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We have the profit function of the firm profit = $p^2 -2p -399$.

We take derivative of it we say that the output supply function is =$2p-2$

I understand that Profit = q*p - TC But why do we say the output supply function is equal to it's first derivative? We are not maximizing here. Simply finding the derivative.

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  • $\begingroup$ Did you mean to state the profit function only in terms of the output price $p$? $\endgroup$ Mar 20 at 12:52

2 Answers 2

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First, this only works in perfect competition so it is not a general result. Second, the reason why it works is that, first derivative wrt quantity is the condition for maximizing profit. Profit maximizing firm would want to supply quantity to a market such that its profit is maximized. So firm supply will be given by q that satisfies this foc for given price.

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  • $\begingroup$ In this exercise, we found the FOC in regards to price. Does it make any difference? $\endgroup$ Mar 20 at 12:52
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    $\begingroup$ @alioshakaramazov no, in that case q will still by implied by optimum p $\endgroup$
    – 1muflon1
    Mar 20 at 12:59
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Hotelling’s Lemma states that for the indirect profit function $\Pi^\star(p,w,r)$, we have that the firm’s output supply is given by:

$q^s = \frac{d \Pi^\star}{dp}$

Here is its proof:

The optimized profit function is given by

$\Pi^\star(p,w,r) = \Pi(p,L^\star(p),K^\star(p))$

$ = p \cdot q(L^\star(p), K^\star(p)) - w L^\star(p) - r K^\star(p)$

Recall the optimality conditions for profit maximization are given by

$\frac{\partial \Pi}{\partial L}|_{L = L^\star} = 0$,

$\frac{\partial \Pi}{\partial K}|_{K = K^\star} = 0$.

Taking the total derivative of the optimized profit $\Pi^\star$ with respect to $p$ at the optimal inputs,

$\frac{d\Pi^\star}{dp} = \frac{\partial \Pi}{\partial L}|_{L=L^\star(p)} \frac{\partial L}{\partial w} + \frac{\partial \Pi}{\partial K}|_{K=K^\star(p)} \frac{\partial K}{\partial w} + \frac{\partial \Pi}{\partial p} = 0 + 0 + \frac{\partial \Pi}{\partial p} = \frac{\partial \Pi}{\partial p}$

$= q(L^\star(p),K^\star(p)) = q^s(p)$.

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