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?

  • $\begingroup$ The second and third line are not equations... $\endgroup$
    – VARulle
    Commented Feb 16, 2023 at 0:37
  • $\begingroup$ Sorry, that was a typo. Fixed now! $\endgroup$ Commented Feb 16, 2023 at 9:11

1 Answer 1


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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