<p>If one pursues your calculations and differentiates the first-order condition,
<span class="math-container">$$
  \frac{\partial}{\partial x_1} f(x_1^*, x_2^*) = w_1
$$</span>
w.r.t. <span class="math-container">$w_2,$</span> this yields
<span class="math-container">$$
\frac{\partial^2 f}{\partial x_1 ^2} (x_1^*, x_2^*) \frac{\partial x^*_1}{\partial w_2} (w)
 + \frac{\partial^2 f}{\partial x_1 \partial x_2} (x_1^*, x_2^*) \frac{\partial x^*_2}{\partial w_2} (w) = 0.
$$</span>
If we have <span class="math-container">$\partial x^*_1 / \partial w_2 (w)=0$</span> and <span class="math-container">$\partial x^*_2 / \partial w_2 (w) \neq 0,$</span> then
<span class="math-container">$$
\frac{\partial^2 f}{\partial x_1 \partial x_2} (x_1^*, x_2^*)=0,
$$</span>
for any <span class="math-container">$(x_1^*, x_2^*) \in S(w).$</span> This condition implies that <span class="math-container">$f(x_1, x_2)=g_1(x_1)+g_2(x_2)$</span> on the set of inputs <span class="math-container">$ X = \bigcup\limits_{w\in \mathbb{R}^2} S(w)$</span>.</p>
<p><strong>EDIT</strong>: Integrating <span class="math-container">$$
\frac{\partial^2 f}{\partial x_1 \partial x_2} (x_1, x_2)=0
$$</span>
wrt <span class="math-container">$x_1$</span>, yield <span class="math-container">$$ \frac{\partial f}{\partial x_2} (x_1, x_2)=h_2(x_2), $$</span> and integration wrt <span class="math-container">$x_2$</span> gives the production function for <span class="math-container">$x \in X$</span>:
<span class="math-container">$$ f(x_1, x_2)=g_1(x_1)+g_2(x_2). $$</span></p>

CC BY-SA 4.0
2023-12-08T14:44:24.293Z
Thanks for your answer. Could I ask some more questions based on your answer? 1). Is $S(w)$ means all the optimal factors $(x_1^*, x_2^*)$?   2). Suppose $f: R^2_+ \rightarrow R$. Do we need to worry about whether $(x_1^*, x_2^*) \in S(w)$ can fill in with $R^2_+$ so that we can say that $f(x_1, x_2)=g_1(x_1)+g_2(x_2)$?
dchao
CC BY-SA 4.0
2023-12-08T20:41:20.377Z
1) Yes, the $S(w)$ denotes the set of all optimal inputs that can be chosen for input prices $w$. The union of all these sets define the set of inputs $X$.
Bertrand
CC BY-SA 4.0
2023-12-08T20:46:55.537Z
2) Usually $X$ does not coincide with $\mathbb{R}^2$, because for the initial production function $f$ some $x$ can never be chosen rationally (in the case where $f$ is not quasi-concave for instance). So we can only recover from the profit function, the convex envelope of $f$.
Bertrand
CC BY-SA 4.0
2023-12-09T04:11:57.253Z
Thanks for you kind response. I think you are correct even I am not sure because of my lacking in some optimization knowledge like convex envelope. I am just not sure whether it is valid we can imply from $\frac{\partial^2 f}{\partial x_1 \partial x_2} (x_1^*, x_2^*)=0$ to $f(x_1, x_2)=g_1(x_1)+g_2(x_2)$. But really thanks to you! I may look back to this question when I understand more on that.
dchao
CC BY-SA 4.0
2023-12-10T20:16:58.737Z
@dchao: I have added an edit to answer your question.
Bertrand
CC BY-SA 4.0
2023-12-07T21:51:54.957Z
2023-12-10T20:16:24.887Z
2023-12-10T20:16:24.887Z
Bertrand
<p>I have a question on the following exercise:</p>
<blockquote>
<p>A competitive profit-maximizing firm has a profit function <span class="math-container">$\pi(w_1, w_2) = \phi_1(w_1) + \phi_2(w_2)$</span>. The price of output is normalized to be 1. <br />
(c). Let <span class="math-container">$f(x_1,x_2)$</span> be the production function that generated the profit function of this form. What can we say aboutthe form of this production function? (Hint: look at the first-order condition.)</p>
</blockquote>
<p>The provided solution states that</p>
<blockquote>
<p>The demand for factor <span class="math-container">$i$</span> is only a function of the <span class="math-container">$i$</span>-th price. Therefore the marginal product of factor <span class="math-container">$i$</span> can only depend on the amount of factor <span class="math-container">$i$</span>. It follows that <span class="math-container">$f(x_1,x_2)=g_1(x_1) + g_2(x_2)$</span>.</p>
</blockquote>
<p><strong>Could anyone help to provide mathematical proof on this?</strong></p>
<p>My process is as follows:
By Hotelling's Lemma,
<span class="math-container">$$
  -x_1(w_1, w_2) = \frac{\partial \pi}{\partial w_1} = \frac{\partial \phi_1(w_1)}{\partial w_1}
$$</span>
Therefore, the factor demand function for good 1 only depends on <span class="math-container">$w_1$</span> and we can rewrite it as <span class="math-container">$x_1(w_1)$</span>. Similarly, factor demand function for good 2 is <span class="math-container">$x_2(w_2)$</span>.</p>
<p>The profit function now is
<span class="math-container">$$
  \pi(w_1,w_2) = max_x f(x_1,x_2) - w_1 x_1 - w_2 x_2.
$$</span>
By first-order condition, when <span class="math-container">$x_1,x_2$</span> achieves optimal,
<span class="math-container">$$
  \frac{\partial}{\partial x_i} f(x_1(w_1), x_2(w_2)) = w_i.
$$</span>
How can we use this condition to derive</p>
<blockquote>
<p>The marginal product of factor <span class="math-container">$i$</span> can only depend on the amount of factor <span class="math-container">$i$</span>
???</p>
</blockquote>
<p>Thank you.</p>

<microeconomics>
A question about an exercise in Hal R. Varian, Microeconomic Analysis (1984), Ch.3 Exercises 3.1 (c), Page 48
dchao
# A question about an exercise in Hal R. Varian, Microeconomic Analysis (1984), Ch.3 Exercises 3.1 (c), Page 48

(Asked by: username_0 on 2023-12-05)

I have a question on the following exercise:

> A competitive profit-maximizing firm has a profit function $\pi(w\_1, w\_2) = \phi\_1(w\_1) + \phi\_2(w\_2)$. The price of output is normalized to be 1.   
> (c). Let $f(x\_1,x\_2)$ be the production function that generated the profit function of this form. What can we say aboutthe form of this production function? (Hint: look at the first-order condition.)

The provided solution states that

> The demand for factor $i$ is only a function of the $i$-th price. Therefore the marginal product of factor $i$ can only depend on the amount of factor $i$. It follows that $f(x\_1,x\_2)=g\_1(x\_1) + g\_2(x\_2)$.

**Could anyone help to provide mathematical proof on this?**

My process is as follows:
By Hotelling's Lemma,
$$
-x\_1(w\_1, w\_2) = \frac{\partial \pi}{\partial w\_1} = \frac{\partial \phi\_1(w\_1)}{\partial w\_1}
$$
Therefore, the factor demand function for good 1 only depends on $w\_1$ and we can rewrite it as $x\_1(w\_1)$. Similarly, factor demand function for good 2 is $x\_2(w\_2)$.

The profit function now is
$$
\pi(w\_1,w\_2) = max\_x f(x\_1,x\_2) - w\_1 x\_1 - w\_2 x\_2.
$$
By first-order condition, when $x\_1,x\_2$ achieves optimal,
$$
\frac{\partial}{\partial x\_i} f(x\_1(w\_1), x\_2(w\_2)) = w\_i.
$$
How can we use this condition to derive

> The marginal product of factor $i$ can only depend on the amount of factor $i$
> ???

Thank you.

**Tags:** microeconomics

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## 1 answers

### Answer by username_1 on 2023-12-07. Score: 5 | Accepted answer

If one pursues your calculations and differentiates the first-order condition,
$$
\frac{\partial}{\partial x\_1} f(x\_1^\*, x\_2^\*) = w\_1
$$
w.r.t. $w\_2,$ this yields
$$
\frac{\partial^2 f}{\partial x\_1 ^2} (x\_1^\*, x\_2^\*) \frac{\partial x^\*\_1}{\partial w\_2} (w)
+ \frac{\partial^2 f}{\partial x\_1 \partial x\_2} (x\_1^\*, x\_2^\*) \frac{\partial x^\*\_2}{\partial w\_2} (w) = 0.
$$
If we have $\partial x^\*\_1 / \partial w\_2 (w)=0$ and $\partial x^\*\_2 / \partial w\_2 (w) \neq 0,$ then
$$
\frac{\partial^2 f}{\partial x\_1 \partial x\_2} (x\_1^\*, x\_2^\*)=0,
$$
for any $(x\_1^\*, x\_2^\*) \in S(w).$ This condition implies that $f(x\_1, x\_2)=g\_1(x\_1)+g\_2(x\_2)$ on the set of inputs $ X = \bigcup\limits\_{w\in \mathbb{R}^2} S(w)$.

**EDIT**: Integrating $$
\frac{\partial^2 f}{\partial x\_1 \partial x\_2} (x\_1, x\_2)=0
$$
wrt $x\_1$, yield $$ \frac{\partial f}{\partial x\_2} (x\_1, x\_2)=h\_2(x\_2), $$ and integration wrt $x\_2$ gives the production function for $x \in X$:
$$ f(x\_1, x\_2)=g\_1(x\_1)+g\_2(x\_2). $$

> **Comments:**
> **username_0**: Thanks for your answer. Could I ask some more questions based on your answer? 1). Is $S(w)$ means all the optimal factors $(x_1^*, x_2^*)$?   2). Suppose $f: R^2_+ \rightarrow R$. Do we need to worry about whether $(x_1^*, x_2^*) \in S(w)$ can fill in with $R^2_+$ so that we can say that $f(x_1, x_2)=g_1(x_1)+g_2(x_2)$?  
> **username_1**: 1) Yes, the $S(w)$ denotes the set of all optimal inputs that can be chosen for input prices $w$. The union of all these sets define the set of inputs $X$.  
> **username_1**: 2) Usually $X$ does not coincide with $\mathbb{R}^2$, because for the initial production function $f$ some $x$ can never be chosen rationally (in the case where $f$ is not quasi-concave for instance). So we can only recover from the profit function, the convex envelope of $f$.  
> **username_0**: Thanks for you kind response. I think you are correct even I am not sure because of my lacking in some optimization knowledge like convex envelope. I am just not sure whether it is valid we can imply from $\frac{\partial^2 f}{\partial x_1 \partial x_2} (x_1^*, x_2^*)=0$ to $f(x_1, x_2)=g_1(x_1)+g_2(x_2)$. But really thanks to you! I may look back to this question when I understand more on that.  
> **username_1**: @dchao: I have added an edit to answer your question.  

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