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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.

dchao
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1 Answers1

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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 $x \in X$: $$ f(x_1, x_2)=g_1(x_1)+g_2(x_2). $$

Bertrand
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  • 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 Dec 08 '23 at 14:44
  • 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 Dec 08 '23 at 20:41
  • 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 Dec 08 '23 at 20:46
  • 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 Dec 09 '23 at 04:11
  • @dchao: I have added an edit to answer your question. – Bertrand Dec 10 '23 at 20:16