If I do the LaGrangian for the Expenditure minimization problem, it comes as p1 = p2 = p3, how do I substitute it back in the constraint and find the Hicksian demand to find e(p,u)?


1 Answer 1


You don’t because the utility function is linear. Trying to use the Lagrangian method on linear utility functions yields non-sensical equations. (The consumer can’t make all prices equal).

The optimization problem is

$\min p_1 x_1 + p_2 x_2 + p_3 x_3$

subject to $x_1 + x_2 + x_3 = \overline{U}$

You have to consider cases depending on which price is the least. From each case you can plug in the Hicksian demands into the expenditure $p_1 x_1 + p_2 x_2 + p_3 x_3$ to get the expenditure function $e(p_1,p_2,p3,\overline{U})$.

  • $p_1 < p_2, p_3$

$\implies x_1 = \overline{U}, x_2 = 0, x_3 = 0$

This yields $e = p_1 \overline{U}$

  • $p_2 < p_1, p_3$

$\implies x_1 = 0, x_2 = \overline{U}, x_3 = 0$

This yields $e = p_2 \overline{U}$

  • $p_3 < p_1, p_2$

$\implies x_1 = 0, x_2 = 0, x_3 = \overline{U}$

This yields $e = p_3 \overline{U}$

  • $p_1 = p_2 < p_3$

$\implies x_1 = \alpha, x_2 = \overline{U} - \alpha, x_3 = 0, 0 \leq \alpha \leq \overline{U}$ (a line in 3D space)

This yields $e = p_1 \overline{U} = p_2 \overline{U}$

  • $p_1 = p_3 < p_2$

$\implies x_1 = \alpha, x_2 = 0, x_3 = \overline{U} - \alpha, 0 \leq \alpha \leq \overline{U}$ (a line in 3D space)

This yields $e = p_2 \overline{U} = p_3 \overline{U}$

  • $p_2 = p_3 < p_1$

$\implies x_1 = 0, x_2 = \alpha, x_3 = \overline{U} - \alpha, 0 \leq \alpha \leq \overline{U}$ (a line in 3D space)

This yields $e = p_2 \overline{U} = p_3 \overline{U}$

  • $p_1 = p_2 = p_3$

$\implies x_1 = \alpha, x_2 = \beta, x_3 = \overline{U} - \alpha - \beta, 0 \leq \alpha \leq \overline{U}, 0 \leq \beta \leq \overline{U} - \alpha$ (a plane in 3D space)

This yields $e = p_1 \overline{U} = p_2 \overline{U} = p_3 \overline{U}$

Edit: In general for this problem, the expenditure function looks like this:

$e(p_1,p_2,p_3,\overline{U}) = p_{min} \overline{U}$, where

$p_{min} = \min\{p_1,p_2,p_3\}$.

Note the optimal expenditure function we got $e(p_1,p_2,p_3,\overline{U})$ is piecewise defined, hence not differentiable. This is the reason why you can’t get it through the Lagrangian method, which relies on derivatives (hence differentiability).

The above always happens when given a linear utility function.

Note: $\#$ parameters $=$ dimension of the optimal bundles manifold


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.