Labor=1 in standard growth model

Question

In every standard growth model, we suppose labor=1.
I learnd that this is because optimization solution of labor is equal to 1.
How to derive that optimazation solution?

For example, consider that standard CASS growth model.
$max \sum_{t=0}^{\infty} \beta^t u(c_t)$ s.t $c_t+i_t=y_t$, $k_{t+1}=(1-\delta)k_t +i_t$, $y_t=A_t k_t^\alpha (E_t L_t)^{1-\alpha} $
I tried to use lagrangian but failed because there is no labor term in utility function.
Every FOC condition shows $MPL=0$. Is that right?
How can I derive the optimal solution for labor if there is no labor disutility term in utility function?
Or it's just a simple answer that there is no disutility of labor we chose the maximum labor time(=1)?

Answer 1

Formally, you would need to set up a Kuhn-Karush-Tucker-Problem with the two inequality constraints $N_t<1$ and $N_t>0$. Plugging in for consumption, you get the Lagrangian

$ L = \sum\limits_{t = 0}^\infty {{\beta ^t}\ln \left( {\left( {1 - \delta } \right){K_t} + K_t^\alpha N_t^{1 - \alpha } - {K_{t + 1}}} \right)} + {\beta ^t}{\lambda _{1,t}}\left( {1 - {N_t}} \right) + {\beta ^t}{\lambda _{2,t}}\left( {0 + {N_t}} \right) $

with the relevant first order conditions:

\begin{gathered} \frac{{\partial L}}{{\partial {N_t}}} = {\beta ^t}\frac{1}{{\left( {\left( {1 - \delta } \right){K_t} + K_t^\alpha N_t^{1 - \alpha } - {K_{t + 1}}} \right)}}\left( {1 - \alpha } \right)\left( {K_t^\alpha N_t^{ - \alpha }} \right) - {\beta ^t}{\lambda _{1,t}} + {\beta ^t}{\lambda _{2,t}} = 0 \hfill \\ {\lambda _{1,t}}\left( {1 - {N_t}} \right) = 0 \hfill \\ {\lambda _{2,t}}\left( {0 + {N_t}} \right) = 0 \hfill \\ \end{gathered}

The last two are the complementary slackness conditions. You need to check the three cases of

1. no constraint is binding (interior solution) or
2. the lower bound is binding ($N=0$) or
3. the upper bound is binding ($N=1$)

We do this in turn:

1. If the inequality constraints were non-binding, the first condition implies

\begin{gathered} \frac{{\partial L}}{{\partial {N_t}}} = {\beta ^t}\frac{1}{{\left( {\left( {1 - \delta } \right){K_t} + K_t^\alpha N_t^{1 - \alpha } - {K_{t + 1}}} \right)}}\left( {1 - \alpha } \right)\left( {K_t^\alpha N_t^{ - \alpha }} \right) = 0\\ \Rightarrow \frac{1}{{N_t^\alpha }} = 0 \\ \Rightarrow {N_t} = \infty \end{gathered}

This is a contradiction.

2. Next, if $N_t=0$, then $\lambda_{1,t}=0$ and $\lambda_{2,t}=-\infty$. The multiplier must be positive, thus this is not a solution.

3. Finally, if $N_t=1$, then $\lambda_{2,t}=0$ and $ {\lambda _{1,t}} = \frac{1}{{\left( {\left( {1 - \delta } \right){K_t} + K_t^\alpha - {K_{t + 1}}} \right)}}\left( {1 - \alpha } \right)\left( {K_t^\alpha } \right) > 0 $ which is the solution.

This formally proves that with no disutility of labor, the maximum feasible amount will be chosen.