Problem 56
Question
One of the formulas for inventory management says that the average weekly cost of ordering, paying for, and holding merchandise is $$ A(q)=\frac{k m}{q}+c m+\frac{h q}{2} $$ where \(q\) is the quantity you order when things run low (shoes, radios, brooms, or whatever the item might be); \(k\) is the cost of placing an order (the same, no matter how often you order); \(c\) is the cost of one item (a constant); \(m\) is the number of items sold each week (a constant); and \(h\) is the weekly holding cost per item (a constant that takes into account things such as space, utilities, insurance, and security). Find \(d A / d q\) and \(d^{2} A / d q^{2}\).
Step-by-Step Solution
Verified Answer
\( \frac{dA}{dq} = -\frac{km}{q^2} + \frac{h}{2} \); \( \frac{d^2A}{dq^2} = \frac{2km}{q^3} \).
1Step 1: Understand the Function
The function given is \[ A(q) = \frac{k m}{q} + c m + \frac{h q}{2} \] Our task is to find the first and second derivatives of this function with respect to the variable \( q \). This function represents the average weekly cost of inventory management.
2Step 1: First Derivative
To find \( \frac{dA}{dq} \), differentiate each term of the function \( A(q) \) with respect to \( q \). - The first term \( \frac{km}{q} \) can be rewritten as \( km q^{-1} \). The derivative is \(-km q^{-2}\).- The second term \( cm \) is a constant, and its derivative is 0.- The third term \( \frac{hq}{2} \) simplifies to \( \frac{h}{2} q \), with derivative \( \frac{h}{2} \). Thus, the first derivative is:\[ \frac{dA}{dq} = -\frac{km}{q^2} + \frac{h}{2} \]
3Step 2: Second Derivative
Now, differentiate the first derivative obtained above to find \( \frac{d^2A}{dq^2} \).- The derivative of \( -\frac{km}{q^2} \) is found by differentiating \( -km q^{-2} \), resulting in \( 2km q^{-3} \).- The derivative of \( \frac{h}{2} \) is 0, as it is a constant.Therefore, the second derivative is:\[ \frac{d^2A}{dq^2} = \frac{2km}{q^3} \]
Key Concepts
First DerivativeSecond DerivativeOptimization in Calculus
First Derivative
The first derivative, often represented as \( \frac{dA}{dq} \), helps us understand how a function changes with respect to a variable. In this scenario, the function \( A(q) \) reflects how the average weekly cost varies with the quantity \( q \). Finding the first derivative means we're looking to understand the rate at which this cost changes as we adjust \( q \).
To compute the first derivative of \( A(q) = \frac{km}{q} + cm + \frac{hq}{2} \), we differentiate each term individually with respect to \( q \):
\[ \frac{dA}{dq} = -\frac{km}{q^2} + \frac{h}{2} \]
This derivative shows us how each change in \( q \) impacts the average weekly cost.
To compute the first derivative of \( A(q) = \frac{km}{q} + cm + \frac{hq}{2} \), we differentiate each term individually with respect to \( q \):
- Rewriting \( \frac{km}{q} \) as \( km q^{-1} \), its derivative becomes \(-km q^{-2}\).
- The term \( cm \) is constant, so its derivative is \( 0 \).
- For \( \frac{hq}{2} \), we simplify it to \( \frac{h}{2} q \), and its derivative is \( \frac{h}{2} \).
\[ \frac{dA}{dq} = -\frac{km}{q^2} + \frac{h}{2} \]
This derivative shows us how each change in \( q \) impacts the average weekly cost.
Second Derivative
The second derivative, denoted as \( \frac{d^2A}{dq^2} \), provides insight into the function's curvature or how the rate of change itself is changing. In inventory management, the second derivative helps determine whether increases or decreases in quantities result in diminishing or escalating average costs.
After finding the first derivative \( \frac{dA}{dq} = -\frac{km}{q^2} + \frac{h}{2} \), we take derivatives again:
\[ \frac{d^2A}{dq^2} = \frac{2km}{q^3} \]
This indicates how sharply the cost curve bends. If it is positive, it suggests that cost increases moderately for larger \( q \) values.
After finding the first derivative \( \frac{dA}{dq} = -\frac{km}{q^2} + \frac{h}{2} \), we take derivatives again:
- The derivative of \( -\frac{km}{q^2} \) involves taking the derivative of \( -km q^{-2} \), yielding \( 2km q^{-3} \).
- Since \( \frac{h}{2} \) is constant, its derivative remains \( 0 \).
\[ \frac{d^2A}{dq^2} = \frac{2km}{q^3} \]
This indicates how sharply the cost curve bends. If it is positive, it suggests that cost increases moderately for larger \( q \) values.
Optimization in Calculus
Optimization in calculus involves finding the maximum or minimum values of a function. For inventory management, this often means minimizing costs.
Using the first and second derivatives, we can determine the best quantity \( q \) to minimize the average weekly cost.
Using the first and second derivatives, we can determine the best quantity \( q \) to minimize the average weekly cost.
- The first derivative \( \frac{dA}{dq} = -\frac{km}{q^2} + \frac{h}{2} \) indicates the rate of change of cost with respect to \( q \), setting it to zero helps find critical points.
- Solving \( -\frac{km}{q^2} + \frac{h}{2} = 0 \) can reveal potential points for optimization.
- The second derivative \( \frac{d^2A}{dq^2} = \frac{2km}{q^3} \) aids in determining if a critical point is a minimum. A positive second derivative confirms a minimum.
Other exercises in this chapter
Problem 56
In Exercises \(53-58,\) find the value of \((f \circ g)^{\prime}\) at the given value of \(x\) . $$ f(u)=u+\frac{1}{\cos ^{2} u}, \quad u=g(x)=\pi x, \quad x=1
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Derivative of multiples Does knowing that a function \(g(t)\) is differentiable at \(t=7\) tell you anything about the differentiability of the function 3\(g\)
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Parallel tangents Find the two points where the curve \(x^{2}+x y+y^{2}=7\) crosses the \(x\) -axis, and show that the tangents to the curve at these points are
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In Exercises \(53-58,\) find the value of \((f \circ g)^{\prime}\) at the given value of \(x\) . $$ f(u)=\frac{2 u}{u^{2}+1}, \quad u=g(x)=10 x^{2}+x+1, \quad x
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