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

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.

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 $$x \in X$$: $$f(x_1, x_2)=g_1(x_1)+g_2(x_2).$$
• 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 $f(x_1, x_2)=g_1(x_1)+g_2(x_2)$? Dec 8, 2023 at 14:44
• 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 $X$. Dec 8, 2023 at 20:41
• 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 $f$. Dec 8, 2023 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. Dec 9, 2023 at 4:11