Problem 163 - Calculus Volume 1

Question

In general, the profit function is the difference between the revenue and cost functions: \(P(x)=R(x)-C(x)\). Suppose the price-demand and cost functions for the production of cordless drills is given respectively \(\quad\) by \(p=143-0.03 x\) and \(C(x)=75,000+65 x,\) where \(x\) is the number of cordless drills that are sold at a price of \(p\) dollars per drill and \(C(x)\) is the cost of producing \(x\) cordless drills. a. Find the marginal cost function. b. Find the revenue and marginal revenue functions. c. Find \(R^{\prime}(1000)\) and \(R^{\prime}(4000)\). Interpret the results. d. Find the profit and marginal profit functions. e. Find \(P^{\prime}(1000)\) and \(P^{\prime}(4000)\). Interpret the results.

Step-by-Step Solution

Answer

a. 65, b. R(x) = 143x - 0.03x²; R'(x) = 143 - 0.06x, c. 83, -97, d. P(x) = 78x - 0.03x² - 75000, e. 18, -162.

Step 1: Identify the Cost Function

The cost function is given as \(C(x) = 75,000 + 65x\). The cost function represents the total cost of producing \(x\) units. In this case, the constant \(75,000\) represents the fixed costs, and \(65x\) represents the variable cost for producing each additional unit.

Step 2: Find the Marginal Cost Function

The marginal cost function \(C'(x)\) is the derivative of the cost function \(C(x)\) with respect to \(x\). Since \(C(x) = 75,000 + 65x\), its derivative is the constant term of the variable cost, thus:\[C'(x) = \frac{d}{dx}(75,000 + 65x) = 65\].This means the marginal cost per drill is \(65\) dollars.

Step 3: Formulate the Price-Demand Function

The price-demand function is given as \(p = 143 - 0.03x\). This function relates the price \(p\) to the quantity \(x\) that is demanded. We need this function to find the revenue function in the next step.

Step 4: Find the Revenue Function

The revenue function \(R(x)\) is given by price \(p\) multiplied by the quantity \(x\):\[R(x) = x \cdot p = x \cdot (143 - 0.03x)\].Simplifying, we get:\[R(x) = 143x - 0.03x^2\].

Step 5: Find the Marginal Revenue Function

The marginal revenue function \(R'(x)\) is the derivative of the revenue function \(R(x)\). Using the derivative:\[R'(x) = \frac{d}{dx}(143x - 0.03x^2) = 143 - 0.06x\].

Step 6: Evaluate Marginal Revenue Function at Specific Points

To find \(R'(1000)\), substitute \(x = 1000\) into \(R'(x)\):\[R'(1000) = 143 - 0.06 \times 1000 = 83\].For \(R'(4000)\), substitute \(x = 4000\):\[R'(4000) = 143 - 0.06 \times 4000 = -97\].Interpretation: Selling the 1001st drill adds \\(83 to revenue, while selling the 4001st drill decreases revenue by \\)97.

Step 7: Find the Profit Function

The profit function \(P(x)\) is the difference between revenue and cost:\[P(x) = R(x) - C(x) = (143x - 0.03x^2) - (75000 + 65x)\].Simplifying:\[P(x) = 78x - 0.03x^2 - 75000\].

Step 8: Find the Marginal Profit Function

The marginal profit function \(P'(x)\) is the derivative of the profit function \(P(x)\):\[P'(x) = \frac{d}{dx}(78x - 0.03x^2 - 75000) = 78 - 0.06x\].

Step 9: Evaluate Marginal Profit Function at Specific Points

To find \(P'(1000)\), substitute \(x = 1000\) into \(P'(x)\):\[P'(1000) = 78 - 0.06 \times 1000 = 18\].For \(P'(4000)\), substitute \(x = 4000\):\[P'(4000) = 78 - 0.06 \times 4000 = -162\].Interpretation: The 1001st drill increases profit by \\(18, while the 4001st drill decreases profit by \\)162.

Key Concepts

Marginal Cost

Marginal cost refers to the additional cost of producing one more unit of a product. It helps businesses understand how changes in production levels affect costs.
In our exercise, the cost function is straightforward, given by \( C(x) = 75,000 + 65x \). Fixed costs here amount to \( 75,000 \), which do not change with production volume, while \( 65x \) represents the variable cost per drill.
To derive the marginal cost, we compute the derivative of the cost function, which is a key concept in calculus. This derivative, denoted as \( C'(x) \), is simply \( 65 \). This implies that every additional drill increases the total cost by $65.
Having a constant marginal cost is typical in scenarios where fixed costs remain unchanged, and variable costs per unit are stable. Understanding this helps businesses with pricing strategies and efficiency improvements.

Marginal Revenue

Marginal revenue is the increase in revenue from selling one more unit of a product. It aids firms in understanding how sales affect their total revenue.
Calculating the revenue function is the first task, which involves the product of price and quantity. Given by \( R(x) = x \cdot (143 - 0.03x) \), the function simplifies to \( R(x)=143x - 0.03x^2 \).
The next step is to find the marginal revenue function, which involves taking the derivative of the revenue function. This calculation yields \( R'(x) = 143 - 0.06x \).
Evaluating points like \( R'(1000) \) and \( R'(4000) \), the results indicate that revenue can either increase or decrease with additional sales, also reflecting demand's elasticity. At \( x=1000 \), the marginal revenue is $83, suggesting added sales are beneficial. However, at \( x=4000 \), it turns negative, hinting that excess production harms revenue.

Profit Function

The profit function is a cornerstone of business economics, detailing the net gain from operations. It links revenue and costs, aiding in strategic decisions.
The exercise provides the profit calculation as \( P(x) = R(x) - C(x) \). Plugging in our functions gives \( P(x) = (143x - 0.03x^2) - (75,000 + 65x) \). Simplified, we have \( P(x) = 78x - 0.03x^2 - 75,000 \).
This quadratic function shows how profit changes with different sales volumes, indicating optimal production levels.
Understanding profit functions allows businesses to maximize financial outcomes. It provides insights into balancing production costs and sales revenues effectively.

Derivative Calculus

Derivative calculus is pivotal in marginal analysis, providing techniques to find rates of change. It is integral to understanding business dynamics.
The derivative of a function evaluates how it changes at any given point. In the context of marginal analysis, it uncovers the marginal cost, marginal revenue, and marginal profit.
Our solution highlights this process vividly. Derivatives, such as \( C'(x) = 65 \) for costs, or \( R'(x) = 143 - 0.06x \) for revenue, are calculated to reflect incremental changes in cost and revenue respective to produced units.
Ultimately, calculus offers analytical precision. Businesses apply these concepts to make informed decisions about optimal production levels and pricing strategies, ensuring competitive advantages.