Problem 15
Question
A reasonably realistic model of a firm's costs is given by the short-run Cobb- Douglas cost curve $$ C(q)=K q^{1 / \alpha}+F $$ where \(a\) is a positive constant, \(F\) is the fixed cost, and \(K\) measures the technology available to the firm. (a) Show that \(C\) is concave down if \(a>1\) (b) Assuming that \(a<1,\) find what value of \(q\) minimizes the average cost.
Step-by-Step Solution
Verified Answer
(a) Show: Concave down if \(a>1\); (b) Minimize AC: \(q = \left( \frac{F}{K (\frac{1}{\alpha} - 1)} \right)^{\frac{\alpha}{1 - \alpha}}\).
1Step 1: Analyze the Concavity Condition
To analyze the concavity of the function, we need to evaluate the second derivative of the cost function, \(C(q) = K q^{1/\alpha} + F\). We first find the first derivative \(C'(q)\). Differentiating, we get:\[ C'(q) = K \cdot \frac{1}{\alpha} \cdot q^{1/\alpha - 1} \]Next, find the second derivative \(C''(q)\):\[ C''(q) = K \cdot \frac{1}{\alpha} \cdot \left( \frac{1}{\alpha} - 1 \right) \cdot q^{1/\alpha - 2} \]The concavity of \(C(q)\) is determined by the sign of \(C''(q)\). For \(a>1\), since \(\frac{1}{\alpha} < 1\), \(C''(q) < 0\), proving that \(C(q)\) is concave down.
2Step 2: Set up the Average Cost Function
The average cost \(AC(q)\) is given by dividing the total cost \(C(q)\) by output \(q\):\[ AC(q) = \frac{C(q)}{q} = \frac{K q^{1/\alpha} + F}{q} = K q^{1/\alpha - 1} + \frac{F}{q} \]
3Step 3: Find the Minimum Average Cost
To find the value of \(q\) that minimizes the average cost \(AC(q)\), take the derivative \(AC'(q)\) and set it to zero:\[ AC'(q) = K \left( \frac{1}{\alpha} - 1 \right) q^{1/\alpha - 2} - \frac{F}{q^2} \]Set \(AC'(q) = 0\) for minimization:\[ K \left( \frac{1}{\alpha} - 1 \right) q^{1/\alpha - 2} - \frac{F}{q^2} = 0 \]Solve for \(q\) by isolating \(q^{1/\alpha - 2}\):\[ K \left( \frac{1}{\alpha} - 1 \right) q^{1/\alpha - 2} = \frac{F}{q^2} \]Simplifying, we find:\[ q^{1/\alpha} = \left( \frac{F}{K (\frac{1}{\alpha} - 1)} \right)^{\frac{\alpha}{1 - \alpha}} \]Thus, the value of \(q\) that minimizes the average cost is:\[ q = \left( \frac{F}{K (\frac{1}{\alpha} - 1)} \right)^{\frac{\alpha}{1 - \alpha}} \]
Key Concepts
Concavity AnalysisAverage Cost MinimizationSecond Derivative Test
Concavity Analysis
The concept of concavity in a function describes the direction in which the curve bends. When we say a graph is concave down, it means that it forms a U-shape that opens downward. For the Cobb-Douglas cost function, understanding its concavity is crucial in analyzing cost behavior and optimization. To determine concavity, we look at the cost function given by \(C(q) = K q^{1/\alpha} + F\). The key tool here is the second derivative, \(C''(q)\). - The first step is finding the first derivative: - \(C'(q) = K \cdot \frac{1}{\alpha} \cdot q^{1/\alpha - 1}\). - Next, calculate the second derivative: - \(C''(q) = K \cdot \frac{1}{\alpha} \cdot \left(\frac{1}{\alpha} - 1\right) \cdot q^{1/\alpha - 2}\). The sign of \(C''(q)\) dictates the concavity:- If \(C''(q) < 0\), the function is concave down; this occurs when \(\alpha > 1\), because it assures that the term \(\frac{1}{\alpha} - 1 \) becomes negative. Understanding this concept helps visualize how costs decrease with changes in production, vital for strategic decision making.
Average Cost Minimization
Minimizing average cost is crucial for efficient resource utilization. It helps firms understand the scale of production that allows cost minimization per unit of output.For the given cost function \(C(q) = K q^{1/\alpha} + F\), the average cost (AC) function is expressed as:- \(AC(q) = \frac{C(q)}{q} = K q^{1/\alpha - 1} + \frac{F}{q}\).To find the point of minimum average cost, you: - Take the derivative of the average cost function: - \(AC'(q) = K \left( \frac{1}{\alpha} - 1 \right) q^{1/\alpha - 2} - \frac{F}{q^2}\) - Set the derivative equal to zero to find critical points: - \[ K \left( \frac{1}{\alpha} - 1 \right) q^{1/\alpha - 2} - \frac{F}{q^2} = 0 \]Solving this equation gives:- \[ q = \left( \frac{F}{K \left( \frac{1}{\alpha} - 1 \right)} \right)^{\frac{\alpha}{1 - \alpha}} \]This value of \(q\) reflects the production level where average costs are minimized, providing an optimal scale for decision making.
Second Derivative Test
The second derivative test is a mathematical tool used to confirm the nature of critical points found in a function. For our cost function, it's a method to ensure whether a point is a minimum or maximum. After identifying critical points from the first derivative of the average cost function, use the second derivative to test them:- If \(AC''(q) > 0\), then the function has a local minimum at that point. - If \(AC''(q) < 0\), then there's a local maximum. - If \(AC''(q) = 0\), the test is inconclusive.In this scenario, after differentiating the average cost and applying the second derivative test, confirm that the computed value of \(q\) indeed minimizes the average cost.The second derivative formula in the context is derived from differentiating the first derivative of the average cost:- It builds on steps finding \(AC'(q)\), thus ensuring accuracy and enlightenment on the curve's shape around \(q\). Understanding the second derivative test empowers one to confirm confidently the efficiency of production levels in minimizing costs.
Other exercises in this chapter
Problem 14
Find the critical points of the function and classify them as local maxima or local minima or neither. $$g(x)=x e^{-3 x}$$
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When production is \(2000,\) marginal revenue is \(\$ 4\) per unit and marginal cost is \(\$ 3.25\) per unit. Do you expect maximum profit to occur at a product
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Use the first derivative to find all critical points and use the second derivative to find all inflection points. Use a graph to identify each critical point as
View solution Problem 15
Find the critical points of the function and classify them as local maxima or local minima or neither. $$h(x)=x+1 / x$$
View solution