<p>In answer to question (1):</p>
<p>Using equation (1) we can substitute for <span class="math-container">$\lambda$</span> in (2) and (3) to obtain:</p>
<p><span class="math-container">$p_2-p_1(x_2^2+x_3^2)^{-1/3}x_2=0 \qquad(5)$</span></p>
<p><span class="math-container">$p_3-p_1 (x_2^2+x_3^2)^{-1/3}x_3=0 \qquad(6)$</span></p>
<p>If we can solve for <span class="math-container">$x_2$</span> and <span class="math-container">$x_3$</span>, we can substitute in (4) and rearrange to solve for <span class="math-container">$x_1$</span>.</p>
<p>What is less obvious is how to solve for <span class="math-container">$x_2$</span> and <span class="math-container">$x_3$</span>, but this can be done as follows.  Since (5) and (6) contain the common term <span class="math-container">$p_1(x_2^2+x_3^2)^{-1/3}$</span>, we can infer from them that:</p>
<p><span class="math-container">$\dfrac{p_2}{x_2}=\dfrac{p_3}{x_3} \qquad(7)$</span></p>
<p>and therefore:</p>
<p><span class="math-container">$x_3=\dfrac{p_3}{p_2}x_2 \qquad(8)$</span></p>
<p>Substituting for <span class="math-container">$x_3$</span> in (5):</p>
<p><span class="math-container">$p_2-p_1\Bigg(x_2^2+\dfrac{p_3^2}{p_2^2}x_2^2\Bigg)^{-1/3}x_2=0 \qquad(9)$</span></p>
<p><span class="math-container">$p_2-p_1x_2^{1/3}\Bigg(1+\dfrac{p_3^2}{p_2^2}\Bigg)=0 \qquad(10)$</span></p>
<p>(10) can then be rearranged to solve for <span class="math-container">$x_2$</span>.  A similar argument from (6) and (7) will solve for <span class="math-container">$x_3$</span>.</p>

CC BY-SA 4.0
2023-02-16T19:52:04.267Z
2023-02-16T19:52:04.267Z
Adam Bailey
<p>I have an optimization problem from microeconomics that yields me the following first-order conditions based on a Lagrangian:</p>
<p><span class="math-container">$ p_1 = \lambda \qquad(1)$</span></p>
<p><span class="math-container">$ p_2 - \lambda (x_2^2+x_3^2)^{-1/3}x_2=0 \qquad(2)$</span></p>
<p><span class="math-container">$ p_3 - \lambda (x_2^2+x_3^2)^{-1/3}x_3=0 \qquad(3)$</span></p>
<p><span class="math-container">$ x_1+(x_2^2+x_3^2)^{2/3}=0 \qquad(4)$</span></p>
<p>I know that in the solution <span class="math-container">$\ x_1$</span> is negative and the other two variables are positive. Supposedly, this system of equations has a solution without the need to add any extra conditions, but sadly I cannot find it. My questions are:</p>
<p>(1) How to solve for <span class="math-container">$ x_1, x_2$</span> and <span class="math-container">$ x_3 $</span> as functions of the prices in this particular case?</p>
<p>(2) What to do more generally when the first-order conditions cancel like they seem to do in the above example?</p>

CC BY-SA 4.0
2023-02-16T00:37:51.457Z
The second and third line are not equations...
VARulle
CC BY-SA 4.0
2023-02-16T09:11:21.700Z
Sorry, that was a typo. Fixed now!
Econometric Novice
<microeconomics><consumer-theory><production-function>
Solving Lagrangian FOCs: a few difficulties
Econometric Novice
# Solving Lagrangian FOCs: a few difficulties

(Asked by: username_0 on 2023-02-15)

I have an optimization problem from microeconomics that yields me the following first-order conditions based on a Lagrangian:

$ p\_1 = \lambda \qquad(1)$

$ p\_2 - \lambda (x\_2^2+x\_3^2)^{-1/3}x\_2=0 \qquad(2)$

$ p\_3 - \lambda (x\_2^2+x\_3^2)^{-1/3}x\_3=0 \qquad(3)$

$ x\_1+(x\_2^2+x\_3^2)^{2/3}=0 \qquad(4)$

I know that in the solution $\ x\_1$ is negative and the other two variables are positive. Supposedly, this system of equations has a solution without the need to add any extra conditions, but sadly I cannot find it. My questions are:

(1) How to solve for $ x\_1, x\_2$ and $ x\_3 $ as functions of the prices in this particular case?

(2) What to do more generally when the first-order conditions cancel like they seem to do in the above example?

**Tags:** microeconomics, consumer-theory, production-function
> **Comments:**
> **username_1**: The second and third line are not equations...  
> **username_0**: Sorry, that was a typo. Fixed now!  

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## 1 answers

### Answer by username_2 on 2023-02-16. Score: 1 | Accepted answer

In answer to question (1):

Using equation (1) we can substitute for $\lambda$ in (2) and (3) to obtain:

$p\_2-p\_1(x\_2^2+x\_3^2)^{-1/3}x\_2=0 \qquad(5)$

$p\_3-p\_1 (x\_2^2+x\_3^2)^{-1/3}x\_3=0 \qquad(6)$

If we can solve for $x\_2$ and $x\_3$, we can substitute in (4) and rearrange to solve for $x\_1$.

What is less obvious is how to solve for $x\_2$ and $x\_3$, but this can be done as follows. Since (5) and (6) contain the common term $p\_1(x\_2^2+x\_3^2)^{-1/3}$, we can infer from them that:

$\dfrac{p\_2}{x\_2}=\dfrac{p\_3}{x\_3} \qquad(7)$

and therefore:

$x\_3=\dfrac{p\_3}{p\_2}x\_2 \qquad(8)$

Substituting for $x\_3$ in (5):

$p\_2-p\_1\Bigg(x\_2^2+\dfrac{p\_3^2}{p\_2^2}x\_2^2\Bigg)^{-1/3}x\_2=0 \qquad(9)$

$p\_2-p\_1x\_2^{1/3}\Bigg(1+\dfrac{p\_3^2}{p\_2^2}\Bigg)=0 \qquad(10)$

(10) can then be rearranged to solve for $x\_2$. A similar argument from (6) and (7) will solve for $x\_3$.


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