Cobb-Douglas Example

Econ 502: Advanced Microeconomics

Utility Maximization

1.1 Problem Statement

The consumer’s problem is: \[\max_{x,y} \quad x^{\alpha}y^{\beta}\] subject to: \[p_x x + p_y y = I\] \[x \geq 0, \quad y \geq 0\]

1.2 Lagrangian Method

Set up the Lagrangian: \[\mathcal{L}(x, y, \lambda) = x^{\alpha}y^{\beta} + \lambda(I - p_x x - p_y y)\]

First-order conditions: \[\begin{align} \frac{\partial \mathcal{L}}{\partial x} &= \alpha x^{\alpha-1}y^{\beta} - \lambda p_x = 0 \tag{1}\\ \frac{\partial \mathcal{L}}{\partial y} &= \beta x^{\alpha}y^{\beta-1} - \lambda p_y = 0 \tag{2}\\ \frac{\partial \mathcal{L}}{\partial \lambda} &= I - p_x x - p_y y = 0 \tag{3} \end{align}\]

1.3 Solving for Demand Functions

From equations (1) and (2): \[\frac{\alpha x^{\alpha-1}y^{\beta}}{\beta x^{\alpha}y^{\beta-1}} = \frac{p_x}{p_y}\]

Simplifying: \[\frac{\alpha y}{\beta x} = \frac{p_x}{p_y}\]

Solving for \(y\): \[y = \frac{\beta p_x}{\alpha p_y} x \tag{4}\]

Substitute equation (4) into the budget constraint (3): \[p_x x + p_y \cdot \frac{\beta p_x}{\alpha p_y} x = I\]

\[p_x x \left(1 + \frac{\beta}{\alpha}\right) = I\]

\[p_x x \cdot \frac{\alpha + \beta}{\alpha} = I\]

Solving for \(x\): \[\boxed{x^*(p_x, p_y, I) = \frac{\alpha I}{(\alpha + \beta)p_x}}\]

Substituting back into equation (4): \[y^* = \frac{\beta p_x}{\alpha p_y} \cdot \frac{\alpha I}{(\alpha + \beta)p_x} = \frac{\beta I}{(\alpha + \beta)p_y}\]

\[\boxed{y^*(p_x, p_y, I) = \frac{\beta I}{(\alpha + \beta)p_y}}\]

1.4 Indirect Utility Function

Substitute the demand functions into the utility function: \[V(p_x, p_y, I) = \left(\frac{\alpha I}{(\alpha+\beta)p_x}\right)^{\alpha} \left(\frac{\beta I}{(\alpha+\beta)p_y}\right)^{\beta}\]

Simplifying: \[V(p_x, p_y, I) = \frac{\alpha^{\alpha}\beta^{\beta}}{(\alpha+\beta)^{\alpha+\beta}} \cdot \frac{I^{\alpha+\beta}}{p_x^{\alpha}p_y^{\beta}}\]

Let \(K = \frac{\alpha^{\alpha}\beta^{\beta}}{(\alpha+\beta)^{\alpha+\beta}}\):

\[\boxed{V(p_x, p_y, I) = K \cdot \frac{I^{\alpha+\beta}}{p_x^{\alpha}p_y^{\beta}}}\]

Expenditure Minimization

2.1 Problem Statement

The dual problem is: \[\min_{x,y} \quad p_x x + p_y y\] subject to: \[x^{\alpha}y^{\beta} = \bar{U}\] \[x \geq 0, \quad y \geq 0\]

where \(\bar{U}\) is a target utility level.

2.2 Lagrangian Method

Set up the Lagrangian: \[\mathcal{L}(x, y, \mu) = p_x x + p_y y + \mu(\bar{U} - x^{\alpha}y^{\beta})\]

First-order conditions: \[\begin{align} \frac{\partial \mathcal{L}}{\partial x} &= p_x - \mu \alpha x^{\alpha-1}y^{\beta} = 0 \tag{5}\\ \frac{\partial \mathcal{L}}{\partial y} &= p_y - \mu \beta x^{\alpha}y^{\beta-1} = 0 \tag{6}\\ \frac{\partial \mathcal{L}}{\partial \mu} &= \bar{U} - x^{\alpha}y^{\beta} = 0 \tag{7} \end{align}\]

2.3 Solving for Hicksian Demand

From equations (5) and (6): \[\frac{p_x}{p_y} = \frac{\mu \alpha x^{\alpha-1}y^{\beta}}{\mu \beta x^{\alpha}y^{\beta-1}} = \frac{\alpha y}{\beta x}\]

Solving for \(y\): \[y = \frac{\beta p_x}{\alpha p_y} x \tag{8}\]

Substitute equation (8) into the utility constraint (7): \[x^{\alpha} \left(\frac{\beta p_x}{\alpha p_y} x\right)^{\beta} = \bar{U}\]

\[x^{\alpha+\beta} \left(\frac{\beta p_x}{\alpha p_y}\right)^{\beta} = \bar{U}\]

\[x^{\alpha+\beta} = \bar{U} \left(\frac{\alpha p_y}{\beta p_x}\right)^{\beta}\]

\[x = \bar{U}^{1/(\alpha+\beta)} \left(\frac{\alpha p_y}{\beta p_x}\right)^{\beta/(\alpha+\beta)}\]

Rearranging: \[\boxed{x^h(p_x, p_y, \bar{U}) = \left(\frac{\alpha}{\beta}\right)^{\beta/(\alpha+\beta)} \left(\frac{p_y}{p_x}\right)^{\beta/(\alpha+\beta)} \bar{U}^{1/(\alpha+\beta)}}\]

Similarly, substituting back: \[\boxed{y^h(p_x, p_y, \bar{U}) = \left(\frac{\beta}{\alpha}\right)^{\alpha/(\alpha+\beta)} \left(\frac{p_x}{p_y}\right)^{\alpha/(\alpha+\beta)} \bar{U}^{1/(\alpha+\beta)}}\]

2.4 Expenditure Function

The expenditure function is: \[E(p_x, p_y, \bar{U}) = p_x x^h + p_y y^h\]

Substituting the Hicksian demands: \[E = p_x \left(\frac{\alpha}{\beta}\right)^{\beta/(\alpha+\beta)} \left(\frac{p_y}{p_x}\right)^{\beta/(\alpha+\beta)} \bar{U}^{1/(\alpha+\beta)} + p_y \left(\frac{\beta}{\alpha}\right)^{\alpha/(\alpha+\beta)} \left(\frac{p_x}{p_y}\right)^{\alpha/(\alpha+\beta)} \bar{U}^{1/(\alpha+\beta)}\]

Simplifying the first term: \[p_x \left(\frac{\alpha}{\beta}\right)^{\beta/(\alpha+\beta)} \left(\frac{p_y}{p_x}\right)^{\beta/(\alpha+\beta)} = \left(\frac{\alpha}{\beta}\right)^{\beta/(\alpha+\beta)} p_x^{\alpha/(\alpha+\beta)} p_y^{\beta/(\alpha+\beta)}\]

Similarly for the second term: \[p_y \left(\frac{\beta}{\alpha}\right)^{\alpha/(\alpha+\beta)} \left(\frac{p_x}{p_y}\right)^{\alpha/(\alpha+\beta)} = \left(\frac{\beta}{\alpha}\right)^{\alpha/(\alpha+\beta)} p_x^{\alpha/(\alpha+\beta)} p_y^{\beta/(\alpha+\beta)}\]

Therefore: \[E = p_x^{\alpha/(\alpha+\beta)} p_y^{\beta/(\alpha+\beta)} \bar{U}^{1/(\alpha+\beta)} \left[\left(\frac{\alpha}{\beta}\right)^{\beta/(\alpha+\beta)} + \left(\frac{\beta}{\alpha}\right)^{\alpha/(\alpha+\beta)}\right]\]

Let: \[C = \left(\frac{\alpha}{\beta}\right)^{\beta/(\alpha+\beta)} + \left(\frac{\beta}{\alpha}\right)^{\alpha/(\alpha+\beta)}\]

Then: \[\boxed{E(p_x, p_y, \bar{U}) = C \cdot p_x^{\alpha/(\alpha+\beta)} p_y^{\beta/(\alpha+\beta)} \bar{U}^{1/(\alpha+\beta)}}\]