Problem 14 - Applied Calculus

Question

The quantity, \(Q,\) of a certain product manufactured depends on the quantity of labor, \(L,\) and of capital, \(K,\) used according to the function $$Q=900 L^{1 / 2} K^{2 / 3}$$ Labor costs \(\$ 100\) per unit and capital costs \(\$ 200\) per unit. What combination of labor and capital should be used to produce 36,000 units of the goods at minimum cost? What is that minimum cost?

Step-by-Step Solution

Answer

Use 90 units of labor and 60 units of capital for a minimum cost of $21,000.

Step 1: Understand the Problem

We need to produce 36,000 units ( Q = 36,000 ) using a cost function that depends on labor ( L ) and capital ( K ). Labor costs $100 per unit and capital costs $200 per unit. Our goal is to find the combination of L and K that minimizes the cost while satisfying the production requirement.

Step 2: Define the Cost Function

The total cost, $C, is given by the sum of the costs of labor and capital:\[ C = 100L + 200K. \]

Step 3: Relate the Production Function

The production function is given as:\[ Q = 900 L^{1/2} K^{2/3}. \]Since we want to produce Q = 36,000 units, we have:\[ 36,000 = 900 L^{1/2} K^{2/3}. \]

Step 4: Simplify the Production Equation

Solve for L^{1/2} K^{2/3} by dividing both sides of the production equation by 900:\[ L^{1/2} K^{2/3} = \frac{36,000}{900} = 40. \]

Step 5: Use Lagrange Multipliers to Minimize Cost

To minimize the cost, consider the Lagrangian function:\[ \mathcal{L}(L, K, \lambda) = 100L + 200K + \lambda (40 - L^{1/2} K^{2/3}). \]Find partial derivatives, set them to zero, and solve:\[ \frac{\partial \mathcal{L}}{\partial L} = 100 - \frac{\lambda}{2} L^{-1/2} K^{2/3} = 0, \]\[ \frac{\partial \mathcal{L}}{\partial K} = 200 - \frac{2\lambda}{3} L^{1/2} K^{-1/3} = 0, \]\[ \frac{\partial \mathcal{L}}{\partial \lambda} = 40 - L^{1/2} K^{2/3} = 0. \]

Step 6: Solve the System of Equations

From the system of partial derivatives, solve the first two equations to eliminate \(\lambda\) and relate L and K:1. \[ 100 = \frac{\lambda}{2} L^{-1/2} K^{2/3}, \]2. \[ 200 = \frac{2\lambda}{3} L^{1/2} K^{-1/3}. \]Dividing the first equation by the second, we find:\[ \frac{3}{2} = \frac{L}{K}. \]Thus, K = \frac{2}{3} L.

Step 7: Substitute and Solve for L and K

Substitute K from Step 6 into the simplified production equation:\[ L^{1/2} \left(\frac{2}{3} L\right)^{2/3} = 40. \]This simplifies to:\[ \left(\frac{2}{3}\right)^{2/3} L = 40, \]Solving for L gives L = 90. Then K = \frac{2}{3} \times 90 = 60.

Step 8: Calculate the Minimum Cost

Now substitute the values of L and K back into the cost function:\[ C = 100 \times 90 + 200 \times 60 = 9000 + 12000 = 21,000. \]

Key Concepts

Cost Minimization

In any production environment, cost minimization is key to increasing efficiency and profitability. For businesses, the goal is to determine the optimal combination of resources that result in the desired output at the lowest possible cost. This involves analyzing the prices and quantities of inputs, such as labor and capital. In our example, labor costs \(100 per unit and capital costs \)200 per unit. The challenge is to use these resources to produce 36,000 units at minimum cost.

To tackle this, we represent the total cost, \( C \), as a combination of labor, \( L \), and capital, \( K \):

• Cost function: \( C = 100L + 200K \)

Finding the values of \( L \) and \( K \) that fulfill our production goals while minimizing costs will efficiently utilize resources and lower expenses.

Lagrange Multipliers

The Lagrange multiplier method is a strategic approach for constrained optimization problems, like our production situation. Since we want to minimize the cost while adhering to a production constraint, introducing a new variable, \( \lambda \), helps us form an equation that balances both objectives.

Construct the Lagrangian function:

• \( \mathcal{L}(L, K, \lambda) = 100L + 200K + \lambda (40 - L^{1/2} K^{2/3}) \)

This formulation combines the cost function and the production constraint. By taking the partial derivatives with respect to \( L \), \( K \), and \( \lambda \), setting them to zero, and solving the resulting equations, we forge a path to determining optimal resource usage. These steps ensure that both costs and production requirements are satisfied.

Production Function

A production function depicts the relationship between input combinations and the maximum output they can generate. For this example, we use a specific function: \( Q = 900 L^{1/2} K^{2/3} \). This function highlights a scenario where the output, \( Q \), depends on labor, \( L \), and capital, \( K \), in a non-linear manner.

In our problem, we aim for a production level of 36,000 units. By setting \( 36,000 = 900 L^{1/2} K^{2/3} \), we express our goal in terms of known quantities of labor and capital. Simplifying gives us \( L^{1/2} K^{2/3} = 40 \). This sets a guideline that any combination of labor and capital must satisfy to produce the desired output.

Partial Derivatives

Partial derivatives are fundamental to optimizing problems constrained by multiple variables. In the Lagrange method, these derivatives help us determine how to adjust each variable to maintain a balance between cost and output when fine-tuning towards optimal solutions.

In this scenario, the partial derivatives of the Lagrangian with respect to \( L \), \( K \), and \( \lambda \) yield the following equations:

• \( \frac{\partial \mathcal{L}}{\partial L} = 100 - \frac{\lambda}{2} L^{-1/2} K^{2/3} = 0 \)
• \( \frac{\partial \mathcal{L}}{\partial K} = 200 - \frac{2\lambda}{3} L^{1/2} K^{-1/3} = 0 \)
• \( \frac{\partial \mathcal{L}}{\partial \lambda} = 40 - L^{1/2} K^{2/3} = 0 \)

Solving these equations allows us to derive the relationship between labor and capital, ultimately helping calculate the amounts needed to achieve minimum production cost. This technique ensures that all components work in unison to meet production goals economically.