2
$\begingroup$

Consider a firm that has a production function given by $F(x)$, where $x \in \mathbb{R}^n_+$. Assume that the function $F$ is strictly increasing in each argument, concave, twice continuously differentiable and homogeneous of degree one. Let the price of the good produced be given and equal to $p$ and let the vector of input prices be $w \in \mathbb{R}^n_+$.

(i) Show that if the firm acts as a price taker profits are zero.

The firm's maximization problem is $\max_x pF(x) -wx $. The first order condition tells us that $p \frac{\partial F}{\partial x_j}(x^*) = w$ for $j = 1,2, ... , n$. So, $\pi^* = p F(x^*) - p\frac{\partial F}{\partial x_j}(x^*)x^* =p [F(x^*) - \frac{\partial F}{\partial x_j}(x^*)x^*]$. For the right most side to be equal to zero, the average product should be equal to the marginal product of $x_j$ for all $j$. But why should it be true?

(ii) Show that if there are $N$ firms with the same production function, that are at an interior point in the input space and facing the same input prices, then given the total amount of all inputs used, the total output of the industry is independent of $N$, regardless of whether they behave as price takes. In what sense does this result justify the usual assumption that there is only one firm, i.e. $N=1$.

I know that this problem has to do with the derivative of a homogeneous of degree one function has a homogeneous of degree zero, but I don't know where I should use it in the proof. I think that there are $N$ identical firms, and these $N$ firms produces outputs until marginal product is equal to marginal cost. Therefore, the total output is not affected by $N$. How can I mathematically present this result? Why does my first statement has to do with this problem?

$\endgroup$
2
  • 2
    $\begingroup$ "$\frac{\partial F}{\partial x_j}(x^*)x^*$" should be a sum ($wx$ should be a dot product). Then use so-called "Euler's identity for homogeneous functions." $\endgroup$ – Michael Oct 3 '20 at 3:49
  • 2
    $\begingroup$ Please ask one question only. $\endgroup$ – Michael Greinecker Oct 3 '20 at 8:41

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.