Problem 68 - 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 \(x=200\) to \(x=300\)

Step-by-Step Solution

Answer

The total profit is approximately 81,092 thousand dollars.

Step 1: Identify the Problem

The task is to find the total profit from producing 200 to 300 units. To find this, we will first calculate the total revenue and total cost over this interval, and then compute the profit by subtracting the total cost from the total revenue.

Step 2: Find Total Revenue

The total revenue from producing units from 200 to 300 is the integral of the marginal revenue function from 200 to 300. So, calculate the integral:\[R(x) = \int_{200}^{300} 700 x^{-1} \, dx = 700 [\ln|x|]_{200}^{300}\]Evaluate this integral:\[R(300) - R(200) = 700 (\ln(300) - \ln(200))\]

Step 3: Calculate Total Cost

Similarly, the total cost from producing 200 to 300 units is the integral of the marginal cost function over the same interval. Calculate the integral:\[C(x) = \int_{200}^{300} 500 x^{-1} \, dx = 500 [\ln|x|]_{200}^{300}\]Evaluate this integral:\[C(300) - C(200) = 500 (\ln(300) - \ln(200))\]

Step 4: Compute Total Profit

The total profit is the difference between the total revenue and total cost over the interval 200 to 300:\[P(x) = (R(300) - R(200)) - (C(300) - C(200))\]Substitute the evaluated integrals:\[P(x) = [700 (\ln(300) - \ln(200))] - [500 (\ln(300) - \ln(200))]\]Simplify:\[P(x) = 200 (\ln(300) - \ln(200))\]

Step 5: Evaluate the Profit

Now, calculate the expression for profit:\[P(x) = 200 (\ln(300) - \ln(200))\]First, compute the natural logarithm values and then the difference:\[\ln(300) \approx 5.70378, \quad \ln(200) \approx 5.29832\]Find the difference and multiply by 200:\[P(x) = 200 \times (5.70378 - 5.29832) = 200 \times 0.40546 \approx 81.092\]Thus, the total profit is approximately $81,092 (thousands of dollars).

Key Concepts

Marginal Revenue

In economics, marginal revenue refers to the additional income generated by selling one more unit of a product. It is crucial for determining the optimal level of production. In our problem, the marginal revenue function is given by \( M R(x) = 700 x^{-1} \). This function represents the incremental revenue in thousands of dollars for each additional unit sold, where \( x \) represents the number of units.

The concept of marginal revenue helps companies decide how many units to produce to maximize profits. It involves understanding how changes in production level affect revenue. Keeping track of these changes assists companies in regulating production to meet demand without sacrificing profits.

Marginal Cost

Marginal cost represents the cost of producing one additional unit of a product. In our problem, the marginal cost function is \( M C(x) = 500 x^{-1} \). Like marginal revenue, it is measured in thousands of dollars per additional unit.

This function shows us the incremental cost associated with producing more units. Marginal cost is important in decision-making as it directly impacts a company's pricing strategies.

• When the marginal cost is lower than marginal revenue, producing more may increase profits.
• If the marginal cost surpasses marginal revenue, it might be wiser to reduce production.

By understanding both marginal costs and revenues, businesses can optimize their production for maximum efficiency.

Profit Calculation

Profit calculation is the heart of economic decisions for businesses because it directly indicates success. To calculate profit, subtract the total cost from the total revenue over a specific period or production level.

In this case, total revenue is determined by integrating the marginal revenue function over the interval of interest. Similarly, total cost is found by integrating the marginal cost function over the same range.

The difference between these two integrals gives us the total profit. In our exercise, the total profit from producing between 200 and 300 units is computed as \( P(x) = 200 (\ln(300) - \ln(200)) \). After evaluating, this results in an approximate profit of $81,092 (thousands of dollars), signifying a healthy yield from this production range.

Integration

Integration is a fundamental concept in calculus used to find areas under curves. In economics, it helps calculate total revenue and total costs from marginal functions. It essentially sums up small changes (like marginal costs or revenues) over a continuous interval to provide an aggregate result.

In our problem, integration of the marginal revenue \( M R(x) = 700 x^{-1} \) and marginal cost \( M C(x) = 500 x^{-1} \) functions helps determine the total revenue and total cost between 200 and 300 units. The integration process is presented as:

• \( R(x) = \int_{200}^{300} 700 x^{-1} \, dx = 700 [\ln|x|]_{200}^{300} \)
• \( C(x) = \int_{200}^{300} 500 x^{-1} \, dx = 500 [\ln|x|]_{200}^{300} \)

This method consolidates the incremental changes over the interval to arrive at a total measure. Integration thus proves invaluable in consolidating numerous small changes into a comprehensive whole, thereby enabling strategic business decisions.