Problem 27 - Applied Calculus

Question

A warehouse selling cement has to decide how often and in what quantities to reorder. It is cheaper, on average, to place large orders, because this reduces the ordering cost per unit. On the other hand, larger orders mean higher storage costs. The warehouse always reorders cement in the same quantity, \(q .\) The total weekly cost, \(C,\) of ordering and storage is given by \(C=\frac{a}{q}+b q, \quad\) where \(a, b\) are positive constants. (a) Which of the terms, \(a / q\) and \(b q,\) represents the ordering cost and which represents the storage cost? (b) What value of \(q\) gives the minimum total cost?

Step-by-Step Solution

Answer

(a) \(\frac{a}{q}\) is ordering cost; \(bq\) is storage cost. (b) Minimum cost occurs at \(q = \sqrt{\frac{a}{b}}\).

Step 1: Identify the Terms

The expression \(C=\frac{a}{q}+bq\) is given for analyzing costs. Here, the term \(\frac{a}{q}\) is inversely related to \(q\), indicating that as the order size \(q\) increases, this cost decreases. This is characteristic of ordering costs, which per unit decrease with larger orders. The term \(bq\) is directly proportional to \(q\), suggesting it increases with order size, indicative of storage costs.

Step 2: Differentiate the Cost Function

To find the value of \(q\) that minimizes the total cost \(C\), we first need to differentiate the cost function with respect to \(q\). The function is \(C = \frac{a}{q} + bq\). Differentiating gives us: \(\frac{dC}{dq} = -\frac{a}{q^2} + b\).

Step 3: Set the Derivative to Zero

For optimization, set the derivative equal to zero and solve for \(q\). Doing so, \(-\frac{a}{q^2} + b = 0\). This simplifies to \(b = \frac{a}{q^2}\).

Step 4: Solve for q

Rearrange \(b = \frac{a}{q^2}\) to solve for \(q\). Multiplying both sides by \(q^2\) gives \(bq^2 = a\). Solving for \(q\), we get \(q = \sqrt{\frac{a}{b}}\). This is the optimal order quantity that minimizes the total cost.

Key Concepts

Cost Function

In calculus optimization, a cost function is used to model the costs associated with different business operations. The problem involving a warehouse shows how costs can be optimized through careful analysis. Here, the total cost function is given as:

• \( C = \frac{a}{q} + bq \)

This cost function consists of two main components:

• Ordering Cost: \( \frac{a}{q} \)
• Storage Cost: \( bq \)

The ordering cost \( \frac{a}{q} \) decreases as the order quantity \( q \) increases. This is because placing larger orders reduces the cost per unit. Conversely, the storage cost \( bq \) grows with the size of the order, as storing more items demands more space and potentially higher expenses.
The objective is to find an order quantity \( q \) that minimizes the total cost. Balancing ordering and storage costs through such a function is crucial in ensuring efficient inventory management.

Derivatives

In this context, derivatives are essential for finding the minimum value of the cost function, which represents the optimal solution to our inventory issue. Using calculus, we differentiate the total cost function \( C = \frac{a}{q} + bq \) with respect to \( q \). This derivative is noted as:

• \( \frac{dC}{dq} = -\frac{a}{q^2} + b \)

Setting the derivative equal to zero is a method used in calculus optimization to find the points where the function reaches minimum or maximum values. Here, solving \( -\frac{a}{q^2} + b = 0 \) gives:

• \( b = \frac{a}{q^2} \)

This equality helps us to find the critical points that will indicate the minimized cost. Rearranging and solving gives the point \( q = \sqrt{\frac{a}{b}} \) as the order quantity for the minimal total cost.
Thus, derivatives act as powerful tools in calculus optimization, particularly in distinguishing an inventory management strategy that yields the least cost.

Inventory Management

Inventory management is a critical aspect of any business handling products or materials. The exercise demonstrated with the warehouse scenario highlights the balance between minimizing costs and maintaining efficient stock levels. The two primary costs in inventory management are:

• Ordering costs: Costs associated with placing orders, which decrease with fewer, larger orders.
• Storage costs: Expenses incurred by holding stock, which increase with order size.

The calculus optimization approach used in the exercise allows the business to determine the best order quantity. By calculating \( q = \sqrt{\frac{a}{b}} \), the company can consistently purchase optimal quantities that minimize the sum of the ordering and holding costs.
Effective inventory management ensures that the supply chain operates smoothly, costs are controlled, and customer demands are met without unnecessary expenditure on excessive stock. Employing optimization techniques in this domain enables businesses to achieve economic efficiency and avoid storage and ordering bottlenecks.