Problem 44 - College Algebra and Calculus: An Applied Approach

Question

The annual inventory cost for a manufacturer is given by \(C=1,008,000 / Q+6.3 Q\) where \(Q\) is the order size when the inventory is replenished. Find the change in annual cost when \(Q\) is increased from 350 to 351 , and compare this with the instantaneous rate of change when \(Q=350\).

Step-by-Step Solution

Answer

The actual change in cost when Q increases from 350 to 351 will be calculated in Step 1. The instantaneous rate of change in cost when Q = 350 will be calculated in Step 3. Comparing these two results in Step 4 will give the final result.

Step 1: Calculate the Difference in Annual Costs

The first part of the problem involves finding the difference in annual costs when Q increases from 350 to 351.\nC(Q) = 1,008,000/Q + 6.3*Q\nSo the cost at Q = 350 is: C(350) = 1,008,000/350 + 6.3*350\nAnd the cost at Q = 351 is: C(351) = 1,008,000/351 + 6.3*351\nThen calculate the difference: ∆C = C(351) - C(350)

Step 2: Calculate the Derivative

The second part of the problem requires the derivative of the function, which represents the instantaneous rate of change.\nDifferentiate the cost function C with respect to Q to get the rate of change: C'(Q) = -1,008,000/Q² + 6.3.

Step 3: Evaluate the Derivative at Q = 350

Next, evaluate the derivative at the point Q = 350. So the instantaneous rate of change when Q = 350 is: C'(350) = -1,008,000/(350)^2 + 6.3.

Step 4: Comparison

Compare both results (∆C and C'(350)) to see how the actual change when Q increases from 350 to 351 compares to the instantaneous rate of change at Q = 350.

Key Concepts

Inventory Cost Analysis

Inventory cost analysis is a crucial aspect for manufacturers as it helps optimize the ordering and storage of inventory. The given function for the inventory cost, \[ C = \frac{1,008,000}{Q} + 6.3 \cdot Q \] combines two main components:

• The term \( \frac{1,008,000}{Q} \) represents the holding cost, which decreases as order size \( Q \) increases. This is because ordering larger quantities reduces the frequency of placing orders, thereby lowering holding costs.
• The term \( 6.3 \cdot Q \) denotes the ordering cost, which increases linearly with \( Q \). This reflects the cost associated with purchasing and managing higher quantities of inventory at once.

Understanding these parts allows the manufacturer to balance their ordering strategy. Finding the ideal order size \( Q \) that minimizes the total cost involves careful calculation to ensure both holding and ordering costs are considered. Thus, inventory cost analysis allows businesses to make financially sound decisions about purchasing and storing goods, directly impacting profitability.

Instantaneous Rate of Change

In the context of differential calculus, the instantaneous rate of change provides insight into how a function behaves at a precise point. For the manufacturer, understanding the instantaneous rate of change of the inventory cost at a specific order size \( Q \) helps in predicting how small shifts in \( Q \) may impact overall costs.It is equivalent to finding the slope of the tangent to the cost curve at a given point. For our inventory cost example:

• The derivative \( C'(Q) = -\frac{1,008,000}{Q^2} + 6.3 \) serves as our tool for calculating the instantaneous rate of change.
• Evaluating this derivative at \( Q = 350 \) means finding the rate at which costs change with respect to changes in order size at that exact point, without making any actual changes to \( Q \).

Conclusively, by determining the instantaneous rate of change, manufacturers gain useful predictive capabilities for their inventory cost management, efficiently informing decision-making and logistical planning.

Derivative Calculation

Calculating the derivative of a function is one of the foundations of differential calculus. It shows us how a function changes as its input changes. In the inventory cost scenario, the derivative \( C'(Q) \) indicates how sensitive the total cost \( C \) is with respect to changes in order size \( Q \).To calculate the derivative of our cost function \( C(Q) = \frac{1,008,000}{Q} + 6.3Q \), follow these steps:

• The derivative of \( \frac{1,008,000}{Q} \) is computed using the rule for derivatives of functions of the form \( \frac{a}{x} \), resulting in \( -\frac{1,008,000}{Q^2} \).
• The derivative of \( 6.3Q \) is straightforward, resulting in 6.3, since the derivative of \( a \cdot x \) is simply \( a \).

Combining these gives the derivative: \[ C'(Q) = -\frac{1,008,000}{Q^2} + 6.3 \] Evaluating this derivative at a specific point, like \( Q = 350 \), provides the actual rate at which the cost is changing at that moment. This calculation is crucial for understanding the dynamic nature of cost in response to order size changes, and is particularly helpful in marginal analysis and optimization.