Problem 53 - 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

#tag_title#Short Answer The optimal values for capital (K) and labor (L) that maximize the Cobb-Douglas production function subject to the given constraint are K = 3 and L = 2.

Step 1: Define the objective function and constraint

The objective function (production function) is given by: $$P = f(K, L) = K^{1/2}L^{1/2}$$ And the constraint is: $$20K + 30L = 300$$

Step 2: Introduce the Lagrange multiplier

Introduce a Lagrange multiplier, λ, and form the following Lagrangian: $$\mathcal{L}(K,L,\lambda) = K^{1/2}L^{1/2} - \lambda(20K+30L-300)$$

Step 3: Compute the partial derivatives of the Lagrangian

Calculate the partial derivatives of the Lagrangian with respect to K, L, and λ: $$\frac{\partial \mathcal{L}}{\partial K} = \frac{1}{2}K^{-1/2}L^{1/2} - 20\lambda$$ $$\frac{\partial \mathcal{L}}{\partial L} = \frac{1}{2}K^{1/2}L^{-1/2} - 30\lambda$$ $$\frac{\partial \mathcal{L}}{\partial\lambda} = -20K -30L + 300$$

Step 4: Solve the system of equations

Set the partial derivatives equal to zero and solve the system of equations: $$\frac{1}{2}K^{-1/2}L^{1/2} - 20\lambda = 0 \tag{1}$$ $$\frac{1}{2}K^{1/2}L^{-1/2} - 30\lambda = 0 \tag{2}$$ $$-20K -30L + 300 = 0 \tag{3}$$ Using equation (1) and (2): $$K^{-1/2}L^{1/2}=40\lambda$$ and $$K^{1/2}L^{-1/2}=60\lambda$$ Now divide the first equation by the second equation: $$\frac{K^{-1/2}L^{1/2}}{K^{1/2}L^{-1/2}}=\frac{40\lambda}{60\lambda} \implies \frac{L}{K} = \frac{2}{3}$$ Now plug \(\frac{L}{K} = \frac{2}{3}\) into the constraint (equation 3): $$20K+30\left(\frac{2}{3}K\right)=300 \Rightarrow K=3$$ And from \(\frac{L}{K} = \frac{2}{3}\) we get \(L = 2\)

Step 5: Find the optimal value of K and L

The optimal values for K and L that maximize the production function subject to the given constraint are K = 3 and L = 2.

Key Concepts

Lagrange multipliers

Lagrange multipliers are a powerful technique for constrained optimization problems. In calculus, when you want to find the extrema (maximum or minimum) of a function subject to a constraint, Lagrange multipliers come in handy.The technique involves introducing an extra variable, called the Lagrange multiplier, which we denote as \(\lambda\). This method helps manage the relationship between the function we are maximizing or minimizing and the constraint it must adhere to.In the context of the Cobb-Douglas production function, the problem was framed with a production function \(P = K^{1/2}L^{1/2}\) and a constraint \(20K + 30L = 300\). The Lagrangian, \(\mathcal{L}(K,L,\lambda)\), combines these by subtracting the constraint multiplied by \(\lambda\) from the production function.Understanding and utilizing Lagrange multipliers is crucial for solving such optimization problems. When you equate the partial derivatives of the Lagrangian to zero, they hint towards the points where the function might reach maximum or minimum production given the budget constraint.

Optimization

Optimization is all about making the best or most effective use of a resource or system. In economics, this often translates to maximizing profit or output, or minimizing costs, under given constraints.In the exercise, we aim to maximize the Cobb-Douglas production function \(K^{1/2}L^{1/2}\), a common representation of output in terms of labor \(L\) and capital \(K\). The function is subject to an economic constraint, which is the budget.The optimization process starts by setting up an equation that links the function and the constraint — often forming a problem solvable by Lagrange multipliers. The goal is to find the optimal combination of input resources, in this case, \(K\) and \(L\), that yield the highest possible production value, all while remaining within budget limitations.Through partial derivatives and solving the resulting system of equations, we're able to determine values of \(K\) and \(L\) that maximize production: \(K=3\) and \(L=2\) in this specific problem.

Constraint

Constraints define the boundaries or limits within which a solution must be found. They are a critical component in optimization problems because they reflect real-world limits, like budgets, material supplies, or time.In our exercise, the constraint is expressed as \(20K + 30L = 300\). This linear equation represents the financial limitation on total resources (capital \(K\) and labor \(L\)) used for production.The constraint is what makes the problem interesting and applicable to real situations. Instead of finding the absolute highest possible output, we need to consider the scenario within which this output can actually occur, based on available resources.By integrating constraints into our optimization calculations using techniques like Lagrange multipliers, we ensure that the solution is realistic and actionable, respecting practical limits while seeking maximum efficiency and output.