Problem 48 - Single Variable Calculus

Question

Optimal Inventory Control The equation $$ A(q)=\frac{k m}{q}+c m+\frac{h q}{2} $$ gives the annual cost of ordering and storing (as yet unsold) merchandise. Here, \(q\) is the size of each order, \(k\) is the cost of placing each order, \(c\) is the unit cost of the product, \(m\) is the number of units of the product sold per year, and \(h\) is the annual cost for storing each unit. Determine the size of each order such that the annual cost \(A(q)\) is as small as possible.

Step-by-Step Solution

Answer

In conclusion, to minimize the annual cost of ordering and storing merchandise, the optimal size of each order should be: $$ q = \sqrt{\frac{2km}{h}} $$ This will allow the business owner to make the most efficient use of their resources and reduce overall expenses.

Step 1: Find the first derivative of A(q) with respect to q

To start, we need to take the derivative of the function with respect to \(q\). Recall that the derivatives of a constant, and constants multiplied by a variable are constant, and the derivatives of a variable to the power of \(n\) is \(n \cdot variable^{n-1}\). Using these rules, we get: $$ \frac{dA(q)}{dq} = -\frac{km}{q^2} + 0 + \frac{h}{2} $$

Step 2: Set the first derivative equal to zero to find the critical points

To find the critical points, we need to set the first derivative equal to zero and solve for \(q\): \[ -\frac{km}{q^2} + \frac{h}{2} = 0 \] Now, solve for \(q\), first by adding \(\frac{km}{q^2}\) to both sides: \[ \frac{km}{q^2} = \frac{h}{2} \] Next, we can multiply both sides by \(2q^2\) to eliminate the fractions: \[ 2km = hq^2 \] Finally, we can solve for \(q\) by dividing both sides by \(h\) and taking the square root: \[ q = \sqrt{\frac{2km}{h}} \]

Step 3: Determine if the critical point found is a minimum

To confirm that this critical point corresponds to a minimum value for the cost function, we need to analyze the second derivative. Specifically, if the second derivative at this point is positive, the point will correspond to a minimum for the cost function. To find the second derivative, we take the derivative of the first derivative with respect to \(q\): $$ \frac{d^2A(q)}{dq^2} = \frac{2km}{q^3} $$ Notice that the second derivative does not depend on the constants, and it is positive for all positive values of \(q\). Thus, the critical point we found in step 2 corresponds to a minimum cost.

Step 4: Conclusion

We found that for the given problem, the optimal size of each order to minimize the annual cost is: $$ q = \sqrt{\frac{2km}{h}} $$ This result gives the business owner the information needed to minimize the annual costs of ordering and storing (as yet unsold) merchandise.

Key Concepts

Calculus

Calculus is a field of mathematics that helps us understand changes. It's about investigating how things grow or shrink. In our exercise, calculus is used to find the most cost-efficient way to order inventory. We evaluate the function representing the annual cost to determine how it changes with different order sizes.

• Calculus concepts like differentiation (finding derivatives) are crucial to solving real-world optimization problems.
• By using calculus, businesses can determine the best order size that minimizes costs and maximizes efficiency.

Calculus breaks down complex relationships into simpler steps. This helps businesses make better decisions by understanding how different factors like order size affect costs. So, calculus is not just about numbers but about making informed choices.

Derivative

A derivative represents the rate at which a quantity changes with respect to another. In simplest terms, it's like the "speed" of change of the function. For this exercise, the derivative of the cost function with respect to the order quantity (\( q \)) tells us how cost changes as order size changes.

• The first derivative, denoted as \( \frac{dA(q)}{dq} \), helps us find intervals of increase or decrease in the cost.
• Understanding where this derivative equals zero (and the behavior around it) can pinpoint where costs may be minimized.

Think of the derivative as a tool for "reading" a function's behavior. It lays the groundwork for identifying the most financially beneficial strategy in inventory management.

Critical Points

Critical points are places on a graph where the derivative is zero or undefined. These points are important because they can represent maximums, minimums, or flat spots on the curve. In this problem, the critical point helps us find the order size that potentially minimizes the cost.

• To find critical points, set the first derivative to zero and solve for the variable, here the order size \( q \).
• In practical terms, this means calculating when the rate of cost change equals zero, indicating a possible minimum.

Critical points provide insight into where the best outcomes lie. By verifying if these points result in the lowest costs, businesses can effectively manage inventory levels, reducing unnecessary expenses.

Second Derivative Test

The Second Derivative Test is used to confirm whether critical points are minima, maxima, or neither. It's like a "double-check" to ensure the solution is not just theoretically correct but also practically applicable.

• The second derivative, \( \frac{d^2A(q)}{dq^2} \), indicates concavity. A positive value suggests the function is concave up, confirming a minimum.
• If the second derivative is negative, the function is concave down, signifying a maximum at that point.

In our context, after finding a critical point, checking the sign of the second derivative helps ensure that the order size indeed minimizes costs. This reliable method provides confidence in business strategy, securing cost savings and operational efficiency.