<p>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).</p>
<p>The optimization problem is</p>
<p><span class="math-container">$\min p_1 x_1 + p_2 x_2 + p_3 x_3$</span></p>
<p>subject to <span class="math-container">$x_1 + x_2 + x_3 = \overline{U}$</span></p>
<p>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 <span class="math-container">$p_1 x_1 + p_2 x_2 + p_3 x_3$</span> to get the expenditure function <span class="math-container">$e(p_1,p_2,p3,\overline{U})$</span>.</p>
<ul>
<li><span class="math-container">$p_1 &lt; p_2, p_3$</span></li>
</ul>
<p><span class="math-container">$\implies x_1 = \overline{U}, x_2 = 0, x_3 = 0$</span></p>
<p>This yields <span class="math-container">$e = p_1 \overline{U}$</span></p>
<ul>
<li><span class="math-container">$p_2 &lt; p_1, p_3$</span></li>
</ul>
<p><span class="math-container">$\implies x_1 = 0, x_2 = \overline{U}, x_3 = 0$</span></p>
<p>This yields <span class="math-container">$e = p_2 \overline{U}$</span></p>
<ul>
<li><span class="math-container">$p_3 &lt; p_1, p_2$</span></li>
</ul>
<p><span class="math-container">$\implies x_1 = 0, x_2 = 0, x_3 = \overline{U}$</span></p>
<p>This yields <span class="math-container">$e = p_3 \overline{U}$</span></p>
<ul>
<li><span class="math-container">$p_1 = p_2 &lt; p_3$</span></li>
</ul>
<p><span class="math-container">$\implies x_1 = \alpha, x_2 = \overline{U} - \alpha, x_3 = 0, 0 \leq \alpha \leq \overline{U}$</span> (a line in 3D space)</p>
<p>This yields <span class="math-container">$e = p_1 \overline{U} = p_2 \overline{U}$</span></p>
<ul>
<li><span class="math-container">$p_1 = p_3 &lt; p_2$</span></li>
</ul>
<p><span class="math-container">$\implies x_1 = \alpha, x_2 = 0, x_3 = \overline{U} - \alpha, 0 \leq \alpha \leq \overline{U}$</span> (a line in 3D space)</p>
<p>This yields <span class="math-container">$e = p_2 \overline{U} = p_3 \overline{U}$</span></p>
<ul>
<li><span class="math-container">$p_2 = p_3 &lt; p_1$</span></li>
</ul>
<p><span class="math-container">$\implies x_1 = 0, x_2 = \alpha, x_3 = \overline{U} - \alpha, 0 \leq \alpha \leq \overline{U}$</span> (a line in 3D space)</p>
<p>This yields <span class="math-container">$e = p_2 \overline{U} = p_3 \overline{U}$</span></p>
<ul>
<li><span class="math-container">$p_1 = p_2 = p_3$</span></li>
</ul>
<p><span class="math-container">$\implies x_1 = \alpha, x_2 = \beta, x_3 = \overline{U} - \alpha - \beta, 0 \leq \alpha \leq \overline{U}, 0 \leq \beta \leq \overline{U} - \alpha$</span> (a plane in 3D space)</p>
<p>This yields <span class="math-container">$e = p_1 \overline{U} = p_2 \overline{U} = p_3 \overline{U}$</span></p>
<p>Edit: In general for this problem, the expenditure function looks like this:</p>
<p><span class="math-container">$e(p_1,p_2,p_3,\overline{U}) = p_{min} \overline{U}$</span>, where</p>
<p><span class="math-container">$p_{min} = \min\{p_1,p_2,p_3\}$</span>.</p>
<p>Note the optimal expenditure function we got <span class="math-container">$e(p_1,p_2,p_3,\overline{U})$</span> 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).</p>
<p>The above always happens when given a linear utility function.</p>
<p>Note: <span class="math-container">$\#$</span> parameters <span class="math-container">$=$</span> dimension of the optimal bundles manifold</p>

CC BY-SA 4.0
2022-11-19T16:07:15.750Z
2022-11-19T17:15:02.027Z
2022-11-19T17:15:02.027Z
Nicolas Torres
<p>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)?</p>

<microeconomics><optimization>
How to find the e(p,u) of u(x) = x1 + x2 + x3
reindeer
# How to find the e(p,u) of u(x) = x1 + x2 + x3

(Asked by: username_0 on 2022-11-19)

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)?

**Tags:** microeconomics, optimization

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## 1 answers

### Answer by username_1 on 2022-11-19. Score: 1 | Accepted answer

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


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