1
$\begingroup$

MWG 5C6 asks: "Suppose a concave prod function f(z) with inputs $(z_1,...,z_L-1)$ and also that $\partial f(z))/\partial z_l \geqslant 0$ for all l and $z\geqslant0$ and that $D^2f(z)$ is negative definite at all z. Use FOC and Implicit Function Theorem (IFT) to prove that an increase in output price increases profit-max level. I did undertand the most part of Solutions, although it´s not clear why does the Implict Function Theorem implies that z is a unique maximizing point that satisfies $p\bigtriangledown f(z)-w=0$.

$\endgroup$
3
$\begingroup$

The uniqueness of z comes from the ND of $D^2f$ I believe. Let $z^*(p)$ be the solution of the problem at price $p$. Define the function, $G(p,z):=p \nabla f(z)-w$. From uniqueness we have $z=z^*(p)$ if and only if $G(p,z)=0$. That is, $G(p,z)=0$ is an implicit equation which tells us how $z^*(p)$ changes when $p$ does. The IFT gives us a way to find the derivative of this function

$$\frac{\partial z^*_i}{\partial p}=\frac{-\frac{\partial G}{\partial p}}{\frac{\partial G}{\partial z_i}}=-\frac{1}{p}H^{-1}\nabla f(z)$$

Where $H$ is the hessian of $f$. So $\sum_i\frac{\partial f}{\partial z_i}\frac{\partial z^*_i}{\partial p}>0$ as $-H^{-1}$ is positive def.

The derivaitve of the profit function wrt to $p$ is $\frac{d}{dp}pf(z^*(p))=f(z(p))+p\sum_i\frac{\partial f}{\partial z_i}\frac{\partial z^*_i}{\partial p}$

$\endgroup$
0

Not the answer you're looking for? Browse other questions tagged or ask your own question.