Question

The ordering and transportation cost \(C\) per unit (in thousands of dollars) of the components used in manufacturing a product is given by $$ C=100\left(\frac{200}{x^{2}}+\frac{x}{x+30}\right), \quad 1 \leq x $$ where \(x\) is the order size (in hundreds). Find the rate of change of \(C\) with respect to \(x\) for each order size. What do these rates of change imply about increasing the size of an order? Of the given order sizes, which would you choose? Explain. (a) \(x=10\) (b) \(x=15\) (c) \(x=20\)

Short answer

The rate of change of \(C\) with respect to \(x\), \(C'\), is negative for each of the given order sizes, which implies that the cost per unit decreases as the order size increases. Therefore, ordering at the largest order size of \(x = 20\) would be the most cost-efficient choice.

Step-by-step solution

Define the Cost Function

The function that describes the cost \(C\) per unit as a function of the order size \(x\) (in hundreds) is given by \(C=100\left(\frac{200}{x^{2}}+\frac{x}{x+30}\right)\).

Compute the Derivative

To find the rate of change of \(C\) with respect to \(x\), compute the derivative of the cost function, i.e., \(C'\). Use the quotient and chain rule to find the derivative. The derivative of \(C\) is then \(C' = 100\left(-\frac{400}{x^{3}} + \frac{30}{(x+30)^{2}}\right)\).

Evaluate the Derivative

Evaluate \(C'\) at \(x = 10, 15, 20\) to find the rates of change of the cost per unit for these specific order sizes. They are: \(C'(10) = 100\left(-\frac{400}{1000} + \frac{30}{1600}\right)\), \(C'(15) = 100\left(-\frac{400}{3375} + \frac{30}{2025}\right)\), \(C'(20) = 100\left(-\frac{400}{8000} + \frac{30}{2500}\right)\).

Interpret the Results

The rates of change represent how the cost per unit changes with the order size. If the rate is negative, the cost per unit decreases as the order size increases. If the rate is positive, the cost per unit increases as the order size increases. The smallest order size that produces a negative rate of change can be considered the most cost-efficient, as it results in a decrease in the unit cost with an increase in the order size.

Key concepts

Derivative Applications

In calculus, derivatives play a crucial role in understanding how a certain quantity changes in response to changes in another related quantity. Specifically, in practical applications such as business, physics, and engineering, derivatives help us determine rates of change. For instance, when examining the cost function of ordering and transporting manufacturing components, as highlighted in the exercise, deriving the rate of change of cost with respect to order size is essential.

The process involves computing the derivative of the cost function, denoted by \(C'\), where \(C\) is the cost and \(x\) is the order size. This rate of change, represented by the derivative, tells us how much extra cost is incurred for each additional hundred units ordered. In such economic contexts, a negative derivative esentially indicates that costs decrease as we increase the order quantity, potentially leading to economies of scale. Conversely, a positive derivative would imply that increasing order quantities would result in increased costs per unit. By comprehending these implications, businesses can make informed decisions regarding inventory and ordering strategies.

Optimization Problems in Calculus

Optimization is about finding the best possible solution within a given set of constraints. In the context of calculus, it typically refers to locating maximum or minimum values of a function. The exercise provided illustrates how this concept can be used to minimize cost. By finding where the derivative of the cost function equals zero, we can determine the order size that either maximizes or minimizes cost.

However, merely setting the derivative equal to zero does not always guarantee a minimum cost. It is necessary to conduct further tests, such as the second derivative test, to ensure that the critical point identified indeed corresponds to a minimum. In the absence of a minimum, one must examine the end behavior of the function or consider any constraints that may impact the decision, such as budget limits or storage capacity, to find the most practical and cost-effective order size.

Cost Function

A cost function in economics and business represents the cost of producing goods or services as a function of various factors, such as labor, materials, and order size. The function presented in the exercise, \(C=100\left(\frac{200}{x^2}+\frac{x}{x+30}\right)\), encapsulates the total cost per unit depending on the order size \(x\). Understanding a cost function's behavior through calculus enables businesses to predict expenses and make decisions that could reduce cost and maximize profit.

A critical insight provided by the cost function is the concept of diminishing returns. As order quantity increases, the cost per unit might initially decrease due to fixed costs being spread over more units. However, beyond a certain point, additional orders may not reduce the cost per unit as significantly, or might even start to increase costs due to inefficiencies or other constraints. Calculus helps identify these inflection points, thus aiding businesses in optimizing their order sizes to align with their financial and operational goals.