<p>The identity <span class="math-container">$y=y$</span> can be surprinsingly useful when both expressions for <span class="math-container">$y$</span> are spuriously different. This can be illustrated by the inverse function theorem. If <span class="math-container">$y=f(x)$</span> is equivalent to <span class="math-container">$x=f^{-1}(y)$</span>, it is identically true that: <span class="math-container">$$ y=f(f^{-1}(y)) \hspace{5mm} or \hspace{5mm} x=f^{-1}(f(x)),$$</span> which yields the classical relationship between <span class="math-container">$f'$</span> and <span class="math-container">$(f^{-1})'$</span>. In the multiple variable case of your example, it is quite similar.</p>

CC BY-SA 4.0
2021-12-09T18:54:39.230Z
2021-12-09T18:54:39.230Z
Bertrand
<p>Suppose the firm has a minimum cost function <span class="math-container">$C(\vec{w}, q)$</span> and sets up the following profit maximisation problem:</p>
<p><span class="math-container">$max_{q} \text{ } pq - C(\vec{w}, q)$</span>. The below FOC characterises the solution:</p>
<p><span class="math-container">$p = C_q(\vec{w}, q)$</span> where the subscript denotes partial derivatives.</p>
<p>If I understand correctly, rearranging the FOC to get the <span class="math-container">$q$</span> as a function of <span class="math-container">$\vec{w}$</span> and <span class="math-container">$p$</span> gives us the supply function <span class="math-container">$S(\vec{w}, p) = q$</span>. To investigate <span class="math-container">$q$</span>'s response to changes in <span class="math-container">$p$</span> and <span class="math-container">$\vec{w}$</span> we substitute the supply function into the FOC <em>from which it was originally derived</em>:</p>
<p><span class="math-container">$$p = C_q(\vec{w}, S(\vec{w}, p))$$</span></p>
<p>Then we would compute partial derivatives and study its signs. What is the intution behind the maths that allow us to rearrange an equation only to substitute the result back into that same equation? If we have an equation <span class="math-container">$y = x + 3$</span> and rearrange it to <span class="math-container">$x = y - 3$</span>, would not there be very little use in substituting the latter into the former to arrive at <span class="math-container">$y = y$</span>?</p>

<mathematical-economics><profit-maximization><producer-theory>
Investigating the firm's supply function
Panhaboth K
# Investigating the firm's supply function

(Asked by: username_0 on 2021-12-09)

Suppose the firm has a minimum cost function $C(\vec{w}, q)$ and sets up the following profit maximisation problem:

$max\_{q} \text{ } pq - C(\vec{w}, q)$. The below FOC characterises the solution:

$p = C\_q(\vec{w}, q)$ where the subscript denotes partial derivatives.

If I understand correctly, rearranging the FOC to get the $q$ as a function of $\vec{w}$ and $p$ gives us the supply function $S(\vec{w}, p) = q$. To investigate $q$'s response to changes in $p$ and $\vec{w}$ we substitute the supply function into the FOC *from which it was originally derived*:

$$p = C\_q(\vec{w}, S(\vec{w}, p))$$

Then we would compute partial derivatives and study its signs. What is the intution behind the maths that allow us to rearrange an equation only to substitute the result back into that same equation? If we have an equation $y = x + 3$ and rearrange it to $x = y - 3$, would not there be very little use in substituting the latter into the former to arrive at $y = y$?

**Tags:** mathematical-economics, profit-maximization, producer-theory


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## 1 answers

### Answer by username_1 on 2021-12-09. Score: 0 | Accepted answer

The identity $y=y$ can be surprinsingly useful when both expressions for $y$ are spuriously different. This can be illustrated by the inverse function theorem. If $y=f(x)$ is equivalent to $x=f^{-1}(y)$, it is identically true that: $$ y=f(f^{-1}(y)) \hspace{5mm} or \hspace{5mm} x=f^{-1}(f(x)),$$ which yields the classical relationship between $f'$ and $(f^{-1})'$. In the multiple variable case of your example, it is quite similar.


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