Question

Find the values of \(K\) and \(L\) that maximize the following production functions subject to the given constraint, assuming \(K \geq 0\) and \(L \geq 0\) $$P=f(K, L)=K^{1 / 2} L^{1 / 2} \text { for } 20 K+30 L=300$$

Short answer

Answer: The values of K and L that maximize the production function are K = 9 and L = 6, and the maximum production value is P = 3√6.

Step-by-step solution

State the Lagrangian function

In order to find the maximum of the production function given the constraint, we need to introduce the Lagrangian function, which is defined as $$\mathcal{L}(K, L, \lambda) = K^{1/2}L^{1/2} - \lambda(20K + 30L - 300)$$ where \(\lambda\) is a Lagrange multiplier.

Calculate the partial derivatives

To find the stationary points of the Lagrangian function, we need to find its partial derivatives with respect to \(K\), \(L\), and \(\lambda\): $$\frac{\partial \mathcal{L}}{\partial K} = \frac{1}{2}K^{-1/2}L^{1/2} - 20\lambda,$$ $$\frac{\partial \mathcal{L}}{\partial L} = \frac{1}{2}K^{1/2}L^{-1/2} - 30\lambda,$$ $$\frac{\partial \mathcal{L}}{\partial \lambda} = 20K + 30L - 300.$$

Set the partial derivatives to zero

The stationary points occur when the partial derivatives are equal to zero. This gives us a system of three equations with three unknowns (\(K\), \(L\), and \(\lambda\)): $$\frac{1}{2}K^{-1/2}L^{1/2} - 20\lambda = 0,$$ $$\frac{1}{2}K^{1/2}L^{-1/2} - 30\lambda = 0,$$ $$20K + 30L - 300 = 0.$$

Solve the system of equations

To solve the system of equations, let's start by dividing the first two equations to eliminate the \(\lambda\) term: $$\frac{\frac{1}{2}K^{-1/2}L^{1/2}}{\frac{1}{2}K^{1/2}L^{-1/2}} = \frac{20\lambda}{30\lambda} \implies \frac{L}{K} = \frac{2}{3}.$$ Now, let's substitute \(L = \frac{2}{3}K\) into the constraint equation \(20K + 30L - 300 = 0\): $$20K + 30\left(\frac{2}{3}K\right) - 300 = 0 \implies K = 9.$$ Substituting \(K = 9\) in the equation \(L = \frac{2}{3}K\) gives us \(L = 6\).

Verify the maximum and find the corresponding production value

Since we only found one stationary point for the problem and the production function is continuous, we have a maximum. Thus, the values of \(K\) and \(L\) that maximize the production function are \(K = 9\) and \(L = 6\). We can then find the maximum production value by substituting these values into the production function \(P = f(K, L) = K^{1/2}L^{1/2}\): $$P = 9^{1/2} \cdot 6^{1/2} = 3 \cdot \sqrt{6}.$$ #Conclusion# The values of \(K\) and \(L\) that maximize the production function are \(K = 9\) and \(L = 6\), and the maximum production value is \(P = 3\sqrt{6}\).

Key concepts

Production Function

A production function describes the relationship between input resources like capital (\( K \)) and labor (\( L \)) and the output of a product or service. The production function used in this problem is expressed as \( P = K^{1/2}L^{1/2} \), which is a form of a Cobb-Douglas production function. This type of function assumes constant returns to scale, meaning if you increase inputs by a certain factor, output increases by the same factor.
In our exercise, we need to maximize the production output \( P \) while taking into account a budget constraint. The constraint here is in the form \( 20K + 30L = 300 \), which means the combined resource cost for capital and labor cannot exceed 300 units. Thus, the goal is to find the values of \( K \) and \( L \) that result in the highest production under this constraint.
Optimizing such a production function helps in real-world scenarios by guiding businesses in allocating their resources efficiently to maximize output.

Lagrange Multiplier

The Lagrange Multiplier is a crucial technique in optimization, especially when dealing with constraints. In this scenario, it helps to find the maximum value of the production function given the constraint. The Lagrangian function for this problem is formed as:
\[ \mathcal{L}(K, L, \lambda) = K^{1/2}L^{1/2} - \lambda(20K + 30L - 300) \]
Here, \( \lambda \) is the Lagrange Multiplier; it represents the rate of change of the objective function in response to the constraint. It effectively measures how much the maximum production value would increase if the resource budget constraint were relaxed.
The Lagrange multiplier technique transforms the constrained optimization problem into a form where both the objective function and constraint are considered together, simplifying the process of finding optimal \( K \) and \( L \) values.

Partial Derivatives

To identify the stationary points of the Lagrangian function, we calculate its partial derivatives. In our case, we compute the derivatives with respect to \( K \), \( L \), and \( \lambda \).
The partial derivatives are:

• \( \frac{\partial \mathcal{L}}{\partial K} = \frac{1}{2}K^{-1/2}L^{1/2} - 20\lambda \)
• \( \frac{\partial \mathcal{L}}{\partial L} = \frac{1}{2}K^{1/2}L^{-1/2} - 30\lambda \)
• \( \frac{\partial \mathcal{L}}{\partial \lambda} = 20K + 30L - 300 \)

Setting these partial derivatives equal to zero gives us the necessary conditions for a maximum. Solving them provides the optimal values of \( K \) and \( L \). This systematic approach ensures all constraints are honored while searching for the maximum production output.
Partial derivatives show the change in the function with respect to each variable while treating others as constants. It is a fundamental tool in calculus for analyzing mathematically how sensitive a function is to changes in inputs.