<p>The first order condition for the problem is</p>
<p><span class="math-container">$$
w = \alpha p_t \left(\frac{K_t}{L_t^*}\right)^{1-\alpha},
$$</span>
where the star indicates the optimally chosen control variable. Solve for <span class="math-container">$L_t^*$</span></p>
<p><span class="math-container">$$
L_t^* = \left(\frac{\alpha}{w}\right)^{\frac{1}{1-\alpha}}p_t^{\frac{1}{1-\alpha}}K_t
$$</span>
and note that by definition <span class="math-container">$\max_x f(x) = f(x^*)$</span> for any objective function <span class="math-container">$f$</span> (as long as the maximum exists). Plugging in the above expression for <span class="math-container">$L_t^*$</span> into the objective function will then get you</p>
<p><span class="math-container">\begin{align}
p_t &amp;\left(\frac{\alpha}{w}\right)^{\frac{\alpha}{1 - \alpha}}p_t^{\frac{\alpha}{1-\alpha}}K_t^\alpha K_t^{1 - \alpha} - w\left(\frac{\alpha}{w}\right)^{\frac{1}{1-\alpha}}p_t^{\frac{1}{1-\alpha}}K_t\\
&amp;=\left(\frac{\alpha}{w}\right)^{\frac{\alpha}{1-\alpha}}p_t^{\frac{1}{1-\alpha}}K_t - \alpha \left(\frac{\alpha}{w}\right)^{\frac{\alpha}{1-\alpha}}p_t^{\frac{1}{1-\alpha}}K_t\\
&amp;=(1 - \alpha) \left(\frac{\alpha}{w}\right)^{\frac{\alpha}{1-\alpha}}p_t^{\frac{1}{1-\alpha}}K_t \\
&amp;= hp_t^{\frac{1}{1-\alpha}}K_t.
\end{align}</span></p>

CC BY-SA 4.0
2023-03-31T15:03:56.270Z
2023-03-31T15:03:56.270Z
Wittgenstein's Poker
<p>I  re-write the <strong>maximization problem</strong> (I neglect the subscript <span class="math-container">$t$</span>):</p>
<p><span class="math-container">$$\max_{L}\left\{ pL^{\alpha}K^{1-\alpha}-wL\right\}\quad (1)$$</span></p>
<p>Taking the <strong>derivative with respect to <span class="math-container">$L$</span></strong> of the expression in <span class="math-container">$(1)$</span> and equating it to zero<br />
we have:</p>
<p><span class="math-container">$$\alpha  pL^{\alpha-1}K^{1-\alpha} -w=0\qquad (2)$$</span></p>
<p>from which we have</p>
<p><span class="math-container">$$L^{\alpha-1}= (\frac {w}{\alpha p}) K^{\alpha-1}$$</span></p>
<p>hence</p>
<p><span class="math-container">$$L= \left(\frac {w}{\alpha p}\right)^{\frac{1}{\alpha -1}}K\qquad (3)$$</span></p>
<p><strong>Substituting <span class="math-container">$(3)$</span> into the function to be maximized</strong> in <span class="math-container">$(1)$</span> yields</p>
<p><span class="math-container">$$\max_{L}\left\{ pL^{\alpha}K^{1-\alpha}-wL\right\}=   p(\frac{\alpha p}{w})^{\frac{\alpha}{1-\alpha}}K^{\alpha} K^{1-\alpha}  -w (\frac{\alpha p}{w})^{\frac{1}{1-\alpha}} K= $$</span>
<span class="math-container">$$p(\frac{\alpha p}{w})^{\frac{\alpha}{1-\alpha}}K - w (\frac{\alpha p}{w})^{\frac{1}{1-\alpha}} K= $$</span>
<span class="math-container">$$=p^{\frac{1}{1-\alpha}} (\frac{\alpha }{w})^{\frac{\alpha}{1-\alpha}}K -p^{\frac{1}{1-\alpha}}w (\frac{\alpha }{w})^{\frac{1}{1-\alpha}}  K=$$</span>                           <span class="math-container">$$=p^{\frac{1}{1-\alpha}}K [(\frac{\alpha}{w})^{\frac{\alpha}{1-\alpha}}-\alpha  \frac{\alpha ^{\frac{1}{1-\alpha}} \frac{1}{\alpha}}{w ^{\frac{\alpha }{1-\alpha}}}]=$$</span>
<span class="math-container">$$ = p^{\frac{1}{1-\alpha}}K [(\frac{\alpha}{w})^{\frac{\alpha}{1-\alpha}}- \alpha   (\frac{\alpha}{w}) ^{\frac{\alpha}{1-\alpha}}]=$$</span>
<span class="math-container">$$ = p^{\frac{1}{1-\alpha}} K(\frac{\alpha}{w})^\frac{\alpha}{1-\alpha}(1-\alpha) =$$</span></p>
<p><span class="math-container">$$= hp^{\frac{1}{1-\alpha}}K$$</span></p>

CC BY-SA 4.0
2023-03-31T15:55:26.353Z
2023-04-01T04:20:27.370Z
2023-04-01T04:20:27.370Z
BakerStreet
<p>I was reading the paper &quot;Optimal Investment Under Uncertainty&quot; (Abel, 1982). At one point the author addresses the following problem:</p>
<p><span class="math-container">$$\max_{L_{t}}\left\{ p_{t}L_{t}^{\alpha}K_{t}^{1-\alpha}-wL_{t}\right\}=hp_{t}^{\frac{1}{1-\alpha}}K_{t}$$</span></p>
<p>Where:</p>
<p><span class="math-container">$$h=(1-\alpha)\left(\frac{\alpha}{w}\right)^{\frac{\alpha}{1-\alpha}}$$</span></p>
<p>I have tried solving the first derivative with respect to L and resubstituting the result in the production function (indirect production function), but I cannot get the result (<span class="math-container">$hp_{t}^{\frac{1}{1-\alpha}}K_{t}$</span>). Can someone explain to me how to proceed?</p>
<p><a href="https://repository.upenn.edu/cgi/viewcontent.cgi?article=1206&amp;context=fnce_papers" rel="nofollow noreferrer">https://repository.upenn.edu/cgi/viewcontent.cgi?article=1206&amp;context=fnce_papers</a></p>
<p>*Pag. 4</p>

<macroeconomics><microeconomics><mathematical-economics><production-function>
Maximization of CD production function
Alessandro
# Maximization of CD production function

(Asked by: username_0 on 2023-03-29)

I was reading the paper "Optimal Investment Under Uncertainty" (Abel, 1982). At one point the author addresses the following problem:

$$\max\_{L\_{t}}\left\{ p\_{t}L\_{t}^{\alpha}K\_{t}^{1-\alpha}-wL\_{t}\right\}=hp\_{t}^{\frac{1}{1-\alpha}}K\_{t}$$

Where:

$$h=(1-\alpha)\left(\frac{\alpha}{w}\right)^{\frac{\alpha}{1-\alpha}}$$

I have tried solving the first derivative with respect to L and resubstituting the result in the production function (indirect production function), but I cannot get the result ($hp\_{t}^{\frac{1}{1-\alpha}}K\_{t}$). Can someone explain to me how to proceed?

<https://repository.upenn.edu/cgi/viewcontent.cgi?article=1206&context=fnce_papers>

\*Pag. 4

**Tags:** macroeconomics, microeconomics, mathematical-economics, production-function


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## 2 answers

### Answer by username_1 on 2023-03-31. Score: 2 | Accepted answer

The first order condition for the problem is

$$
w = \alpha p\_t \left(\frac{K\_t}{L\_t^\*}\right)^{1-\alpha},
$$
where the star indicates the optimally chosen control variable. Solve for $L\_t^\*$

$$
L\_t^\* = \left(\frac{\alpha}{w}\right)^{\frac{1}{1-\alpha}}p\_t^{\frac{1}{1-\alpha}}K\_t
$$
and note that by definition $\max\_x f(x) = f(x^\*)$ for any objective function $f$ (as long as the maximum exists). Plugging in the above expression for $L\_t^\*$ into the objective function will then get you

\begin{align}
p\_t &\left(\frac{\alpha}{w}\right)^{\frac{\alpha}{1 - \alpha}}p\_t^{\frac{\alpha}{1-\alpha}}K\_t^\alpha K\_t^{1 - \alpha} - w\left(\frac{\alpha}{w}\right)^{\frac{1}{1-\alpha}}p\_t^{\frac{1}{1-\alpha}}K\_t\\
&=\left(\frac{\alpha}{w}\right)^{\frac{\alpha}{1-\alpha}}p\_t^{\frac{1}{1-\alpha}}K\_t - \alpha \left(\frac{\alpha}{w}\right)^{\frac{\alpha}{1-\alpha}}p\_t^{\frac{1}{1-\alpha}}K\_t\\
&=(1 - \alpha) \left(\frac{\alpha}{w}\right)^{\frac{\alpha}{1-\alpha}}p\_t^{\frac{1}{1-\alpha}}K\_t \\
&= hp\_t^{\frac{1}{1-\alpha}}K\_t.
\end{align}


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### Answer by username_2 on 2023-03-31. Score: 2

I re-write the **maximization problem** (I neglect the subscript $t$):

$$\max\_{L}\left\{ pL^{\alpha}K^{1-\alpha}-wL\right\}\quad (1)$$

Taking the **derivative with respect to $L$** of the expression in $(1)$ and equating it to zero  
we have:

$$\alpha pL^{\alpha-1}K^{1-\alpha} -w=0\qquad (2)$$

from which we have

$$L^{\alpha-1}= (\frac {w}{\alpha p}) K^{\alpha-1}$$

hence

$$L= \left(\frac {w}{\alpha p}\right)^{\frac{1}{\alpha -1}}K\qquad (3)$$

**Substituting $(3)$ into the function to be maximized** in $(1)$ yields

$$\max\_{L}\left\{ pL^{\alpha}K^{1-\alpha}-wL\right\}= p(\frac{\alpha p}{w})^{\frac{\alpha}{1-\alpha}}K^{\alpha} K^{1-\alpha} -w (\frac{\alpha p}{w})^{\frac{1}{1-\alpha}} K= $$
$$p(\frac{\alpha p}{w})^{\frac{\alpha}{1-\alpha}}K - w (\frac{\alpha p}{w})^{\frac{1}{1-\alpha}} K= $$
$$=p^{\frac{1}{1-\alpha}} (\frac{\alpha }{w})^{\frac{\alpha}{1-\alpha}}K -p^{\frac{1}{1-\alpha}}w (\frac{\alpha }{w})^{\frac{1}{1-\alpha}} K=$$ $$=p^{\frac{1}{1-\alpha}}K [(\frac{\alpha}{w})^{\frac{\alpha}{1-\alpha}}-\alpha \frac{\alpha ^{\frac{1}{1-\alpha}} \frac{1}{\alpha}}{w ^{\frac{\alpha }{1-\alpha}}}]=$$
$$ = p^{\frac{1}{1-\alpha}}K [(\frac{\alpha}{w})^{\frac{\alpha}{1-\alpha}}- \alpha (\frac{\alpha}{w}) ^{\frac{\alpha}{1-\alpha}}]=$$
$$ = p^{\frac{1}{1-\alpha}} K(\frac{\alpha}{w})^\frac{\alpha}{1-\alpha}(1-\alpha) =$$

$$= hp^{\frac{1}{1-\alpha}}K$$


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