Problem 54 - Calculus Early Transcendentals

Question

Production functions Economists model the output of manufacturing systems using production functions that have many of the same properties as utility functions. The family of Cobb-Douglas production functions has the form \(P=f(K, L)=C K^{a} L^{1-a},\) where K represents capital, L represents labor, and C and a are positive real numbers with \(0

Step-by-Step Solution

Answer

#Answer# The optimal allocation for capital (K) and labor (L) that maximizes production is K = 4 and L = 2.

Step 1: Write down the production function and constraint

We are given the Cobb-Douglas production function: $$P = 10K^{\frac{1}{3}}L^{\frac{2}{3}}$$ and the constraint equation: $$30K + 60L = 360$$

Step 2: Create the Lagrangian function

The Lagrangian function, L, combines the production function and the constraint with a Lagrange multiplier, λ: $$\mathcal{L}(K, L, \lambda) = 10K^{\frac{1}{3}}L^{\frac{2}{3}} - \lambda(30K + 60L - 360)$$

Step 3: Compute the partial derivatives

Now, we need to compute the partial derivatives of the Lagrangian function with respect to K, L, and λ: $$\frac{\partial\mathcal{L}}{\partial K} = \frac{10}{3}K^{-\frac{2}{3}}L^{\frac{2}{3}} - 30\lambda = 0$$ $$\frac{\partial\mathcal{L}}{\partial L} = \frac{20}{3}K^{\frac{1}{3}}L^{-\frac{1}{3}} - 60\lambda = 0$$ $$\frac{\partial\mathcal{L}}{\partial \lambda} = 30K + 60L - 360 = 0$$

Step 4: Solve the partial derivative equations

We need to solve these equations to find the values of K, L, and λ. First, let's solve the first two equations for λ: $$\lambda = \frac{1}{30} \cdot \frac{10}{3}K^{-\frac{2}{3}}L^{\frac{2}{3}} = \frac{1}{60} \cdot \frac{20}{3}K^{\frac{1}{3}}L^{-\frac{1}{3}}$$ Now we can combine the expressions for λ and solve for K in terms of L: $$30K^{-\frac{2}{3}}L^{\frac{2}{3}} = 20K^{\frac{1}{3}}L^{-\frac{1}{3}}$$ $$L = K^{\frac{1}{2}}$$

Step 5: Substitute the expression for L in the constraint equation

Now we can substitute the expression for L in terms of K into the constraint equation: $$30K + 60(K^{\frac{1}{2}}) = 360$$

Step 6: Solve for K

Solve the equation above for K: $$K = 4$$

Step 7: Find the value of L

Substitute the value of K into the expression for L: $$L = (4)^{\frac{1}{2}} = 2$$

Step 8: Interpret the results

The optimal allocation for capital (K) and labor (L) that maximizes production is K = 4 and L = 2.

Key Concepts

Lagrange multipliers

Lagrange multipliers are a valuable tool in optimization, especially when dealing with functions subject to constraints. Imagine you're trying to maximize or minimize a function, but you have some restrictions you can't ignore. For instance, consider a production function that you want to maximize given a limited budget for resources.

The main idea behind Lagrange multipliers is to transform a constrained problem into an unconstrained one by introducing an auxiliary variable called the Lagrange multiplier, commonly denoted as \( \lambda \). This multiplier adjusts the objective function by incorporating the constraint, allowing us to find critical points where the gradient of the function aligns with the gradient of the constraint.

In our specific problem, the Lagrangian function is created by combining the production function and the budget constraint. It shows how changes in capital and labor, through derivatives, relate to the optimal allocation of resources.

Optimization under constraints

Optimization under constraints is the process of finding maximum or minimum values of a function while adhering to certain restrictions. In many real-world scenarios, resources are finite, so we must consider limitations when optimizing.

The Cobb-Douglas production function is a classic example used in economics to illustrate optimization under constraints. Here, the goal is to determine the best allocation of capital and labor to maximize output.

• Constraints often take the form of linear equations, representing budgets or availability limits, such as \( 30K + 60L = 360 \) in our exercise.
• The solution requires balancing the production function against these constraints, ensuring resources are used efficiently.

In practice, this involves creating and solving equations for critical variables, like in our step-by-step solution, where we solved for optimal values of \( K \) and \( L \).

Partial derivatives

Partial derivatives allow us to examine how a function changes with respect to one variable at a time, holding others constant. They are crucial for understanding the behavior and relationships within multivariable functions, such as the Cobb-Douglas production function.

In the problem of maximizing production, we start by finding the partial derivatives of the Lagrangian with respect to each variable: \( K \), \( L \), and \( \lambda \).

• These derivatives indicate the rate of change of the Lagrangian function along each respective dimension.
• The critical points, which we find by setting these partial derivatives to zero, represent potential optimal solutions under the given constraint.

Through these calculations, we derive expressions such as those for \( \lambda \), and ultimately solve for the specific values of \( K \) and \( L \) that maximize the production function while satisfying the budget constraint.