Problem 66
Question
Inventory Replenishment 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.
Step-by-Step Solution
Verified Answer
The rates of change at \(x=200\), \(x=250\), and \(x=300\) can be calculated using the derivative of the cost function. The value of the derivative at these points reveals whether the cost is increasing or decreasing at those order sizes.
1Step 1: Differentiate the Cost Function
Differentiate the function \(C=\left(375,000+6 x^{2}\right) / x\). Use power rule and quotient rule steps as applicable to get the derivative \(\frac{dC}{dx}\).
2Step 2: Find Rate of Change at x=200
Substitute \(x=200\) in the derivative \(\frac{dC}{dx}\) obtained from step 1. This will give the rate of change of cost at this order size.
3Step 3: Find Rate of Change at x=250
Substitute \(x=250\) in the derivative \(\frac{dC}{dx}\). This will give the rate of change of cost at this order size.
4Step 4: Find Rate of Change at x=300
Substitute \(x=300\) in the derivative \(\frac{dC}{dx}\). This will give the rate of change of cost at this order size.
5Step 5: Interpret the results
Negative rate of change means the cost is decreasing as the order size increases, while a positive rate signifies an increasing cost.
Key Concepts
Inventory ReplenishmentDifferentiationRate of ChangeQuotient RulePower Rule
Inventory Replenishment
Inventory replenishment is an essential part of supply chain management. It involves restocking the inventory to ensure that there is a sufficient supply of products or components available to meet customer demand or production needs. In mathematical terms, we analyze how changes in order size can impact costs. In this context, the cost per unit is given by the formula \(C=\frac{375,000+6 x^{2}}{x}\), where \(C\) is measured in dollars and \(x\) is the order size.
- Understanding this cost function helps businesses plan their inventory purchases efficiently.
- The primary goal is to minimize costs while maintaining adequate inventory levels.
Differentiation
Differentiation in calculus is the process of finding the rate at which a function changes at any given point. For our inventory cost function \(C\), we want to determine \(\frac{dC}{dx}\), the derivative, to understand how the cost per unit changes with varying order sizes.
- This derivative tells us the sensitivity of the cost to changes in order size.
- It answers how quickly and in what direction (increasing or decreasing) the cost changes.
Rate of Change
The rate of change is a concept that describes how one quantity changes in relation to another. For our exercise, the rate of change refers to \(\frac{dC}{dx}\), which quantifies how the cost per unit \(C\) changes as the order size \(x\) changes.
- A positive rate of change means that as the order size increases, the cost also increases.
- A negative rate of change indicates that an increase in order size leads to a decrease in cost.
Quotient Rule
The quotient rule is a technique used in differentiation when dealing with functions that are divided by each other. In our exercise, the cost function \(C\) is represented as a quotient \(\frac{375,000+6 x^{2}}{x}\). To differentiate it, we apply the quotient rule which states:
\[\frac{d}{dx}\left(\frac{u}{v}\right) = \frac{v\frac{du}{dx} - u\frac{dv}{dx}}{v^2}\]where \(u = 375,000+6x^2\) and \(v = x\).
\[\frac{d}{dx}\left(\frac{u}{v}\right) = \frac{v\frac{du}{dx} - u\frac{dv}{dx}}{v^2}\]where \(u = 375,000+6x^2\) and \(v = x\).
- This method enables us to handle more complex differentiation involving ratios effectively.
- Using the quotient rule helps in accurately determining how each component of the function impacts the overall rate of change.
Power Rule
The power rule is a fundamental tool in calculus for differentiating functions of the form \(x^n\). It simplifies the differentiation process by reducing it to a straightforward procedure: if \(f(x) = x^n\), then \(f'(x) = nx^{n-1}\). In the context of our problem, the term \(6x^2\) within the cost function \(C\) can be differentiated using the power rule.
- By applying the power rule, differentiating terms like \(6x^2\) resulting in \(12x\).
- The power rule accelerates the differentiation process for polynomial terms, making complex functions more manageable.
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