Problem 38 - Calculus with Concepts in Calculus

Question

A company has a daily fixed cost of $$\$ 5000$$. If the company produces \(x\) units daily, then the daily cost in dollars for labor and materials is \(3 x\). The daily cost of equipment maintenance is $$x^{2} / 2,500,000.$$ What daily production minimizes the total daily cost per unit of production? (Hint: The cost per unit is the total \(\operatorname{cost} C(x)\) divided by \(x .\) )

Step-by-Step Solution

Answer

The production level that minimizes the cost per unit is approximately 111,803 units.

Step 1: Define Total Daily Cost

First, let's define the total daily cost function, which is denoted by \(C(x)\). The total cost is the sum of fixed costs, labor and materials cost, and maintenance cost: \[C(x) = 5000 + 3x + \frac{x^2}{2,500,000}.\]

Step 2: Define Cost Per Unit Function

The cost per unit function \(c(x)\) is the total cost \(C(x)\) divided by the number of units \(x\): \[c(x) = \frac{C(x)}{x} = \frac{5000 + 3x + \frac{x^2}{2,500,000}}{x}.\]

Step 3: Simplify the Cost Per Unit Function

Simplify \(c(x)\) by dividing each term within the total cost by \(x\): \[c(x) = \frac{5000}{x} + 3 + \frac{x}{2,500,000}.\]

Step 4: Find the Derivative of Cost Per Unit

To find the production level that minimizes \(c(x)\), take the derivative of \(c(x)\) with respect to \(x\): \[c'(x) = -\frac{5000}{x^2} + \frac{1}{2,500,000}.\]

Step 5: Set the Derivative to Zero

Find critical points by setting \(c'(x) = 0\): \[-\frac{5000}{x^2} + \frac{1}{2,500,000} = 0.\] Solve for \(x\): \[\frac{5000}{x^2} = \frac{1}{2,500,000}.\]

Step 6: Solve for Critical Point

Multiply both sides by \(2,500,000x^2\): \[5000 \cdot 2,500,000 = x^2.\] Solve for \(x\): \[x = \sqrt{5000 \times 2,500,000}.\]

Step 7: Calculate the Optimal Production Level

Compute \(x\): \[x = \sqrt{12,500,000,000}.\] Simplify the square root: \[x \approx 111,803.39.\] Therefore, \(x \approx 111,803\) units (rounded to the nearest unit).

Step 8: Verify the Minimum

Check the behavior of \(c'(x)\) around \(x = 111,803\) or evaluate the second derivative to make sure it's a minimum. The calculated critical point indeed results in a minimum as verified by the increase in \(c(x)\) beyond \(x = 111,803\).

Key Concepts

Cost Function

In the context of optimization problems, a cost function represents the total expense associated with producing a specific number of goods. It incorporates various costs incurred during production. In this problem, the company has three types of costs: fixed costs, labor and material costs, and maintenance costs. Each of these components contributes to the total daily cost function.

• Fixed Costs: A constant amount the company must pay daily, regardless of production. Here, it's \$5000.
• Labor and Material Costs: This cost is variable and depends on the number of units produced. It's given by \(3x\).
• Maintenance Costs: This involves a quadratic relation in terms of units produced, expressed as \(\frac{x^2}{2,500,000}\).

Together, these components shape the total daily cost function \(C(x) = 5000 + 3x + \frac{x^2}{2,500,000}\). This function is crucial for determining how costs fluctuate with changes in production levels.

Derivative

The derivative is a powerful tool in calculus used to find rates of change and identify local optimizations, such as minima or maxima. In optimization exercises, finding the derivative is essential to determining points where the cost function can be minimized or maximized. Once the total cost per unit function \(c(x)\) was defined and simplified to \(c(x) = \frac{5000}{x} + 3 + \frac{x}{2,500,000}\), we proceeded to calculate its derivative.

• The derivative of \(\frac{5000}{x}\) with respect to \(x\) is \(-\frac{5000}{x^2}\).
• The derivative of the constant \(3\) is 0.
• The derivative of \(\frac{x}{2,500,000}\) is \(\frac{1}{2,500,000}\).

The resulting derivative, \(c'(x) = -\frac{5000}{x^2} + \frac{1}{2,500,000}\), helps identify the critical points, which are pivotal in finding the minimum cost per unit.

Critical Points

Critical points occur where the derivative of a function is zero or undefined. These points are essential for finding local maxima or minima in optimization problems. After finding the derivative \(c'(x) = -\frac{5000}{x^2} + \frac{1}{2,500,000}\), the next step is to set it equal to zero to find the critical points that potentially minimize the cost per unit.

• By setting \(-\frac{5000}{x^2} + \frac{1}{2,500,000} = 0\), we find the value of \(x\) that equates these components.
• Solving this algebraic equation gives the critical value: \(x = \sqrt{12,500,000,000}\).
• This critical point, \(x \approx 111,803\), is where the cost per unit is potentially minimized.

Once the critical point is identified, further analysis, such as using the second derivative test, confirms that this point corresponds to a minimum.

Quadratic Functions

Quadratic functions form a fundamental part of many optimization problems, including this one. A quadratic function typically has the form \(ax^2 + bx + c\), and its graph is a parabola. In this problem, the maintenance cost \(\frac{x^2}{2,500,000}\) reflects a quadratic relationship.

• The presence of the \(x^2\) term indicates that costs do not increase linearly with production, illustrating economies or diseconomies of scale.
• Analyzing quadratic functions helps evaluate how variables interact as production changes, impacting the entire cost function and its optimization.
• The vertex of the parabola, found through calculus, identifies where the cost per unit is minimized.

Understanding the role of quadratic functions is crucial for grasping how changes in production levels affect costs and achieving optimal results.