<p>Start from the first order conditions:
<span class="math-container">$$
\begin{align*}
&amp;F_k = r + (1- p)\\
&amp;F_L = w\\
\end{align*}
$$</span>
Now totally differentiate with respect to <span class="math-container">$p$</span>:
<span class="math-container">$$
\begin{align*}
&amp;F_{KK} dK + F_{KL} dL = - dp\\
&amp;F_{LK} dK + F_{LL} dL = \frac{\partial w}{\partial p} dp\\
\end{align*}
$$</span>
This is equivalent to:
<span class="math-container">$$
\begin{align*}
&amp;dK = \frac{- dp - F_{KL} dL}{F_{KK}}\\
&amp;dL = \frac{\frac{\partial w}{\partial p}dp - F_{LK} dK}{F_{LL}}
\end{align*}
$$</span>
substituting the second into the first gives:
<span class="math-container">$$
\begin{align*}
&amp;dK = \frac{(- dp)F_{LL} - F_{KL}(\frac{\partial w}{\partial p} dp - F_{LK} dK)}{F_{KK} F_{LL}},\\
\to &amp;\left(1 - \frac{F_{KL} F_{LK}}{F_{LL} F_{KK}}\right) dK = \frac{-F_{LL}dp- F_{KL}\frac{\partial w}{\partial p} dp}{F_{KK} F_{LL}},\\
\to &amp;dK = \frac{-F_{LL} - F_{KL}\frac{\partial w}{\partial p}}{F_{LL} F_{KK} - (F_{LK})^2} dp
\end{align*}
$$</span>
This gives:
<span class="math-container">$$
\frac{\partial K}{\partial p} = \frac{-F_{LL} - F_{KL}\frac{\partial w}{\partial p}}{F_{LL} F_{KK} - (F_{LK})^2}
$$</span>
The expression for <span class="math-container">$\frac{\partial L}{\partial p}$</span> can be derived in a similar way.</p>

CC BY-SA 4.0
2021-05-24T15:38:03.503Z
2021-05-24T15:38:03.503Z
tdm
<p>There is the following system of four equations and four endogenous variables <span class="math-container">$(K,L,w,q)$</span>. Assume <span class="math-container">$F$</span> is a concave function.</p>
<p><span class="math-container">$\partial F(K,L)/\partial K = r + (1-p)$</span></p>
<p><span class="math-container">$\partial F(K,L)/\partial L = w$</span></p>
<p><span class="math-container">$F(K,L)-rK-wL-q-(1-p)K = 0$</span></p>
<p><span class="math-container">$v(w,p)=\bar{v}$</span></p>
<p>It seems to linearize the system around the equilibrium values to get <span class="math-container">$\frac{\partial K}{\partial p}$</span> and <span class="math-container">$\frac{\partial L}{\partial p}$</span> and it seems that these equations become</p>
<p><span class="math-container">$\frac{\partial K}{\partial p} = \frac{-F_{LL}-F_{KL}\partial w/\partial p}{(F_{KK}F_{LL}-F^2_{KL})}$</span></p>
<p><span class="math-container">$\frac{\partial L}{\partial p} = \frac{F_{KK}\partial w/\partial p + F_{KL}}{(F_{KK}F_{LL}-F^2_{KL})}$</span></p>
<p>I tried to Taylar expansion but I could not get these equations. How do we get these last equations? These equations are from p.85 <a href="https://link.springer.com/article/10.1007/s11166-006-0172-y" rel="nofollow noreferrer">Kousky et al. (2006)</a></p>

<microeconomics><production-function>
Linearization and the effect of a change
hrkshr
# Linearization and the effect of a change

(Asked by: username_0 on 2021-05-24)

There is the following system of four equations and four endogenous variables $(K,L,w,q)$. Assume $F$ is a concave function.

$\partial F(K,L)/\partial K = r + (1-p)$

$\partial F(K,L)/\partial L = w$

$F(K,L)-rK-wL-q-(1-p)K = 0$

$v(w,p)=\bar{v}$

It seems to linearize the system around the equilibrium values to get $\frac{\partial K}{\partial p}$ and $\frac{\partial L}{\partial p}$ and it seems that these equations become

$\frac{\partial K}{\partial p} = \frac{-F\_{LL}-F\_{KL}\partial w/\partial p}{(F\_{KK}F\_{LL}-F^2\_{KL})}$

$\frac{\partial L}{\partial p} = \frac{F\_{KK}\partial w/\partial p + F\_{KL}}{(F\_{KK}F\_{LL}-F^2\_{KL})}$

I tried to Taylar expansion but I could not get these equations. How do we get these last equations? These equations are from p.85 [Kousky et al. (2006)](https://link.springer.com/article/10.1007/s11166-006-0172-y)

**Tags:** microeconomics, production-function

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## 1 answers

### Answer by username_1 on 2021-05-24. Score: 3 | Accepted answer

Start from the first order conditions:
$$
\begin{align\*}
&F\_k = r + (1- p)\\
&F\_L = w\\
\end{align\*}
$$
Now totally differentiate with respect to $p$:
$$
\begin{align\*}
&F\_{KK} dK + F\_{KL} dL = - dp\\
&F\_{LK} dK + F\_{LL} dL = \frac{\partial w}{\partial p} dp\\
\end{align\*}
$$
This is equivalent to:
$$
\begin{align\*}
&dK = \frac{- dp - F\_{KL} dL}{F\_{KK}}\\
&dL = \frac{\frac{\partial w}{\partial p}dp - F\_{LK} dK}{F\_{LL}}
\end{align\*}
$$
substituting the second into the first gives:
$$
\begin{align\*}
&dK = \frac{(- dp)F\_{LL} - F\_{KL}(\frac{\partial w}{\partial p} dp - F\_{LK} dK)}{F\_{KK} F\_{LL}},\\
\to &\left(1 - \frac{F\_{KL} F\_{LK}}{F\_{LL} F\_{KK}}\right) dK = \frac{-F\_{LL}dp- F\_{KL}\frac{\partial w}{\partial p} dp}{F\_{KK} F\_{LL}},\\
\to &dK = \frac{-F\_{LL} - F\_{KL}\frac{\partial w}{\partial p}}{F\_{LL} F\_{KK} - (F\_{LK})^2} dp
\end{align\*}
$$
This gives:
$$
\frac{\partial K}{\partial p} = \frac{-F\_{LL} - F\_{KL}\frac{\partial w}{\partial p}}{F\_{LL} F\_{KK} - (F\_{LK})^2}
$$
The expression for $\frac{\partial L}{\partial p}$ can be derived in a similar way.


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