Problem 28 - 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 the ordering cost, \(bq\) is the storage cost; (b) The optimal order quantity is \(q = \sqrt{\frac{a}{b}}\).

Step 1: Identify Cost Components

Examine the expression for the total weekly cost \(C = \frac{a}{q} + bq\). Recognize that the term \(\frac{a}{q}\) involves dividing a constant by \(q\). This suggests it is inversely proportional to \(q\), which is characteristic of the ordering cost (as larger orders reduce ordering frequency). The term \(bq\) is directly proportional to \(q\), which is typical for storage costs (as more items require more storage space).

Step 2: Define the Problem

We need to find the value of \(q\) that minimizes the total cost \(C\). This typically involves finding the derivative of \(C\) with respect to \(q\), setting it to zero, and solving for \(q\).

Step 3: Differentiate the Cost Function

Compute the derivative of the cost function \(C\) with respect to \(q\): \[\frac{dC}{dq} = \frac{d}{dq}\left(\frac{a}{q} + bq\right) = -\frac{a}{q^2} + b.\]

Step 4: Solve for Critical Points

Set the derivative equal to zero to find the critical points:\[-\frac{a}{q^2} + b = 0.\]Rearrange and solve for \(q\):\[b = \frac{a}{q^2}\]\[bq^2 = a\]\[q^2 = \frac{a}{b}\]\[q = \sqrt{\frac{a}{b}}.\]

Step 5: Verify Minimum Cost Condition

Check the second derivative to confirm that this value of \(q\) is a minimum. The second derivative of \(C\) with respect to \(q\) is:\[\frac{d^2C}{dq^2} = \frac{2a}{q^3}.\]Since \(\frac{2a}{q^3} > 0\) for \(q > 0\), the critical point is a minimum.

Key Concepts

Cost Minimization

Cost minimization is a key component of many business strategies, as it focuses on reducing expenses while maintaining the necessary output level. In our context of a cement warehouse, the goal is to find the optimal order size that minimizes total costs, which include both ordering and storage costs.

Ordering costs are associated with the frequency of placing orders. Larger orders mean fewer transactions and therefore, lower ordering costs per unit. However, this gives rise to an increase in storage costs, as larger inventories require more space.

The total cost function given by the warehouse is \(C = \frac{a}{q} + bq\), where \(\frac{a}{q}\) captures the inverse relationship of ordering cost to order size, and \(bq\) represents the storage cost, which increases linearly with the amount stored.

Therefore, cost minimization in this scenario involves balancing between these two types of costs to find the order size \(q\) that results in the lowest possible total cost.

Differentiation

Differentiation is a mathematical process used to find the rate at which a function changes at any given point. In this exercise, differentiation helps us identify the order size \(q\) that minimizes the total cost \(C(q)\).

By computing the first derivative of the cost function \(C = \frac{a}{q} + bq\), we aim to find where this rate of change is zero, which helps in identifying critical points where the minimum cost might occur: \[ \frac{dC}{dq} = -\frac{a}{q^2} + b. \]

This derivative indicates how the total cost changes concerning any changes in the order size. Setting this derivative equal to zero allows us to solve for \(q\), highlighting where potential cost minimization occurs.

Once the differentiation and simplification are complete, solving for \(q\) will help determine our cost-minimizing strategy.

Critical Points

Critical points are values at which a function's derivative is zero or undefined, indicating potential maximum, minimum, or saddle points. In optimization problems, finding these points is crucial for determining the extremum, whether it be a highest or lowest value.

To find the critical point of the total cost function \(C = \frac{a}{q} + bq\), we set its first derivative equal to zero and solve for \(q\): \[-\frac{a}{q^2} + b = 0.\] Solving this yields:\[q^2 = \frac{a}{b} \Rightarrow q = \sqrt{\frac{a}{b}}.\]

This specific value of \(q\) represents the order quantity at which the total cost is potentially minimized. Checking the second derivative ensures that this critical point is indeed a minimum rather than a maximum or inflection point, crucial for confirming our cost-saving strategy.

Economic Order Quantity

The Economic Order Quantity (EOQ) model is a common approach used by businesses to determine the optimal order size that minimizes total inventory costs, encompassing both ordering and holding costs.

In this exercise, the EOQ represents the order size \(q\) that minimizes the cost function \(C = \frac{a}{q} + bq\). With the formula derived for \(q\), we see that:\[q = \sqrt{\frac{a}{b}}.\]

This formula results directly from balancing the trade-off between ordering cost and holding cost. The EOQ is a cornerstone of inventory management theory, providing a simplified yet effective method for cost control.

By applying this model, the cement warehouse can ensure they are ordering the optimal amount of cement to minimize expenses associated with inventory management, keeping both storage and ordering costs in check efficiently.