The expression \(500x^{-1}\) tells us that as more units are produced, the more the cost per additional unit decreases, assuming \(x\) is increasing. Understanding marginal cost helps businesses decide how many units to produce by comparing it to the marginal revenue, ensuring that the cost of producing additional units does not outweigh the revenue gained from selling them.

Profit Function

The profit function is a measure of the profit earned by the company over a certain range of units produced. In our example, it's crucial to identify profit as the difference between total revenue and total costs. The original step-by-step solution treats the profit function as the integral of the difference between marginal revenue and marginal cost, specifically, \( P(x) = \int (MR(x) - MC(x)) \, dx \). This reflects how profits are accumulated over a range—from \(x=100\) to \(x=200\) in this case—by considering the continuous addition of profits per unit. The focus here is on maximizing one’s overall earnings by ensuring that the difference between marginal revenue and marginal cost is positive across the interval.

Definite Integral

Definite integrals are used in calculus to calculate the total accumulation over an interval. In this exercise, we study how definite integrals are applied to determine total profit. The definite integral \( \int_{100}^{200} 200x^{-1} \, dx \) calculates the total net profit as it is evaluated over the specific interval of \(x = 100\) to \(x = 200\). By integrating \(200x^{-1}\), which represents the difference between marginal revenue and marginal cost, the solution provides the accumulated profit. The set upper and lower bounds (100 and 200) specify the limits of integration, effectively calculating the complete profit from producing units within this range. The process of evaluating this integral involves calculating the natural log terms, allowing us to use the properties of logarithms to simplify and find the final numeric profit value. This approach highlights the power of integrals in practical applications like revenue and cost analysis.