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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?

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    $\begingroup$ 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. $\endgroup$ Dec 18, 2021 at 6:36
  • $\begingroup$ Thanks @LudwigNagasena. Are you saying then that the cost structure (relative prices of w and r) has no bearing on total output? $\endgroup$ Dec 18, 2021 at 14:01
  • $\begingroup$ 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. $\endgroup$ Dec 18, 2021 at 21:35

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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.

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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.

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  • $\begingroup$ 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... $\endgroup$ Dec 18, 2021 at 15:08
  • $\begingroup$ @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 $\endgroup$
    – 1muflon1
    Dec 18, 2021 at 15:11
  • $\begingroup$ 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. $\endgroup$ Dec 18, 2021 at 19:10
  • $\begingroup$ @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 $\endgroup$
    – 1muflon1
    Dec 18, 2021 at 19:11

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