Problem 163 - Calculus Volume 1

Question

[T] 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 by p = 143 ? 0.03x and C(x) = 75,000 + 65x, 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?(1000) and R?(4000). Interpret the results. d. Find the profit and marginal profit functions. e. Find P?(1000) and P?(4000). Interpret the results.

Step-by-Step Solution

Answer

MC = 65, MR = 143 - 0.06x, \( R'(1000) = 83 \), \( R'(4000) = -97 \). Profit: \( P'(1000) = 18 \), \( P'(4000) = -162 \).

Step 1: Find Revenue Function

To find the revenue function, use the formula \( R(x) = p imes x \), where \( p \) is the price-demand function given as \( p = 143 - 0.03x \). Substitute \( p \) into the revenue formula to get \( R(x) = (143 - 0.03x) \times x = 143x - 0.03x^2 \).

Step 2: Calculate Marginal Cost Function

The marginal cost (MC) is the derivative of the cost function \( C(x) = 75,000 + 65x \). Calculate \( C'(x) \) to find the marginal cost: \( C'(x) = 65 \).

Step 3: Calculate Marginal Revenue Function

The marginal revenue function (\( R'(x) \)) is the derivative of the revenue function, \( R(x) = 143x - 0.03x^2 \). Differentiate: \( R'(x) = 143 - 0.06x \).

Step 4: Evaluate Marginal Revenue at Specific Quantities

To find \( R'(1000) \), substitute \( x = 1000 \) into \( R'(x) = 143 - 0.06x \): \( R'(1000) = 143 - 0.06\times 1000 = 83 \). For \( R'(4000) \), substitute \( x = 4000 \): \( R'(4000) = 143 - 0.06\times 4000 = -97 \). This means selling 1000 units would add \(83 per drill in revenue, while selling 4000 units decreases revenue by \)97 per drill.

Step 5: Derive Profit Function

The profit function \( P(x) = R(x) - C(x) \). Substitute the given functions: \( P(x) = (143x - 0.03x^2) - (75,000 + 65x) = 78x - 0.03x^2 - 75,000 \).

Step 6: Calculate Marginal Profit Function

The marginal profit function \( P'(x) \) is the derivative of the profit function \( P(x) = 78x - 0.03x^2 - 75,000 \). Differentiate: \( P'(x) = 78 - 0.06x \).

Step 7: Evaluate Marginal Profit at Specific Quantities

To find \( P'(1000) \), substitute \( x = 1000 \) into \( P'(x) = 78 - 0.06x \): \( P'(1000) = 78 - 0.06\times 1000 = 18 \). For \( P'(4000) \), substitute \( x = 4000 \): \( P'(4000) = 78 - 0.06\times 4000 = -162 \). Selling 1000 drills increases profit by \(18 per drill, whereas selling 4000 drills decreases profit by \)162 per drill.

Key Concepts

Marginal Cost

The marginal cost is a critical concept in economics. It tells you how much extra cost you incur for producing one additional unit of a product. In mathematical terms, it's the derivative of the cost function.
For this exercise, the total cost function is given by:

• \( C(x) = 75,000 + 65x \)

This function describes the production cost for \( x \) cordless drills. The marginal cost function is derived by differentiating \( C(x) \) with respect to \( x \):

• \( C'(x) = 65 \)

This means that every additional drill produced costs an extra \( 65 \) dollars. Understanding this helps businesses decide how many units to produce to maximize their profit without unnecessary extra cost.

Revenue Function

The revenue function is crucial for measuring a business's total earnings based on the number of items it sells. It's calculated by multiplying the number of units sold by the price per unit.
Here, the price-demand function is provided as:

• \( p = 143 - 0.03x \)

Where \( p \) stands for price per cordless drill, and \( x \) is the number of drills sold. The revenue function is determined by:

• \( R(x) = p \, \times \, x = (143 - 0.03x) \times x = 143x - 0.03x^2 \)

This equation reflects how changes in quantity affect overall revenues, considering both the price and quantity sold. Understanding this relationship is essential for businesses to forecast earnings and strategize sales targets.

Marginal Revenue

Marginal revenue describes the additional income from selling one more unit of a product. It's an essential factor in determining optimal pricing and sales strategy.
To find the marginal revenue function from the revenue function, use its derivative:

• \( R'(x) = 143 - 0.06x \)

This equation gives you insight into how revenue changes with additional sales. For instance, evaluating at specific quantities provides:

• \( R'(1000) = 83 \)
• \( R'(4000) = -97 \)

The positive value at 1000 units means selling one more drill adds \( 83 \, \text{dollars}\) to revenue. A negative value at 4000 units indicates revenue loss per additional unit due to overproduction or market saturation. Balancing these insights aids in effective sales planning.

Profit Function

The profit function tells a business how much it earns after accounting for costs. It's the difference between revenue and costs.
For the given scenario:

• \( P(x) = R(x) - C(x) = (143x - 0.03x^2) - (75,000 + 65x) \)
• \( P(x) = 78x - 0.03x^2 - 75,000 \)

This equation calculates the net earnings from selling \( x \) cordless drills, considering both gains from sales and expenses from production. By understanding the profit function, businesses can predict net earnings and adjust production strategies to maximize profit margins.

Marginal Profit

Marginal profit measures how profit changes with the sale of one additional unit. This concept is key in determining the optimal level of production where profit can be maximized.
To find the marginal profit function, differentiate the profit function:

• \( P'(x) = 78 - 0.06x \)

Evaluate it at specific points to understand implications:

• \( P'(1000) = 18 \)
• \( P'(4000) = -162 \)

At \( x = 1000 \), a positive \( 18 \) indicates that selling an additional drill increases profit. However, at \( x = 4000 \), the negative \( 162 \) suggests each extra unit reduces profit, signaling excessive output or declining demand. These insights help in aligning production levels with market conditions to optimize profit.