Utility Maximization

Classwork 2

Utility maximization with the Cobb-Douglas utility function

The Cobb-Douglas utility function is

\[ \begin{align} u(x_{1}, x_{2}) = x_{1}^{\alpha}x_{2}^{\beta}. \end{align} \] Since utility functions are only defined up to a monotonic transformation, it is convenient to take logs of this expression and work with:

\[ \begin{align} \log u(x_{1}, x_{2}) = \alpha \log x_{1} + \beta \log x_{2}. \end{align} \]

• By the way, a monotonic transformation is a way of transforming one set of numbers into another set of numbers in a way that the order of the numbers is preserved.

  • Therefore, a monotonic transformation of a utility function preserves the order of the utility values.

  • If bundle \(A\) is preferred to bundle \(B\), then a monotonic transformation of the utility function will still show that \(A\) is preferred to \(B\).

Let’s find the demand functions for \(x_1\) and \(x_2\) for the Cobb-Douglas utility function. The problem we want to solve is

\[ \begin{align} \max_{x_{1}, x_{2}}&\; \log u(x_{1}, x_{2}) = \alpha \log x_{1} + \beta \log x_{2} \tag{1}\\ \text{such that}&\quad Y \;\geq\; p_{1}x_{1} + p_{2}x_{2} \tag{2} \end{align} \]

There are at least two ways to solve this problem.

1. Substituing out the budget contraint in the maximization problem

Since “the more, the better” holds for this consumer, the budget constraint holds with equality at the optimum. Using this fact, we can then substitute out \(x_{2}\) in problem (1) using the budget constraint (2) with equality.

\[ \begin{align} \max_{x_{1}}&\; \log u(x_{1}) = \alpha \log x_{1} + \beta \log \left(\frac{Y - p_{1}x_{1}}{p_{2}}\right) \tag{3} \end{align} \] The first order condition for problem (3) is

\[ \begin{align} \frac{\alpha}{x_{1}} + \frac{\beta}{\left(\frac{Y - p_{1}x_{1}}{p_{2}}\right)}\left(-\frac{p_{1}}{p_{2}}\right) \,=\, 0. \tag{4} \end{align} \] Re-arranging the first-order condition (4) and the budget constraint (2) with equality gives the demand for \(x_{1}\) and \(x_{2}\):

\[ \begin{align} x_{1}^{*} = \frac{\alpha}{\alpha + \beta} \frac{Y}{p_{1}},\\ x_{2}^{*} = \frac{\beta}{\alpha + \beta} \frac{Y}{p_{2}}. \end{align} \]

2. Lagrange’s method

Now for Lagrange’s method for the problem (1) with constraint (2). Set up the Lagrangian: \[ \mathcal{L} = \alpha\log x_{1} + \beta\log x_{2} + \lambda(Y - p_{1}x_{1} - p_{2}x_{2}). \tag{5} \] Differentiate (5) with respect to each of the choice variables, \(x_{1}, x_{2}\), and \(\lambda\) to get the three first-order conditions:

\[ \begin{align} \frac{\partial\mathcal{L}}{\partial x_{1}} &= \frac{\alpha}{x_{1}} - \lambda p_{1} = 0, \tag{6}\\ \frac{\partial\mathcal{L}}{\partial x_{2}} &= \frac{\beta}{x_{2}} - \lambda p_{2} = 0, \tag{7}\\ \frac{\partial\mathcal{L}}{\partial x_{1}} &= Y - p_{1}x_{1} - p_{2}x_{2} = 0. \tag{8}\\ \end{align} \] Now we are to solve the system of equations—three unknowns (\(x_{1}, x_{2}\), and \(\lambda\)), three equations (6), (7), and (8).

Rearranging (6) and (7) gives: \[ \begin{align} x_{1} = \frac{\alpha}{\lambda p_{1}},\tag{8}\\ x_{2} = \frac{\beta}{\lambda p_{2}}.\tag{9} \end{align} \]

Now we need to find \(\lambda\).

Rearranging (6) and (7) also gives: \[ \begin{align} \alpha &= \lambda p_{1}x_{1},\\ \beta &= \lambda p_{2}x_{2}. \end{align} \] Adding the equations above gives:

\[ \alpha + \beta = \lambda(p_{1}x_{1} + p_{2}x_{2}) = \lambda Y, \] which gives us

\[ \lambda = \frac{\alpha + \beta}{Y}.\tag{10} \]

Substitute (10) back into (8) and (9) and solve for \(x_{1}\) and \(x_{2}\) to get

\[ \begin{align} x_{1}^{*} = \frac{\alpha}{\alpha + \beta} \frac{Y}{p_{1}},\\ x_{2}^{*} = \frac{\beta}{\alpha + \beta} \frac{Y}{p_{2}}, \end{align} \]

just as before.