Question

Inventory Replenishment The ordering and transportation cost \(C\) for the components used in manufacturing a product is $$ C=100\left(\frac{200}{x^{2}}+\frac{x}{x+30}\right), \quad x \geq 1 $$ where \(C\) is measured in thousands of dollars and \(x\) is the order size in hundreds. Find the rate of change of \(C\) with respect to \(x\) when \((a) x=10,(b) x=15,\) and \((c) x=20 .\) What do these rates of change imply about increasing order size?

Short answer

The rate of change of \(C\) with respect to \(x\) when \(x = 10, 15, 20\) is \(-0.3 \), \(-0.16 \), and \(-0.11 \) thousand dollars per hundred units respectively. This suggests that although the costs are decreasing with higher order size (\(x\)), the decrease is slowing as \(x\) grows larger.

Step-by-step solution

Compute the derivative of \(C\) with respect to \(x\)

The first step is to find the derivative \(C'(x)\), which represents the rate of change of \(C\) with respect to \(x\). Due to the fact that the function \(C(x)\) consists of two terms, each of which is a quotient, we will need to use the quotient rule for differentiation which is \((f/g)' = (f'g - fg')/g^{2}\), with \(f(x)\) and \(g(x)\) being differentiable functions. The derivative of \(C\) with respect to \(x\) is calculated as follows: \[C'(x) = 100\left(\frac{-400}{x^{3}}+\frac{1}{x+30} - \frac{x}{(x+30)^{2}}\right)\] This simplifies to: \[C'(x) = \frac{-100x^{3}+3000x^{2}-36000x}{x^{2}(x+30)^{2}}\]

Evaluate \(C'(x)\) at \(x = 10\)

After calculating the derivative \(C'(x)\), we can start determining the rates of change by substituting the specified values of \(x\) into \(C'(x)\). Beginning with \(x = 10\), \[C'(10) = \frac{-100*10^{3}+3000*10^{2}-36000*10}{10^{2}(10+30)^{2}} = -0.3\]. So, when \(x = 10\), the rate of change of \(C\) with respect to \(x\) is \(-0.3\) thousand dollars per hundred units.

Evaluate \(C'(x)\) at \(x = 15\) and \(x = 20\)

Similarly, substitute \(x = 15\) and \(x = 20\) into \(C'(x)\) and compute the result. This gives: \[C'(15) = \frac{-100*15^{3}+3000*15^{2}-36000*15}{15^{2}(15+30)^{2}} = -0.16\]. And for \(C'(x = 20)\), \[C'(20) = \frac{-100*20^{3}+3000*20^{2}-36000*20}{20^{2}(20+30)^{2}} = -0.11\]. So, the rate of change of \(C\) with \(x\) is \(-0.16\) thousand dollars per hundred units when \(x = 15\), and \(-0.11\) thousand dollars per hundred units when \(x = 20\).

Interpret the result

The rate of change at these specific values suggests that the cost \(C\) is decreasing faster with higher order quantities. Specifically, the rate of cost decrease is fastest at \(x = 10\) and decreases as \(x\) gets larger. This implies that increasing the order size would lead to slower cost reductions.

Key concepts

Understanding Ordering Cost

Ordering cost refers to the expenses incurred each time an inventory order is placed. These costs can include expenses such as the paperwork processing fees, communication charges, and labor costs for handling the order.

In the context of this problem, the ordering cost is part of the overall cost function, given by the equation: \[ C = 100\left(\frac{200}{x^{2}}+\frac{x}{x+30}\right) \]This complex equation involves two main terms that incorporate both ordering and transportation costs.

Managing ordering costs effectively is crucial for a business because it directly impacts profitability. By optimizing the order size (in hundreds, denoted by \(x\)), a company can reduce these costs, making operations more efficient.

Exploring Transportation Cost

Transportation cost forms a significant portion of the total inventory replenishment cost. It includes all expenses required to move the inventory from the supplier to the manufacturing unit. These can involve:

• Freight charges
• Handling fees
• Insurance during transit

In the equation provided, the transportation cost is embedded in the term \(\frac{x}{x+30}\). The term \(x+30\) suggests that as order size \(x\) increases, the cost is influenced by an additional factor.

Thus, understanding how different order sizes modulate these transportation expenses is fundamental. The equation illustrates that as the order size increases, these costs might settle or continue scaling at a rate governed by \(x+30\). Knowing how transportation costs interact with ordering decisions is vital for logistical planning.

The Power of Differentiation

In calculus, differentiation is a powerful tool used to determine the rate at which one quantity changes with respect to another. Here, we use differentiation to find out how the cost \(C\) changes concerning the order quantity \(x\).

The derivative of a function gives us its rate of change, which helps in understanding how the cost dynamics evolve with varying \(x\). Applying the quotient rule, which is:\[ (f/g)' = \frac{f'g - fg'}{g^{2}} \]we differentiate the given expression for \(C\). Finding \(C'(x)\) is crucial because it helps identify the optimum order size that minimizes costs. By solving this derivative and substituting different values for \(x\), businesses can make informed decisions by analyzing the cost-benefit of different order sizes.

Differentiation thus plays an essential role in strategic inventory management.

Understanding Rate of Change

The rate of change tells us how rapidly or slowly a particular quantity varies with changes in another variable. For the given problem, we want to know how fast or slow the total cost \(C\) changes as the order size \(x\) fluctuates.

By computing \(C'(x)\), we evaluated the rate at which costs change for values \(x = 10\), \(x = 15\), and \(x = 20\), obtaining:

• \(C'(10) = -0.3\)
• \(C'(15) = -0.16\)
• \(C'(20) = -0.11\)

These negative values indicate that costs are decreasing as more units are ordered. However, the rate of decrease diminishes with larger order sizes, implying a slowing reduction in costs, suggesting diminishing returns.

Ultimately, understanding this rate of change aids in deciding the optimal level of stock to order, balancing cost savings with operational constraints.