# Solving Lagrangian FOCs: a few difficulties

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?

• The second and third line are not equations... Feb 16 at 0:37
• Sorry, that was a typo. Fixed now! Feb 16 at 9:11

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 $$x_3$$.