<p>Formally, you would need to set up a Kuhn-Karush-Tucker-Problem with the two inequality constraints <span class="math-container">$N_t&lt;1$</span> and <span class="math-container">$N_t&gt;0$</span>. Plugging in for consumption, you get the Lagrangian</p>
<p><span class="math-container">$
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)
$</span></p>
<p>with the relevant first order conditions:</p>
<p><span class="math-container">\begin{gathered}
  \frac{{\partial L}}{{\partial {N_t}}} = {\beta ^t}\frac{1}{{\left( {\left( {1 - \delta } \right){K_t} + K_t^\alpha N_t^{1 - \alpha } - {K_{t + 1}}} \right)}}\left( {1 - \alpha } \right)\left( {K_t^\alpha N_t^{ - \alpha }} \right) - {\beta ^t}{\lambda _{1,t}} + {\beta ^t}{\lambda _{2,t}} = 0 \hfill \\
  {\lambda _{1,t}}\left( {1 - {N_t}} \right) = 0 \hfill \\
  {\lambda _{2,t}}\left( {0 + {N_t}} \right) = 0 \hfill \\
\end{gathered}</span></p>
<p>The last two are the complementary slackness conditions. You need to check the three cases of</p>
<ol>
<li>no constraint is binding (interior solution) or</li>
<li>the lower bound is binding (<span class="math-container">$N=0$</span>) or</li>
<li>the upper bound is binding (<span class="math-container">$N=1$</span>)</li>
</ol>
<p>We do this in turn:</p>
<ol>
<li>If the inequality constraints were non-binding, the first condition implies</li>
</ol>
<p><span class="math-container">\begin{gathered}
  \frac{{\partial L}}{{\partial {N_t}}} = {\beta ^t}\frac{1}{{\left( {\left( {1 - \delta } \right){K_t} + K_t^\alpha N_t^{1 - \alpha } - {K_{t + 1}}} \right)}}\left( {1 - \alpha } \right)\left( {K_t^\alpha N_t^{ - \alpha }} \right) = 0\\
 \Rightarrow \frac{1}{{N_t^\alpha }} = 0 \\
\Rightarrow {N_t} = \infty
\end{gathered}</span></p>
<p>This is a contradiction.</p>
<ol start="2">
<li><p>Next, if <span class="math-container">$N_t=0$</span>, then <span class="math-container">$\lambda_{1,t}=0$</span> and <span class="math-container">$\lambda_{2,t}=-\infty$</span>. The multiplier must be positive, thus this is not a solution.</p>
</li>
<li><p>Finally, if <span class="math-container">$N_t=1$</span>, then <span class="math-container">$\lambda_{2,t}=0$</span> and
<span class="math-container">$
{\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) &gt; 0
$</span>
which is the solution.</p>
</li>
</ol>
<p>This formally proves that with no disutility of labor, the maximum feasible amount will be chosen.</p>

CC BY-SA 4.0
2022-04-13T07:21:23.803Z
2022-04-13T07:21:23.803Z
jpfeifer
<p>In every standard growth model, we suppose labor=1.<br />
I learnd that this is because optimization solution of labor is equal to 1.<br />
How to derive that optimazation solution?</p>
<p>For example, consider that standard CASS growth model.<br />
<span class="math-container">$max \sum_{t=0}^{\infty} \beta^t u(c_t)$</span> s.t <span class="math-container">$c_t+i_t=y_t$</span>, <span class="math-container">$k_{t+1}=(1-\delta)k_t +i_t$</span>, <span class="math-container">$y_t=A_t k_t^\alpha (E_t L_t)^{1-\alpha} $</span><br />
I tried to use lagrangian but failed because there is no labor term in utility function.<br />
Every FOC condition shows <span class="math-container">$MPL=0$</span>. Is that right?<br />
How can I derive the optimal solution for labor if there is no labor disutility term in utility function?<br />
Or it's just a simple answer that there is no disutility of labor we chose the maximum labor time(=1)?</p>

CC BY-SA 4.0
2022-04-13T16:57:43.513Z
It should be Ramsey-Cass-Koopmans standard growth model. Let’s give due credit :)
erik
<macroeconomics><labor-economics><economic-growth>
Labor=1 in standard growth model
guest
# Labor=1 in standard growth model

(Asked by: username_0 on 2022-04-13)

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)?

**Tags:** macroeconomics, labor-economics, economic-growth
> **Comments:**
> **username_1**: It should be Ramsey-Cass-Koopmans standard growth model. Let’s give due credit :)  

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## 1 answers

### Answer by username_2 on 2022-04-13. Score: 2 | Accepted answer

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.


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