<p>Observe that if <span class="math-container">$x_1^{*} &gt; x_2^{*}$</span>, then <span class="math-container">$$\pi(x_1^{*}, x_2^{*}) = p {x_2^*}^{\alpha} - w_1x_1^{*} - w_2x_2^* &lt; p {x_2^*}^{\alpha} - w_1x_2^{*} - w_2x_2^* = \pi(x_2^*, x_2^*)$$</span></p>
<p>Due to symmetry, we can conclude that the optimal <span class="math-container">$x_1$</span> and <span class="math-container">$x_2$</span> will be equal.</p>
<p>From the profit function <span class="math-container">$\pi(w_1, w_2) = \max_{x \geq 0} [px^\alpha - w_1x - w_2x]$</span>, we get</p>
<p><span class="math-container">$$x_i^*(p,w_1, w_2) = \left(\frac{p \alpha}{w_1 + w_2}\right)^{\frac{1}{1-\alpha}}$$</span> for <span class="math-container">$i=1,2$</span></p>
<p>Calculate <span class="math-container">$\Pi^*(p,w_1,w_2)$</span> using the factor demands we found above.</p>

CC BY-SA 4.0
2023-04-25T10:03:40.737Z
2023-04-25T10:03:40.737Z
solow supremacy
<p>Given the production function <span class="math-container">$f:\mathbb{R}^2_+\rightarrow\mathbb{R}$</span> defined as <span class="math-container">$f(x_1,x_2) = (\min(x_1,x_2))^\alpha$</span>, where <span class="math-container">$\alpha &gt; 0$</span>, we can solve the profit maximisation problem of the competitive firm in two steps.</p>
<p>In step 1, we solve the cost minimisation problem:
<span class="math-container">\begin{eqnarray*}\min_{x_1\geq 0, x_2\geq 0} &amp; w_1x_1+w_2x_2\\ \text{s.t. } &amp; (\min(x_1,x_2))^\alpha \geq y\end{eqnarray*}</span>
where <span class="math-container">$w_1 &gt; 0, w_2 &gt; 0$</span> and <span class="math-container">$y \geq 0$</span>.</p>
<p>Solving the problem gives conditional input demands as:
<span class="math-container">\begin{eqnarray*}(x_1^c,x_2^c)(w_1,w_2,y) = (y^{\frac{1}{\alpha}}, y^{\frac{1}{\alpha}})\end{eqnarray*}</span></p>
<p>and the cost function is
<span class="math-container">\begin{eqnarray*}c(w_1,w_2,y) = (w_1+w_2)y^{\frac{1}{\alpha}}\end{eqnarray*}</span></p>
<p>In step 2, we'll solve the profit maximisation problem which is
<span class="math-container">\begin{eqnarray*}\max_{y\geq 0} &amp; \ py - (w_1+w_2)y^{\frac{1}{\alpha}}\end{eqnarray*}</span>
where <span class="math-container">$p&gt;0$</span>, <span class="math-container">$w_1&gt;0$</span> and <span class="math-container">$w_2&gt;0$</span></p>
<p>Here we'll consider three cases for <span class="math-container">$\alpha$</span>:</p>
<ul>
<li><p><span class="math-container">$0 &lt; \alpha &lt; 1$</span> (Case of Decreasing returns to scale): Solving the profit maximisation problem in this case will give the supply function as:
<span class="math-container">\begin{eqnarray*}y^s(w_1,w_2,p) = \left(\frac{\alpha p}{ w_1+w_2}\right)^{\frac{\alpha}{1-\alpha}}\end{eqnarray*}</span>
and the input demands are
<span class="math-container">\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*}</span>
Profit function is
<span class="math-container">\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*}</span></p>
</li>
<li><p><span class="math-container">$\alpha = 1$</span> (Case of Constant returns to scale): In this case, supply is
<span class="math-container">\begin{eqnarray*}y^s(w_1,w_2,p) \in \begin{cases} \{0\} &amp; \text{if } p &lt; w_1+w_2 \\ \mathbb{R}_+ &amp; \text{if } p = w_1+w_2 \\ \emptyset &amp; \text{if } p &gt; w_1+w_2 \end{cases} \end{eqnarray*}</span>
and the input demands are
<span class="math-container">\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*}</span>
Here, profit function is only defined for the case where <span class="math-container">$p\leq w_1+w_2$</span> and is given by
<span class="math-container">\begin{eqnarray*}\pi(w_1,w_2,p) = 0\end{eqnarray*}</span></p>
</li>
<li><p><span class="math-container">$\alpha &gt; 1$</span> (Case of Increasing returns to scale): Here the solution to the profit maximisation problem does not exist i.e. supply does not exist.</p>
</li>
</ul>

CC BY-SA 4.0
2023-04-25T12:28:12.533Z
2023-04-25T12:44:42.640Z
2023-04-25T12:44:42.640Z
Amit
<p>Given</p>

<p><span class="math-container">$$f(x_1,x_2) = \min\{x_1,x_2\}^\alpha$$</span></p>

<p>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.</p>

<p>I know that I for example have to use Lagrange. Thus </p>

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

<p>Can you help me in the right direction?</p>

<p>Thanks in advance! </p>

CC BY-SA 4.0
2020-03-30T12:32:21.270Z
"I know that I for example have to use Lagrange." How do you know this? Is it explicitly stated in the exercise?
Giskard
CC BY-SA 4.0
2020-03-30T12:34:17.097Z
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.
Mathias
CC BY-SA 4.0
2020-03-30T13:22:56.583Z
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.
Giskard
CC BY-SA 4.0
2020-03-30T13:57:15.263Z
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.
Mathias
CC BY-SA 4.0
2020-03-30T15:03:01.047Z
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.
Giskard
<profit-maximization>
Given $f(x_1,x_2) = \min\{x_1,x_2\}^\alpha$ find the profit-maximizing factor demands, supply function and profit fuction
Mathias
# Given $f(x_1,x_2) = \min\{x_1,x_2\}^\alpha$ find the profit-maximizing factor demands, supply function and profit fuction

(Asked by: username_0 on 2020-03-30)

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!

**Tags:** profit-maximization

> **Comments:**
> **username_1**: "I know that I for example have to use Lagrange." How do you know this? Is it explicitly stated in the exercise?  
> **username_0**: 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.  
> **username_1**: 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.  
> **username_0**: 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.  
> **username_1**: 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.  

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## 2 answers

### Answer by username_2 on 2023-04-25. Score: 2

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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### Answer by username_3 on 2023-04-25. Score: 2

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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