Question

The variable cost for the production of a calculator is \(\$ 14.25\) and the initial investment is \(\$ 110,000\). Find the total cost \(C\) as a function of \(x\), the number of units produced. Then use differentials to approximate the change in the cost for a one-unit increase in production when \(x=50,000\). Make a sketch showing \(d C\) and \(\Delta C\). Explain why \(d C=\Delta C\) in this problem.

Short answer

The cost function is \(C(x) = \$14.25x + \$110,000\). The differential indicates that for a one-unit increase in production when \(x=50,000\), the cost increases by \$14.25. Since the cost function is linear, the differential cost equals the change in cost, i.e., \(dC = \Delta C\).

Step-by-step solution

Determine the cost function

Given that the variable cost per calculator is \$14.25, for \(x\) units produced the variable cost becomes \$14.25x. Adding the initial investment (\$110,000), the cost function \(C(x)\) becomes: \(C(x) = \$14.25x + \$110,000\).

Calculate the differential of the cost

The function \(C(x)\) is linear, so its derivative \(C'(x)\) will give the rate of change in cost for a one-unit increase in production. The derivative of a constant is zero and the derivative of \(14.25x\) w.r.t \(x\) is \(14.25\). Therefore, \(C'(x) = \$14.25\). The differential of the cost is \(dC = C'(x) dx\). Considering that a one-unit increase refers to \(dx = 1\), the change in cost is \(dC = C'(50,000) dx = 14.25 * 1 = \$14.25\).

Analyze the diagram and elucidate the equality

To draw the diagram, mark the production of 50,000 units on the x-axis and correspondingly, its cost on the y-axis. Then draw a tangent line at this point on the cost curve. Here, both \(dC\) and \(\Delta C\) represent the change in cost for a unit increase in production. Because the cost function is linear (straight line), the change in cost (\(\Delta C\)) over any interval will match the differential (\(dC\)). This is essentially due to the fact that the tangent line to any point on a straight line coincides with the straight line itself.

Key concepts

Variable Cost

Variable cost is a fundamental concept in both economics and accounting, referring to expenses that change in proportion to the activity of a business. In the context of manufacturing, variable costs increase with each additional unit produced, as they are directly related to production levels.

For example, if the cost to produce one calculator is \( \(14.25 \), then the variable cost for producing \( x \) calculators is the product of \( \)14.25 \) and \( x \), represented mathematically as \( \(14.25x \). This does not include fixed costs, or initial investments, such as the purchase of equipment, which remain constant regardless of the production volume.

In the exercise provided, the initial investment, which is a fixed cost, is \( \)110,000 \). Combining this with the variable cost gives us the total cost function \( C(x) = \(14.25x + \)110,000 \). Understanding this distinction between variable and fixed costs is crucial when calculating the total cost for any level of production.

Differentials in Calculus

Differentials in calculus are incredibly useful for estimating changes in functions over tiny intervals. They revolve around the concept of the derivative, which measures how a function changes as its input changes.

The differential, denoted as \( dC \), represents an infinitesimally small change in a function, in this case, the cost function \( C(x) \). The formula for the differential is \( dC = C'(x) dx \), where \( C'(x) \) is the derivative of \( C(x) \) with respect to \( x \), and \( dx \) is a small change in \( x \).

In a practical scenario, like the one-unit production increase in the exercise, we apply differentials to approximate the change in cost. Because we're dealing with a linear function, finding the derivative is straightforward. We simply take the coefficient of \( x \), which is \( \(14.25 \), representing a consistent rate of change in cost per unit produced. Then, by setting \( dx = 1 \), which symbolizes one additional unit produced, the change in the cost function is simply \( dC = \)14.25 \).

Tangent Line Approximation

The tangent line approximation is a technique used to estimate the value of a function at a particular point using the tangent line at that point. It's particularly useful for complex functions where calculating exact values is difficult.

In the setting of our exercise, we are examining a linear cost function. Since a straight line is its own tangent at any point, when we talk about the approximation of the change in cost for a linear function like the one we have (\( C(x) = \(14.25x + \)110,000 \)), the tangent line approximation exactly equals the actual change in cost.

Therefore, when \( x = 50,000 \) units, drawing a tangent line at this point on the cost graph will simply result in the same line as the cost function itself. This means that for any increase in units, the change in cost, or \( dC \), will precisely match the actual change in cost, or \( \Delta C \), for a linear function. This concept helps in visualizing the relationship between a function and its tangent, especially when the calculations for \( dC \) and \( \Delta C \) are the same due to the linearity of the function.