Problem 52 - Applied Calculus: For Business, Economics, and the Social and Life Sciences

Question

A manufacturer finds that in producing \(x\) units per day (for \(0

Step-by-Step Solution

Answer

The level of production that minimizes total cost is \( \bigg( \frac{500}{3} \bigg)^{1/3} \) units per day.

Step 1: Identify the components of the total cost

The total cost per day consists of three parts:- A fixed cost of $$1200- A production cost of $$1.20 per unit produced- An ordering cost of \(\frac{100}{x^2}\) dollars per day

Step 2: Express the total cost as a function

The total cost function, C(x), is the sum of the three costs identified in Step 1:\[ C(x) = 1200 + 1.20x + \frac{100}{x^2} \]

Step 3: Find the derivative of the cost function

To minimize the total cost, first find the derivative of the function C(x):\[ C'(x) = \frac{d}{dx}(1200 + 1.20x + \frac{100}{x^2}) = 1.20 - \frac{200}{x^3} \]

Step 4: Set the derivative equal to zero and solve for x

Set \(C'(x) = 0\) to find the critical points:\[ 1.20 - \frac{200}{x^3} = 0 \]Solve for x:\[ 1.20 = \frac{200}{x^3} \]\[ x^3 = \frac{200}{1.20} \]\[ x^3 = \frac{500}{3} \]\[ x = \bigg( \frac{500}{3} \bigg)^{1/3} \]

Step 5: Determine if the critical point is a minimum

To determine if the critical point is a minimum, take the second derivative of \(C(x)\):\[ C''(x) = \frac{d}{dx}(1.20 - \frac{200}{x^3}) = \frac{600}{x^4} \]Since \( C''(x) > 0 \) for all \( x > 0 \), the critical point is a minimum.

Step 6: Calculate the exact level of production

Evaluate the critical point found in Step 4:\[ x = \bigg( \frac{500}{3} \bigg)^{1/3} \]Calculate the cube root to find x.

Key Concepts

fixed cost

Fixed costs are expenses that do not change with the level of goods or services produced. In this exercise, the manufacturer has a fixed cost of \(1200 per day for wages. This means that no matter how many units are produced, the wages paid per day remain constant at \)1200. These costs are crucial because they must be covered regardless of production volume. Understanding fixed costs helps businesses determine the minimum revenue needed to avoid losses. In this context, the fixed cost is simply added to other variable costs to get the total cost.

production cost

Production cost varies with the number of units produced. Here, the production cost is \(1.20 per unit. This means if a manufacturer produces 10 units, the cost would be \)12. If they produce 50 units, the cost would be $60. Production costs include expenses like materials, labor (that varies with production), and other variable costs associated with manufacturing an additional unit. The formula for production cost in this exercise is simple multiplication: \(1.20x\) where \(x\) is the number of units produced.

ordering cost

Ordering costs are associated with placing orders and acquiring raw materials. In this exercise, the ordering cost is given by \( \frac{100}{x^2} \), where \ x \ is the number of units produced. This cost decreases as production increases because ordering becomes more efficient. For example, producing fewer units means higher ordering costs per unit, but producing more units reduces the ordering cost per unit. It is crucial to balance ordering and production to minimize total costs.

cost function

A cost function represents the total cost associated with producing a given number of units. In this exercise, the cost function is \( C(x) = 1200 + 1.20x + \frac{100}{x^2} \). It combines fixed costs, production costs, and ordering costs. The objective is to minimize this function to find the production level that results in the least total cost. By analyzing and manipulating the cost function, businesses can determine the most cost-efficient production strategy. Understanding the components and behavior of a cost function is vital for cost management and optimization.

calculus in economics

Calculus is a powerful tool in economics, especially for optimization problems like cost minimization. In this exercise, calculus is used to find the minimum total cost. The first derivative of the cost function, \( C'(x) \), is set to zero to find the critical points. These points indicate where the function's slope changes. The second derivative, \( C''(x) \), determines if the critical point is a minimum or maximum. If the second derivative is positive, the point is a minimum. This process helps identify the optimal production level, balancing fixed, production, and ordering costs in the most efficient manner.