Question

Find the number of units \(x\) that produces the minimum average cost per unit \(\bar{C}\). \(C=0.125 x^{2}+20 x+5000\)

Short answer

200 units

Step-by-step solution

Determine the average cost function

The average cost function \(\bar{C}\) is the total cost \(C\) divided by the quantity \(x\). So, \(\bar{C} = C/x = (0.125 x^{2}+20 x+5000) / x = 0.125x + 20 + 5000/x\).

Find the derivative of the average cost function

We differentiate the function \(\bar{C}\) with respect to \(x\) to get \(\bar{C}'\), using the fact that the derivative of \(x^n\) (n \neq 0) is \(n*x^{(n-1)}\) and the derivative of a constant is 0. It gives us: \(\bar{C}' = 0.125 + 0 - 5000/x^2\).

Find the value of \(x\) that minimizes the average cost

At the minimum average cost, the derivative of the average cost function we found in step 2 would be equal to zero. So, to find the value of \(x\) that minimizes our average cost, we solve \(\bar{C}' = 0\), or \(0.125 - 5000/x^2=0\). Solving gives \(x = sqrt(5000/0.125) = 200\).

Verify that it is a minimum value

We can confirm that this value of \(x\) produces a minimum (as opposed to a maximum or a point of inflection) by showing that the second derivative is positive. The second derivative of \(\bar{C}\) is found by differentiating \(\bar{C}'\), which gives \(0 + 10000/x^3\). Since \(x = 200 > 0\), it follows that \(\bar{C}''(200) > 0\), thus confirming that the average cost is minimized when \(x = 200\) units are produced.

Key concepts

Calculus

Calculus is a fundamental branch of mathematics focusing on rates of change and accumulation. It comprises two main areas: differentiation and integration. Differentiation deals with finding the rate at which quantities change, while integration is about finding the total amount of something over an interval. In this problem, calculus is primarily used to find minimum points by employing derivatives. These points help determine the optimal number of units that minimize the average cost per unit in a production setting. By understanding calculus concepts, students can apply these mathematical techniques to optimize real-world problems.

Derivative

A derivative is a tool in calculus that measures how a function changes as its input changes. Simply put, it tells us the slope of the function at any given point. In the problem at hand, the derivative of the average cost function is calculated to find when it reaches its lowest point. We calculate the first derivative \[ \bar{C}' = 0.125 - \frac{5000}{x^2} \] This expression indicates how the average cost per unit changes as production levels change. By setting the derivative equal to zero and solving for \( x \), we find the critical points, where possible minimum or maximum costs occur. The derivative essentially helps us pinpoint the exact production level where costs are minimized.

Second Derivative

The second derivative is the derivative of the derivative of a function. It tells us about the curvature or concavity of the original function. In economic terms, it reveals whether a minimum or maximum point from the first derivative produces a cost-efficient outcome or not. In this problem, the second derivative of \[ \bar{C}'' = \frac{10000}{x^3} \] is calculated to verify whether the critical point found from the first derivative is indeed a minimum. A positive value for the second derivative indicates the original function is concave upwards, confirming a minimum point. Thus, when \( x = 200 \), the positive second derivative confirms that we have found the point of minimized average cost.

Cost Function

A cost function is a mathematical relationship describing the total cost incurred by a firm based on the quantity of goods produced. In our example, the cost function is given by \[ C = 0.125x^2 + 20x + 5000 \] This reflects both variable costs, represented by terms involving \( x \), and fixed costs shown by constants. The average cost function, \( \bar{C} \), is derived by dividing the total cost \( C \) by the quantity \( x \). This gives us insight into how cost behaves per unit of production. By applying calculus to this function, we determine the most efficient production level, minimizing costs. Understanding the cost function helps firms make pricing and production decisions efficiently.