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In lecturer's notes we have a utility function

$U_i = X_i^A * (1-L_i)^{1-a},$ $ 0< a< 1$, $ i = 1,2 $

$MP_1 = w_1 $

$MP_2 = w_2$

$X_1 + X_2 = w_1 * L_1 + w_2 * L_2$

And we need to form a Lagrangian function to optimize for these variables. In his notes our Lagrangian function is:

$\zeta = X_1^a * (1-L_1)^{1-a} - \lambda(\bar{u} - X_2^a * (1-L_2)^{1-a} - \mu [X_1 + X_2 - w_1 * L_1 - w_2 * L_2] $

and he differentiates according to that lagrangian function. However I suppose that:

$ X_1^a * (1-L_1)^{1-a} + X_2^a * (1-L_2)^{1-a} = \bar{u} $

So his lagrangian functions becomes something like:

$\zeta = X_1^a * (1-L_1)^{1-a} - \lambda(X_1^a * (1-L_1)^{1-a} - \mu [X_1 + X_2 - w_1 * L_1 - w_2 * L_2] $

which does not make any sense for me. I feel like it should have been something like:

$\zeta = X_1^a * (1-L_1)^{1-a} - \lambda(\bar{u} - X_1^a * (1-L_1)^{1-a} - \mu [X_1 + X_2 - w_1 * L_1 - w_2 * L_2] $

So we can optimize it both for $X_1$ and $X_2$ I would be so happy if you could help me with the theory and logic behind it.

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  • $\begingroup$ Please clarify your specific problem or provide additional details to highlight exactly what you need. As it's currently written, it's hard to tell exactly what you're asking. $\endgroup$
    – Giskard
    Commented Apr 2, 2022 at 6:29
  • $\begingroup$ @Giskard It's really difficult for me to specify the problem than I already did above, but I will give it another try. I cannot see why we used both the $X_1^a$ and the $\lambda * X_1^a - (1- L_1)^{1-a}$ in the lagrangian function. How can a variable can become its own constraint. $\endgroup$ Commented Apr 2, 2022 at 7:38
  • $\begingroup$ What goal function are you trying to maximize? $\endgroup$
    – Giskard
    Commented Apr 2, 2022 at 9:20
  • $\begingroup$ @Giskard $U_i$ I suppose $\endgroup$ Commented Apr 3, 2022 at 15:20
  • $\begingroup$ It seems like that is indexed notation denoting two functions. $\endgroup$
    – Giskard
    Commented Apr 3, 2022 at 15:46

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