# I need to prove how an increase in output p increases profit-max. Can someone help to understand why IFT implies that z is a unique maximizing point? [closed]

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 $$p\bigtriangledown f(z)-w=0$$.

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 $$\frac{d}{dp}pf(z^*(p))=f(z(p))+p\sum_i\frac{\partial f}{\partial z_i}\frac{\partial z^*_i}{\partial p}$$