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 approximately 8.37 units of labor and 23.77 units of capital for a minimum cost of 6,592.

Step 1: Write the Cost Function

The cost to produce units is given by the sum of the costs for labor and capital. Therefore, the cost function is:\[ C = 100L + 200K \]

Step 2: Implement the Quantity Constraint

We use the production constraint that states the quantity of products, \( Q \), is 36,000 units:\[ 900 L^{1/2} K^{2/3} = 36000 \]

Step 3: Simplify the Constraint Equation

Solve the equation from Step 2 for one of the variables to simplify it:1. Divide both sides by 900: \[ L^{1/2} K^{2/3} = 40 \]2. To express one variable in terms of the other, let's solve for \( K \): \[ K^{2/3} = \frac{40}{L^{1/2}} \] Thus, \[ K = \left(\frac{40}{L^{1/2}}\right)^{3/2} \]

Step 4: Substitute into the Cost Function

Substitute the expression for \( K \) into the cost function:\[ C = 100L + 200 \left(\frac{40}{L^{1/2}}\right)^{3/2} \]Simplify the expression further to find the minimum cost.

Step 5: Differentiate the Cost Function

Differentiate the cost function with respect to \( L \) and set the derivative equal to zero to find the critical points. Let\[ f(L) = 100L + 200 \left(\frac{40}{L^{1/2}}\right)^{3/2} \]Find \( \frac{df}{dL} \) and solve \( \frac{df}{dL} = 0 \) to find the value of \( L \).

Step 6: Solve for the Optimal Quantity of L

Solve the equation from Step 5 for \( L \):1. Compute the derivative: \[ f'(L) = 100 - 200 \cdot \frac{60}{L^{5/2}} \]2. Set \( f'(L) = 0 \). Solve for \( L \): \[ 100 = \frac{12000}{L^{5/2}} \] \[ L^{5/2} = 120 \] \[ L \approx 8.37 \]

Step 7: Calculate the Corresponding K

Substitute the value of \( L \) back into the equation for \( K \):\[ K = \left(\frac{40}{(8.37)^{1/2}}\right)^{3/2} \]Calculate \( K \), which approximately gives \( K \approx 23.77 \).

Step 8: Calculate the Minimum Cost

Substitute the values of \( L \) and \( K \) back into the cost function to get the minimum cost:\[ C = 100(8.37) + 200(23.77) \]Calculate \(C\) to find the minimum cost.

Key Concepts

Production Function

A production function is a mathematical equation that describes the relationship between inputs and outputs in the production process. In this particular exercise, the production function is given by:\[Q = 900 L^{1/2} K^{2/3} \]This function tells us how the quantity of output, \(Q\), is dependent on the inputs used, specifically labor \(L\) and capital \(K\). The exponents \(1/2\) and \(2/3\) signify the marginal productivity of labor and capital, meaning how output changes with a change in each input.
- The exponent \(1/2\) on \(L\) indicates that there are diminishing returns to labor. As more labor is used, the additional output produced from extra labor decreases.
- Similarly, the exponent \(2/3\) for \(K\) signifies diminishing marginal returns to capital.
Overall, the production function helps to represent the technological capabilities of a firm in transforming labor and capital into goods.

Labor and Capital

Labor and capital are the two crucial inputs in the production function. They determine how much a company can produce. In economic theory, converting these inputs into outputs at the least possible cost is vital.

• **Labor** is the human work/time put into the production process. In our exercise, labor costs \(\\(100\) per unit, which indicates the cost associated with hiring workers.
• **Capital** encompasses tools, machinery, and buildings. Here, the unit cost is \(\\)200\), higher than labor, reflecting the expense of using these tangible resources.

To achieve cost minimization, the right balance of labor and capital must be determined by considering their respective costs and productivity contributions. This balancing act is crucial for managing a firm’s resources efficiently while meeting output levels.

Differential Calculus

Differential calculus is a branch of mathematics focused on rates of change and slopes of curves. In optimization problems like ours, it is used to find values that minimize or maximize certain quantities.
In our problem, we differentiate the cost function, with respect to labor, \(L\), to find its critical points:\[ f(L) = 100L + 200 \left(\frac{40}{L^{1/2}}\right)^{3/2} \]Differentiating helps us determine where the slope of the cost function equals zero, signifying a potential minimum or maximum point:\[ f'(L) = 100 - 200 \cdot \frac{60}{L^{5/2}} \]Setting \( f'(L) = 0 \) allows us to solve for \(L\). This is a crucial step in finding where production costs can potentially be minimized, thus aiding in effective decision-making for businesses aiming to produce given quantities economically.

Optimization Problem

Optimization involves finding the best solution—achieving maximum output with minimum input, or the opposite. Our exercise displays this by asking us to produce a set amount of goods, 36,000 units, at the least total cost.
To solve this optimization problem, you need to:

• Define the cost function that adds total labor and capital costs.
• Use the production constraint—ensuring 36,000 units are produced.
• Mathematically manipulate the constraint to express one variable in terms of the other, simplifying the function to a single variable equation.
• Differentiate and solve for the critical points where cost is minimized.

By resolving the above steps, we achieve an efficient input mix of labor and capital that results in the least expense to meet the production target. This process exemplifies the power of optimization in economic resource management.