Question

Given the following monotonically transformed utility function faced by the consumer
lnU(X1X2) = ∝lnX1+βlnX2
The price of good X1 is P1 and the price of good X2 is P2.
Construct the corresponding Langregian function
Derive the optimal demand (Marshallian demand) function for X1 and for X2.

Answer 1

Answer on Question #51481 – Math – Differential Equations

Question

Given the following monotonically transformed utility function faced by the consumer:

$$\ln U(x_1, x_2) = \alpha \ln x_1 + \beta \ln x_2$$

The price of good $x_1$ is $p_1$ and the price of good $x_2$ is $p_2$.

1) Construct the corresponding Lagrangian function

2) Derive the optimal demand (Marshallian demand) function for $x_1$ and for $x_2$.

Solution

1) Lagrangian function:

Let $I$ be income, so we can write the budget constraint:

$$I = x_1 p_1 + x_2 p_2$$ $$I - x_1 p_1 - x_2 p_2 = 0$$

We need to solve the problem of optimization:

$$\max(\ln U(x_1, x_2))$$

So we can construct the Lagrangian function:

$$L(x_1, x_2, \lambda) = \alpha \ln x_1 + \beta \ln x_2 + \lambda (I - x_1 p_1 - x_2 p_2)$$

2) Marshallian demand:

$$[x_1]: \frac{\partial L}{\partial x_1} = \frac{\alpha}{x_1} - \lambda p_1 = 0 \quad \Rightarrow \quad \frac{\alpha}{x_1} = \lambda p_1 \quad \Rightarrow \quad x_1 p_1 = \frac{\alpha}{\lambda}$$ $$[x_2]: \frac{\partial L}{\partial x_2} = \frac{\beta}{x_2} - \lambda p_2 = 0 \quad \Rightarrow \quad \frac{\beta}{x_2} = \lambda p_2 \quad \Rightarrow \quad x_2 p_2 = \frac{\beta}{\lambda}$$ $$[\lambda]: \frac{\partial L}{\partial \lambda} = I - x_1 p_1 - x_2 p_2 = 0$$

After substitution in the budget constraint:

$$I - \frac{\alpha}{\lambda} - \frac{\beta}{\lambda} = 0 \quad \Rightarrow \quad \frac{\alpha + \beta}{\lambda} = I$$ $$\frac{1}{\lambda} = \frac{I}{\alpha + \beta}$$

Substituting back in the original first-order conditions give us Marshallian demand function for $x_1$ and for $x_2$:

$$x^* = \left(x_1 = \frac{\alpha}{\alpha + \beta} \frac{I}{p_1}; x_2 = \frac{\beta}{\alpha + \beta} \frac{I}{p_2}\right)$$

www.AssignmentExpert.com