<p>Consider the following problem:</p>
<p><span class="math-container">\begin{eqnarray*} \max_{y, k, l} &amp; py - wl- rk \\ \text{s.t. } &amp; y\leq k^\alpha l^{1-\alpha} \\ \text{and } &amp; k, l, y \geq 0\end{eqnarray*}</span></p>
<p>where <span class="math-container">$p &gt;0$</span>, <span class="math-container">$w&gt;0$</span> and <span class="math-container">$r&gt;0$</span> are given.</p>
<p>This is how we can solve the above profit maximization problem in two steps:</p>
<p>First solve the cost minimisation problem of the firm:</p>
<p><span class="math-container">\begin{eqnarray*} \min_{k, l} &amp; wl + rk \\ \text{s.t. } &amp; y\leq k^\alpha l^{1-\alpha} \\ \text{and } &amp; k, l\geq 0\end{eqnarray*}</span></p>
<p>Solution to this problem is known as conditional input demand function and satisfy the following two conditions:</p>
<p><span class="math-container">$\text{MRTS} = \dfrac{\text{MP}_L}{\text{MP}_K} = \dfrac{(1-\alpha) k}{\alpha l} = \dfrac{w}{r}$</span></p>
<p><span class="math-container">$y =  k^\alpha l^{1-\alpha}$</span></p>
<p>Solving these two equations gives us the conditional input demands:
<span class="math-container">\begin{eqnarray*} (l^c, k^c) (w, r, y) = \left( \left(\frac{(1-\alpha)r}{\alpha w}\right)^\alpha y, \left(\frac{\alpha w}{(1-\alpha) r}\right)^{1-\alpha} y \right) \end{eqnarray*}</span></p>
<p>Using this we get the optimal cost of supplying <span class="math-container">$y$</span> units of output:</p>
<p><span class="math-container">\begin{eqnarray*}C(w, r, y) = cy  \end{eqnarray*}</span>
where <span class="math-container">$\displaystyle c = w\left(\frac{(1-\alpha)r}{\alpha w}\right)^\alpha  + r \left(\frac{\alpha w}{(1-\alpha) r}\right)^{1-\alpha}$</span></p>
<p>Now we can use this cost function to solve the profit maximisation problem we wanted to solve. This is the second step:
<span class="math-container">\begin{eqnarray*} \max_{y} &amp; \ py - cy \\ \text{s.t. } &amp; y \geq 0\end{eqnarray*}</span></p>
<p>and get the supply as the solution:
<span class="math-container">\begin{eqnarray*} y^s(w,p,r) \in \begin{cases} \{0\} &amp; \text{if } p &lt; c \\ \mathbb{R}_+ &amp; \text{if } p = c \\ \emptyset &amp; \text{if } p &gt; c\end{cases} \end{eqnarray*}</span></p>
<p>This solve the problem.</p>
<p>To obtain the final labor demand and capital demand, we can use the supply and simply plug it into the conditional input demands we found above.</p>
<p>If <span class="math-container">$r$</span> increases, then <span class="math-container">$c$</span> increases. Now consider these  two possibilities:</p>
<ul>
<li>If <span class="math-container">$p$</span> stays bigger than <span class="math-container">$c$</span> even after the increase in <span class="math-container">$r$</span>, then there is no solution to the optimization problem in either case.</li>
<li>If <span class="math-container">$p$</span> becomes smaller than <span class="math-container">$c$</span> as a result of increase in <span class="math-container">$r$</span> but was initially bigger than <span class="math-container">$c$</span>, then even though the solution now exists but the demand for inputs and the supply of output will be <span class="math-container">$0$</span>.</li>
</ul>
<p>Likewise, we can analyse other possibilities.</p>

CC BY-SA 4.0
2022-05-31T06:05:41.907Z
2022-05-31T06:14:18.493Z
2022-05-31T06:14:18.493Z
Amit
<p>You can use envelope theorem to answer this.</p>
<p>you have your objective function where r is the parameter which changes on optimum values you are interested in:</p>
<p><span class="math-container">$$\mathcal{L}(K,L,r) = f(K,L,r) - \lambda \left( g(K(r),L(r))-c \right) $$</span></p>
<p>Next define the value function as:</p>
<p><span class="math-container">$$ V(K,L,r) = f(K^*(r),L^*(r),r)$$</span></p>
<p>then by envelope theorem we have:</p>
<p><span class="math-container">$$\frac{\partial V}{ \partial r} = \frac{\partial \mathcal{L}^*(K^*(r),L^*(r),\lambda(r),r)}{ \partial r}$$</span></p>
<p>If we apply this to your problem:</p>
<p><span class="math-container">$$\mathcal{L} = wL+rK - \lambda (K^{\alpha}L^{1-\alpha} - \bar{Y})$$</span></p>
<p>So our FOCs are:</p>
<p><span class="math-container">$$ w - (1-\alpha)\lambda K^{\alpha}L^{-\alpha} = 0$$</span></p>
<p><span class="math-container">$$ r - (\alpha)\lambda K^{\alpha-1}L^{1-\alpha} = 0$$</span></p>
<p><span class="math-container">$$K^{\alpha}L^{1-\alpha} = \bar{Y}$$</span></p>
<p>So optimum values of <span class="math-container">$K$</span> and <span class="math-container">$L$</span> are:</p>
<p><span class="math-container">$$K^* = \left(\frac{w}{r} \frac{\alpha}{(1-\alpha)}\right)^{(1-\alpha)} \bar{Y}$$</span></p>
<p><span class="math-container">$$L^* = \left(\frac{r}{w} \frac{1-\alpha}{(\alpha)}\right)^{\alpha} \bar{Y}$$</span></p>
<p>So the value function is:</p>
<p><span class="math-container">$$V(K,L,r) = w\left(\frac{r}{w} \frac{1-\alpha}{(\alpha)}\right)^{\alpha} \bar{Y}+r\left(\frac{w}{r} \frac{\alpha}{(1-\alpha)}\right)^{(1-\alpha)} \bar{Y}$$</span></p>
<p>By taking <span class="math-container">$\frac{\partial V}{\partial r}$</span> you can find out how the cost will be affected by <span class="math-container">$r$</span>.</p>
<p>You could do the same for output but then your objective function has to be the output and you need to use costs as constraints.</p>

CC BY-SA 4.0
2021-12-18T15:08:20.663Z
Interesting. Actually, from this, as Y=wl+rK (zero profit condition), depending on how total costs are affected, we can see how Y would change as TR=TC. But then again, it seems that total costs could increase, resulting in increased production, which is counter-intuitive...
Kwame Brown
CC BY-SA 4.0
2021-12-18T15:11:37.217Z
@KwameBrown you are right you can also use TR=TC to directly see how that affect output. BTW if you think the answer above answered your question consider accepting it
1muflon1
CC BY-SA 4.0
2021-12-18T19:10:16.093Z
If you use the zero profit condition, you will simply derive the relationship between $r$, $w$ and $\alpha$ that would lead to zero profit. You can also get this relationship (alongside the equation for $K/L$) by simply taking FOC of the profit function.
LudwigNagasena
CC BY-SA 4.0
2021-12-18T19:11:29.860Z
@LudwigNagasena that might be possible with nice simplistic functions but the example above is general that will work for any problem when functions are differentiable
1muflon1
CC BY-SA 4.0
2021-12-18T14:25:33.220Z
2021-12-18T14:25:33.220Z
1muflon1
<p>Let us assume a very basic model of profit maximization of firms,
wherein their output depends on labour and capital:
<span class="math-container">$$
Y=f\left(K,L\right)=K^{\alpha}L^{1-\alpha}
$$</span>
Their aim is to maximize profits,
<span class="math-container">$$
\pi=K^{\alpha}L^{1-\alpha}-wL-rK
$$</span>
where <span class="math-container">$w$</span> and <span class="math-container">$r$</span> are given as the price of labour and capital respectively.
The dual of this problem can be written as minimizing total costs,
subject to a attaining a given level of output, <span class="math-container">$\bar{Y}$</span> :
<span class="math-container">$$
\text{min}wL+rK
$$</span>
subject to:
<span class="math-container">$$
Y\leq\bar{Y}
$$</span>
Here, we solve this problem in two stages:</p>
<ol>
<li>Determine profit maximizing level of output (given prices)</li>
<li>Conditional on the level of output, choose <span class="math-container">$L$</span> and <span class="math-container">$K$</span> optimally,</li>
</ol>
<p>We obtain <span class="math-container">$L^{*}$</span> and <span class="math-container">$K^{*}$</span>using Lagrangean methods. My question
is this: if say <span class="math-container">$r$</span> were to increase from these equilibrium values,
what can we <strong>definitively</strong> say?</p>
<p>We know that if this were to
happen, <span class="math-container">$K^{*}$</span> would immediately go down, and <span class="math-container">$L^{*}$</span> would immediately
go up (substitution effect) Can it ever be that <span class="math-container">$L^{*}$</span> goes up enough
that <span class="math-container">$Y^{*}$</span>=<span class="math-container">$f(K^{*},L^{*})$</span> goes up? In other words, can it be
that if the cost of a single input rises, total output can ever <strong>increase</strong>?
If not, is that a consequence of the dimishing marginal productivity
of each of these inputs?</p>

CC BY-SA 4.0
2021-12-18T06:36:29.677Z
Your production function has constant returns to scale, i.e. $f(aK,aL)=af(K,L)$, hence if there are $K$ and $L$ such that $\pi(K, L) = 0$, the firm can produce *any* output whatsoever and if $\pi(K,L)>0$ the firm would like to produce as much as possible leading to no solution. The firm can choose any $Y*$ regardless of $w$ and $r$. Hence, we need some additional constraints for the problem to make sense.
LudwigNagasena
CC BY-SA 4.0
2021-12-18T14:01:42.493Z
Thanks @LudwigNagasena. Are you saying then that the cost structure (relative prices of w and r) has no bearing on total output?
Kwame Brown
CC BY-SA 4.0
2021-12-18T21:35:27.300Z
The prices of $w$ and $r$ have bearing on total output, but there are only three possibilities in your case: $\pi(K*,L*) = 0$ and the firm can produce any output, $\pi(K*,L*) > 0$ and the firm would like to produce as much output as possible, $\pi(K*,L*) < 0$ and the firm would like to produce nothing. Regarding the output going up when the price of an input increases, it is possible in some cases but no homothetic function can have such behavior.
LudwigNagasena
<labor-economics>
Comparative statics when price of an input increases
Kwame Brown
# Comparative statics when price of an input increases

(Asked by: username_0 on 2021-12-18)

Let us assume a very basic model of profit maximization of firms,
wherein their output depends on labour and capital:
$$
Y=f\left(K,L\right)=K^{\alpha}L^{1-\alpha}
$$
Their aim is to maximize profits,
$$
\pi=K^{\alpha}L^{1-\alpha}-wL-rK
$$
where $w$ and $r$ are given as the price of labour and capital respectively.
The dual of this problem can be written as minimizing total costs,
subject to a attaining a given level of output, $\bar{Y}$ :
$$
\text{min}wL+rK
$$
subject to:
$$
Y\leq\bar{Y}
$$
Here, we solve this problem in two stages:

1. Determine profit maximizing level of output (given prices)
2. Conditional on the level of output, choose $L$ and $K$ optimally,

We obtain $L^{\*}$ and $K^{\*}$using Lagrangean methods. My question
is this: if say $r$ were to increase from these equilibrium values,
what can we **definitively** say?

We know that if this were to
happen, $K^{\*}$ would immediately go down, and $L^{\*}$ would immediately
go up (substitution effect) Can it ever be that $L^{\*}$ goes up enough
that $Y^{\*}$=$f(K^{\*},L^{\*})$ goes up? In other words, can it be
that if the cost of a single input rises, total output can ever **increase**?
If not, is that a consequence of the dimishing marginal productivity
of each of these inputs?

**Tags:** labor-economics
> **Comments:**
> **username_1**: Your production function has constant returns to scale, i.e. $f(aK,aL)=af(K,L)$, hence if there are $K$ and $L$ such that $\pi(K, L) = 0$, the firm can produce *any* output whatsoever and if $\pi(K,L)>0$ the firm would like to produce as much as possible leading to no solution. The firm can choose any $Y*$ regardless of $w$ and $r$. Hence, we need some additional constraints for the problem to make sense.  
> **username_0**: Thanks @LudwigNagasena. Are you saying then that the cost structure (relative prices of w and r) has no bearing on total output?  
> **username_1**: The prices of $w$ and $r$ have bearing on total output, but there are only three possibilities in your case: $\pi(K*,L*) = 0$ and the firm can produce any output, $\pi(K*,L*) > 0$ and the firm would like to produce as much output as possible, $\pi(K*,L*) < 0$ and the firm would like to produce nothing. Regarding the output going up when the price of an input increases, it is possible in some cases but no homothetic function can have such behavior.  

---

## 2 answers

### Answer by username_2 on 2022-05-31. Score: 2

Consider the following problem:

\begin{eqnarray\*} \max\_{y, k, l} & py - wl- rk \\ \text{s.t. } & y\leq k^\alpha l^{1-\alpha} \\ \text{and } & k, l, y \geq 0\end{eqnarray\*}

where $p >0$, $w>0$ and $r>0$ are given.

This is how we can solve the above profit maximization problem in two steps:

First solve the cost minimisation problem of the firm:

\begin{eqnarray\*} \min\_{k, l} & wl + rk \\ \text{s.t. } & y\leq k^\alpha l^{1-\alpha} \\ \text{and } & k, l\geq 0\end{eqnarray\*}

Solution to this problem is known as conditional input demand function and satisfy the following two conditions:

$\text{MRTS} = \dfrac{\text{MP}\_L}{\text{MP}\_K} = \dfrac{(1-\alpha) k}{\alpha l} = \dfrac{w}{r}$

$y = k^\alpha l^{1-\alpha}$

Solving these two equations gives us the conditional input demands:
\begin{eqnarray\*} (l^c, k^c) (w, r, y) = \left( \left(\frac{(1-\alpha)r}{\alpha w}\right)^\alpha y, \left(\frac{\alpha w}{(1-\alpha) r}\right)^{1-\alpha} y \right) \end{eqnarray\*}

Using this we get the optimal cost of supplying $y$ units of output:

\begin{eqnarray\*}C(w, r, y) = cy \end{eqnarray\*}
where $\displaystyle c = w\left(\frac{(1-\alpha)r}{\alpha w}\right)^\alpha + r \left(\frac{\alpha w}{(1-\alpha) r}\right)^{1-\alpha}$

Now we can use this cost function to solve the profit maximisation problem we wanted to solve. This is the second step:
\begin{eqnarray\*} \max\_{y} & \ py - cy \\ \text{s.t. } & y \geq 0\end{eqnarray\*}

and get the supply as the solution:
\begin{eqnarray\*} y^s(w,p,r) \in \begin{cases} \{0\} & \text{if } p < c \\ \mathbb{R}\_+ & \text{if } p = c \\ \emptyset & \text{if } p > c\end{cases} \end{eqnarray\*}

This solve the problem.

To obtain the final labor demand and capital demand, we can use the supply and simply plug it into the conditional input demands we found above.

If $r$ increases, then $c$ increases. Now consider these two possibilities:

* If $p$ stays bigger than $c$ even after the increase in $r$, then there is no solution to the optimization problem in either case.
* If $p$ becomes smaller than $c$ as a result of increase in $r$ but was initially bigger than $c$, then even though the solution now exists but the demand for inputs and the supply of output will be $0$.

Likewise, we can analyse other possibilities.


---

### Answer by username_3 on 2021-12-18. Score: 0

You can use envelope theorem to answer this.

you have your objective function where r is the parameter which changes on optimum values you are interested in:

$$\mathcal{L}(K,L,r) = f(K,L,r) - \lambda \left( g(K(r),L(r))-c \right) $$

Next define the value function as:

$$ V(K,L,r) = f(K^\*(r),L^\*(r),r)$$

then by envelope theorem we have:

$$\frac{\partial V}{ \partial r} = \frac{\partial \mathcal{L}^\*(K^\*(r),L^\*(r),\lambda(r),r)}{ \partial r}$$

If we apply this to your problem:

$$\mathcal{L} = wL+rK - \lambda (K^{\alpha}L^{1-\alpha} - \bar{Y})$$

So our FOCs are:

$$ w - (1-\alpha)\lambda K^{\alpha}L^{-\alpha} = 0$$

$$ r - (\alpha)\lambda K^{\alpha-1}L^{1-\alpha} = 0$$

$$K^{\alpha}L^{1-\alpha} = \bar{Y}$$

So optimum values of $K$ and $L$ are:

$$K^\* = \left(\frac{w}{r} \frac{\alpha}{(1-\alpha)}\right)^{(1-\alpha)} \bar{Y}$$

$$L^\* = \left(\frac{r}{w} \frac{1-\alpha}{(\alpha)}\right)^{\alpha} \bar{Y}$$

So the value function is:

$$V(K,L,r) = w\left(\frac{r}{w} \frac{1-\alpha}{(\alpha)}\right)^{\alpha} \bar{Y}+r\left(\frac{w}{r} \frac{\alpha}{(1-\alpha)}\right)^{(1-\alpha)} \bar{Y}$$

By taking $\frac{\partial V}{\partial r}$ you can find out how the cost will be affected by $r$.

You could do the same for output but then your objective function has to be the output and you need to use costs as constraints.

> **Comments:**
> **username_0**: Interesting. Actually, from this, as Y=wl+rK (zero profit condition), depending on how total costs are affected, we can see how Y would change as TR=TC. But then again, it seems that total costs could increase, resulting in increased production, which is counter-intuitive...  
> **username_3**: @KwameBrown you are right you can also use TR=TC to directly see how that affect output. BTW if you think the answer above answered your question consider accepting it  
> **username_1**: If you use the zero profit condition, you will simply derive the relationship between $r$, $w$ and $\alpha$ that would lead to zero profit. You can also get this relationship (alongside the equation for $K/L$) by simply taking FOC of the profit function.  
> **username_3**: @LudwigNagasena that might be possible with nice simplistic functions but the example above is general that will work for any problem when functions are differentiable  

---

