Question

The ordering and transportation cost \(C\) per unit for the components used in manufacturing a product is \(C=\left(375,000+6 x^{2}\right) / x, \quad x \geq 1\) where \(C\) is measured in dollars and \(x\) is the order size. Find the rate of change of \(C\) with respect to \(x\) when (a) \(x=200\), (b) \(x=250\), and (c) \(x=300\). Interpret the meaning of these values.

Short answer

The rates of change of cost per unit with respect to order size are -9.375 dollars/unit when \(x=200\), -6 dollars/unit when \(x=250\), and -4.167 dollars/unit when \(x=300\). These values indicate that as we increase the order size, the cost per unit decreases.

Step-by-step solution

Differentiate the Cost Function

The cost function is \(C=\left(375,000+6 x^{2}\right) / x\). We need to find its derivative, to see how \(C\) changes respect to \(x\). First, let's split the function into two terms: \(C = 375,000/x + 6x\). Now, we can use the power rule to differentiate each term of \(C\) separately: \(C' = -375,000/x^2 + 6\). This formula gives the rate of change of the cost per unit with respect to the order size.

Compute Rates of Change

Now we calculate \(C'\) at \(x=200\), \(x=250\), and \(x=300\). For \(x=200\), \(C' = -375,000/(200^2) + 6 = -9.375\). For \(x=250\), \(C' = -375,000/(250^2) + 6 = -6\). For \(x=300\), \(C' = -375,000/(300^2) + 6 = -4.167\). These values are in dollars per unit order size.

Interpret Results

The negative values of the rate of change suggest that as the order size increases, the cost per unit decreases. So, more specifically, when the order size is 200, the cost per unit decreases by $9.375 for each additional unit ordered. When the order size is 250, the cost per unit decreases by $6 for each additional unit ordered. When the order size is 300, the cost per unit decreases by $4.167 for each additional unit ordered.

Key concepts

Cost Function Differentiation

Understanding how business costs change as production scales is pivotal for economic decision-making. This is where cost function differentiation comes into play. A cost function expresses the cost of producing an item relative to the quantity produced, and by differentiating this function, businesses can determine the rate at which costs are changing with respect to production levels.

Differentiating the cost function provides the marginal cost - the cost of producing one more unit of a good. The derivative of cost with respect to the number of goods, symbolized as a prime notation like \(C'\), denotes how the total cost is expected to change as production quantity varies: essentially, how the costs will adjust if we were to produce one more (or one less) unit.

In our example, differentiating the cost function \(C=(375,000 + 6x^2) / x\) gives \(C' = -375,000/x^2 + 6\), where the first term becomes diminutive as the order size \(x\) gets larger. This diminishing term reflects economies of scale, indicating that higher production can dilute fixed costs across more units, reducing the cost per unit.

Calculus in Economics

Calculus, particularly the field of derivatives, is a powerful tool in the arsenal of economists. It allows the analysis of how one economic quantity affects another and how these relationships evolve with changes in market conditions or business strategies.

In the context of understanding cost functions, calculus helps to explore the responsiveness of costs to changes in output levels. It is precisely this responsiveness—how quickly and to what extent costs adjust to changes in production—that is crucial for businesses to optimize production and pricing strategies.

For example, by studying the derivative of the cost function from our problem, which reflects the rate of change of the cost per unit, managers can make informed decisions on how much to produce to minimize costs and maximize profit. It provides a quantitative, actionable insight into the cost dynamics as production capacity is adjusted.

Applying Derivatives

Applying derivatives, in mathematical terms, means utilizing the derivative of a function to solve practical problems. Derivatives measure how a function's output value changes as the input value changes, and this concept is central to various fields, including physics, engineering, and, notably, economics.

The exercise involves applying derivatives to find how the unit cost changes with respect to the order size in a manufacturing scenario. By calculating the derivative \(C'\) of the given cost function and evaluating it at different production levels \(x=200, 250, 300\), we gain tangible figures (\(C'=-9.375, -6, -4.167\)) that reflect the rate of cost reduction per additional unit produced at those levels.

Thus, derivatives allow businesses to predict and plan for optimal production rates. They help in fine-tuning the balance between unit cost and order size for the most cost-effective operation. Real-life applications of derivatives include optimizing inventory levels, setting appropriate pricing strategies, and even informing investment decisions within financial markets.