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Given

$$f(x_1,x_2) = \min\{x_1,x_2\}^\alpha$$

I have to find the profit-maximizing demand functions, supply function and profit function but I am not sure how to when the function is given as it is.

I know that I for example have to use Lagrange. Thus

$$ L = p \cdot f(x_1,x_2) - w_1 \cdot x_1 - w_2 \cdot x_2 - \lambda \left(p_1x_1 + p_2x_2 - m \right) $$ and then afterwards find the first order conditions. But how I find the derivative in regards to both $x_1$ and $x_2$ when I have a function such as $f(x_1,x_2) = \min\{x_1,x_2\}^\alpha$ ?

Can you help me in the right direction?

Thanks in advance!

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    $\begingroup$ "I know that I for example have to use Lagrange." How do you know this? Is it explicitly stated in the exercise? $\endgroup$
    – Giskard
    Mar 30, 2020 at 12:32
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    $\begingroup$ Actually it is not. I "just" thought I had to use Lagrange as this is normally what I would do. Is there any other way? I know that for cost minimization there are other ways. $\endgroup$
    – Mathias
    Mar 30, 2020 at 12:34
  • $\begingroup$ And do you know that all profit maximization problems imply a cost minimization problem as well? If $x_1^*,x_2^*$ are the solution to the profit maximization problem, they are also to solution to the cost minimization problem assuming an output of $y = f(x_1^*,x_2^*)$. So if you can gleam further insights from the cost minimization problem, do so. $\endgroup$
    – Giskard
    Mar 30, 2020 at 13:22
  • $\begingroup$ For cost minimization I know that in optimum we have that $$ TRS = \frac{\partial y / \partial x_1}{\partial y / \partial x_2} = \frac{w_1}{w_2}$$ and because we keep a constant output given by $$f(x_1,x_2) = \min \{x_1,x_2\}^\alpha$$ we can solve this equation system. But how do I for example find $$\frac{ \partial y}{ \partial x_1}$$ where $$ y = f(x_1,x_2)$$. Thanks in advance. $\endgroup$
    – Mathias
    Mar 30, 2020 at 13:57
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    $\begingroup$ How about using the definition of differentiation? The solution is also quite intuitive, you can get it without differentiation. You can also look around the site, there are similar questions about the $\min$ function. It is also likely that you have covered something like this in class. Anyhow, this seems like the self study part, so I am just going to go. I hope your work will be fruitful. $\endgroup$
    – Giskard
    Mar 30, 2020 at 15:03

2 Answers 2

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Observe that if $x_1^{*} > x_2^{*}$, then $$\pi(x_1^{*}, x_2^{*}) = p {x_2^*}^{\alpha} - w_1x_1^{*} - w_2x_2^* < p {x_2^*}^{\alpha} - w_1x_2^{*} - w_2x_2^* = \pi(x_2^*, x_2^*)$$

Due to symmetry, we can conclude that the optimal $x_1$ and $x_2$ will be equal.

From the profit function $\pi(w_1, w_2) = \max_{x \geq 0} [px^\alpha - w_1x - w_2x]$, we get

$$x_i^*(p,w_1, w_2) = \left(\frac{p \alpha}{w_1 + w_2}\right)^{\frac{1}{1-\alpha}}$$ for $i=1,2$

Calculate $\Pi^*(p,w_1,w_2)$ using the factor demands we found above.

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Given the production function $f:\mathbb{R}^2_+\rightarrow\mathbb{R}$ defined as $f(x_1,x_2) = (\min(x_1,x_2))^\alpha$, where $\alpha > 0$, we can solve the profit maximisation problem of the competitive firm in two steps.

In step 1, we solve the cost minimisation problem: \begin{eqnarray*}\min_{x_1\geq 0, x_2\geq 0} & w_1x_1+w_2x_2\\ \text{s.t. } & (\min(x_1,x_2))^\alpha \geq y\end{eqnarray*} where $w_1 > 0, w_2 > 0$ and $y \geq 0$.

Solving the problem gives conditional input demands as: \begin{eqnarray*}(x_1^c,x_2^c)(w_1,w_2,y) = (y^{\frac{1}{\alpha}}, y^{\frac{1}{\alpha}})\end{eqnarray*}

and the cost function is \begin{eqnarray*}c(w_1,w_2,y) = (w_1+w_2)y^{\frac{1}{\alpha}}\end{eqnarray*}

In step 2, we'll solve the profit maximisation problem which is \begin{eqnarray*}\max_{y\geq 0} & \ py - (w_1+w_2)y^{\frac{1}{\alpha}}\end{eqnarray*} where $p>0$, $w_1>0$ and $w_2>0$

Here we'll consider three cases for $\alpha$:

  • $0 < \alpha < 1$ (Case of Decreasing returns to scale): Solving the profit maximisation problem in this case will give the supply function as: \begin{eqnarray*}y^s(w_1,w_2,p) = \left(\frac{\alpha p}{ w_1+w_2}\right)^{\frac{\alpha}{1-\alpha}}\end{eqnarray*} and the input demands are \begin{eqnarray*}(x_1^d,x_2^d)(w_1,w_2,p) = \left(\left(\frac{\alpha p}{ w_1+w_2}\right)^{\frac{1}{1-\alpha}}, \left(\frac{\alpha p}{ w_1+w_2}\right)^{\frac{1}{1-\alpha}}\right)\end{eqnarray*} Profit function is \begin{eqnarray*}\pi(w_1,w_2,p) = p\left(\frac{\alpha p}{ w_1+w_2}\right)^{\frac{\alpha}{1-\alpha}} - (w_1+w_2)\left(\frac{\alpha p}{ w_1+w_2}\right)^{\frac{1}{1-\alpha}}\end{eqnarray*}

  • $\alpha = 1$ (Case of Constant returns to scale): In this case, supply is \begin{eqnarray*}y^s(w_1,w_2,p) \in \begin{cases} \{0\} & \text{if } p < w_1+w_2 \\ \mathbb{R}_+ & \text{if } p = w_1+w_2 \\ \emptyset & \text{if } p > w_1+w_2 \end{cases} \end{eqnarray*} and the input demands are \begin{eqnarray*}(x_1^d,x_2^d)(w_1,w_2,p) = (x_1^c,x_2^c)(w_1,w_2,y^s(w_1,w_2,p))\end{eqnarray*} Here, profit function is only defined for the case where $p\leq w_1+w_2$ and is given by \begin{eqnarray*}\pi(w_1,w_2,p) = 0\end{eqnarray*}

  • $\alpha > 1$ (Case of Increasing returns to scale): Here the solution to the profit maximisation problem does not exist i.e. supply does not exist.

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