Question

A manufacturer has determined that the total cost \(C\) of operating a factory is \(C=0.5 x^{2}+10 x+7200\), where \(x\) is the number of units produced. At what level of production will the average cost per unit be minimized? (The average cost per unit is \(C / x\).)

Short answer

The average cost per unit will be minimized when the level of production is 120 units.

Step-by-step solution

Express the average cost as a function

The average cost function \(A(x)\) is given by the total cost function \(C(x)\) divided by the number of units produced \(x\). Therefore: \(A(x) = \frac{C(x)}{x} = \frac{0.5x^{2} + 10x + 7200}{x}\). Simplify this function by distributing the division operation. \(A(x) = 0.5x + 10 + \frac{7200}{x}\)

Find the derivative of the average cost function

To find the minimum of the average cost function, find its derivative and set it to zero. The derivative of \(A(x)\) is computed as follows: \[A'(x) = 0.5 - \frac{7200}{x^{2}}\]

Set the derivative of the function to zero

Setting the derivative to zero will give the production level that minimizes the average cost. Solve the equation \(A'(x) = 0\) for \(x\): \[0.5 - \frac{7200}{x^{2}} = 0\] Solving for \(x\) gives: \[x = \sqrt{\frac{7200}{0.5}} = \sqrt{14400} = 120\]. Therefore, producing 120 units minimizes the average cost per unit.

Key concepts

Cost Function

In the world of economics, particularly in a manufacturing context, understanding the cost function is crucial. The cost function, often denoted as \(C(x)\), represents the total cost associated with producing \(x\) units of a product. It's a mathematical model that includes all variable and fixed costs involved in production.
In the example given, the cost function is \(C(x) = 0.5x^{2} + 10x + 7200\). Each part of this equation has its own meaning:

• The term \(0.5x^2\) represents the variable costs that increase quadratically with production level, often due to economies of scale or resource limits.
• The linear term \(10x\) reflects costs that grow directly with each additional unit produced, such as materials and direct labor.
• The constant \(7200\) stands for fixed costs, like rent and administrative expenses, that do not change with production volume.

Understanding how these components work together helps manufacturers make informed decisions about how to optimize their operations.

Derivative

Derivatives play a key role in calculus, particularly when working with optimization and rate of change problems. The derivative of a function provides the rate at which that function changes at any given point. In optimization, it helps identify maxima, minima, or the point where change is constant.
For the average cost function \(A(x)\), we first derive the expression \(A(x) = 0.5x + 10 + \frac{7200}{x}\). To find minima, taking its derivative \(A'(x)\) is a crucial step:

• Differentiate \(0.5x\) to get \(0.5\) as it is a simple linear term.
• The constant \(10\) differentiates to \(0\).
• Application of the quotient rule or the power rule to \(\frac{7200}{x}\) results in \(-\frac{7200}{x^2}\).

The derivative \(A'(x) = 0.5 - \frac{7200}{x^2}\) helps us find the critical points to solve the average cost minimization problem.

Production Level

Determining the optimal production level is a fundamental question in manufacturing and cost management. The production level \(x\) is the number of units produced. It's critical in calculating both the total and average costs.
In this context, we examine how the production level impacts the average cost per unit. By finding the derivative of the average cost function \(A'(x)\) and setting it to zero, we solve the equation to find \(x\). This value of \(x\) represents the optimal production level where average costs are minimized.

• An increase in production level initially reduces average costs due to spread in fixed costs.
• Beyond a certain point, variable costs might rise significantly, increasing average cost.
• Here, solving \(A'(x) = 0\) resulted in \(x = 120\). Hence, producing 120 units is optimal for minimizing average cost in our problem.

Choosing the optimal production level, like 120 units here, aligns cost management with strategic production goals.

Optimization Problem

Optimization problems are all about finding the best possible solution among a set of values, often within certain constraints. In economic terms, it's about minimizing costs or maximizing profits.
With average cost minimization, the challenge is to find the production level that results in the lowest possible average cost. This is achieved through calculus by:

• Expressing the average cost function based on the total cost function.
• Calculating its derivative to understand how costs change as production changes.
• Finding the critical points by setting the derivative to zero. This indicates the points where costs are stationary and might be minimum or maximum.

In our exercise, setting \(A'(x) = 0.5 - \frac{7200}{x^2}\) to zero and solving gives the optimal production point, \(x = 120\).
This solution demonstrates using calculus in practical scenarios to make economically sound decisions and drive financial performance by minimizing average costs.