<p>The uniqueness of z comes from the ND of <span class="math-container">$D^2f$</span> I believe. Let <span class="math-container">$z^*(p)$</span> be the solution of the problem at price <span class="math-container">$p$</span>. Define the function, <span class="math-container">$G(p,z):=p \nabla f(z)-w$</span>. From uniqueness we have <span class="math-container">$z=z^*(p)$</span> if and only if <span class="math-container">$G(p,z)=0$</span>. That is, <span class="math-container">$G(p,z)=0$</span> is an implicit equation which tells us how <span class="math-container">$z^*(p)$</span> changes when <span class="math-container">$p$</span> does. The IFT gives us a way to find the derivative of this function</p>
<p><span class="math-container">$$\frac{\partial z^*_i}{\partial p}=\frac{-\frac{\partial G}{\partial p}}{\frac{\partial G}{\partial z_i}}=-\frac{1}{p}H^{-1}\nabla f(z)$$</span></p>
<p>Where <span class="math-container">$H$</span> is the hessian of <span class="math-container">$f$</span>.
So <span class="math-container">$\sum_i\frac{\partial f}{\partial z_i}\frac{\partial z^*_i}{\partial p}&gt;0$</span> as <span class="math-container">$-H^{-1}$</span> is positive def.</p>
<p>The derivaitve of the profit function wrt to <span class="math-container">$p$</span> is <span class="math-container">$\frac{d}{dp}pf(z^*(p))=f(z(p))+p\sum_i\frac{\partial f}{\partial z_i}\frac{\partial z^*_i}{\partial p}$</span></p>

CC BY-SA 4.0
2021-03-25T18:37:48.807Z
2021-03-26T00:11:10.350Z
2021-03-26T00:11:10.350Z
user28714
<p>MWG 5C6 asks: &quot;Suppose a concave prod function f(z) with inputs <span class="math-container">$(z_1,...,z_L-1)$</span> and also that <span class="math-container">$\partial f(z))/\partial z_l \geqslant 0$</span> for all l and <span class="math-container">$z\geqslant0$</span> and that <span class="math-container">$D^2f(z)$</span> 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 <span class="math-container">$p\bigtriangledown f(z)-w=0$</span>.</p>

<microeconomics><production-function><mwg>
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?
Daniele Muniz
# 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?

(Asked by: username_0 on 2021-03-20)

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

**Tags:** microeconomics, production-function, mwg

---

## 1 answers

### Answer by username_1 on 2021-03-25. Score: 3

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


---

