<p>Is the current production level reached at minimum cost? If not, reach it by lower cost. Revenue is unchanged, hence profit increases, thus it was not maximized.</p>

CC BY-SA 4.0
2022-10-31T17:11:26.003Z
Yes, so the proof I wrote is correct? I actually saw Varian's solution manual saying, in the contradiction part: $\exists \ x^{**} : f(x^{**}) \geq f(x^*)$ and $w \cdot x^{*} > w \cdot x^{**}$. I got confused with the $f(x^{**}) \geq f(x^*)$ part as I thought $q^{*} = f(x^*)$ is fixed.
Rick_Morty
CC BY-SA 4.0
2022-10-31T21:13:14.927Z
I am unwilling to comment beyond my proof.
Giskard
CC BY-SA 4.0
2022-10-31T17:04:22.863Z
2022-10-31T17:04:22.863Z
Giskard
<p>Hey hope this helps you out. The maximization problem is:</p>
<p><span class="math-container">$$\max_{x_1,x_2,y} py - w_1x_1 - w_2x_2$$</span></p>
<p><span class="math-container">$s.t. y=f(x_1,x_2)$</span></p>
<p>In order to maximize, we could first choose to maximize with respect to <span class="math-container">$x_1$</span> then to <span class="math-container">$x_2$</span> and later to <span class="math-container">$y$</span>. So we are going to use this in our advantage. First, we are going to maximize with respect to <span class="math-container">$x_1,x_2$</span> and later on with respect to <span class="math-container">$y$</span>.</p>
<p>First Step:</p>
<p><span class="math-container">$$\max_{x_1,x_2} py - w_1x_1 - w_2x_2$$</span></p>
<p><span class="math-container">$s.t. y=f(x_1,x_2)$</span></p>
<p>Because we are not choosing <span class="math-container">$y$</span>, its like having a constant value. Then <span class="math-container">$py$</span> is constant we could take it off for a moment, and our problem would be:</p>
<p><span class="math-container">$$\max_{x_1,x_2} - w_1x_1 - w_2x_2= -(w_1x_1 + w_2x_2)$$</span></p>
<p><span class="math-container">$s.t. y=f(x_1,x_2)$</span></p>
<p>Because we are maximizing <span class="math-container">$-(w_1x_1 + w_2x_2)$</span> we could write it like minimizing <span class="math-container">$w_1x_1 + w_2x_2$</span>. We have arrived to our minimizing problem. And we would get <span class="math-container">$C^*(w_1,w_2,y)$</span>.</p>
<p>Second Step:</p>
<p><span class="math-container">$$\max_{y} py - C^*(w_1,w_2,y)$$</span></p>
<p><span class="math-container">$$\max_{y} py- C^*(w_1,w_2,y)\equiv \max_{x_1,x_2,y} pf(x_1,x_2) - w_1x_1 - w_2x_2$$</span></p>
<p>And these problems are equivalent. We have just altered the order of maximizing. That's why the maximization of profit problem implies minimizing the cost!!</p>

CC BY-SA 4.0
2023-12-15T23:06:51.527Z
2023-12-16T04:37:14.607Z
2023-12-16T04:37:14.607Z
MG123
<ol>
<li><p><strong>How do I show that profit maximization implies cost minimization (in pure competition)?</strong></p>
<p>Suppose we only consider inputs <span class="math-container">$l,k$</span> whose prices are <span class="math-container">$w,r$</span> and output price <span class="math-container">$p$</span>. Profit is <span class="math-container">$\pi = pf(k,l) - wl - rk$</span> where <span class="math-container">$f$</span> is the production function. Let's assume further that <span class="math-container">$q^* = f(k^*, l^*)$</span> maximizes profit.</p>
<p><strong>To show that this minimizes cost, we need to show that <span class="math-container">$(k^*, l^*) = \text{arg}\min (wl+rk)$</span> subject to <span class="math-container">$f(k,l) = q^*$</span>.</strong></p>
<p>Is this all we have to show? It seems too trivial, so I am confused.</p>
<p>But in case that's what we are to show: Assume on the contrary that <span class="math-container">$(k', l') = \text{arg}\min (wl+rk)$</span> subject to <span class="math-container">$f(k,l) = q^*$</span>.</p>
<p>Then <span class="math-container">$wl' + rk' &lt; wl^* + rk^* \implies \pi(l',k',q^*) = pq^* - (wl' + rk') &gt; pq^* - (wl' + rk') = \pi(l^*, k^*, q^*)$</span> contradicting the fact that <span class="math-container">$(k^*, l^*)$</span> maximizes profit in the long run.</p>
<p><strong>Is the proof correct? The fixing of <span class="math-container">$q^{*}$</span> is what I am concerned about.</strong></p>
</li>
<li><p>Is it true for monopoly? Does profit maximization imply cost minimization in monopoly?</p>
</li>
</ol>

CC BY-SA 4.0
2022-11-02T23:45:16.910Z
The argument doesn't use the assumption of perfect competition in the output market, so it holds for any market structure. Just replace $p$ by $p(q^*)$.
VARulle
CC BY-SA 4.0
2022-11-03T05:10:19.327Z
@VARulle Yes! (1) My main doubt is if the assumption of $q^{*} = f(k,l)$ is correct. [Varian's original solution seems a bit different](https://i.imgur.com/FOFk9Fc.png). Can you tell me why he used $\geq$ instead of $=$? (2) Does *"profit maximization imply cost minimization"* here refer to "cost minimization **at the same** ${q^{*}}$ at which profit is maximized"? Thank you!
Rick_Morty
CC BY-SA 4.0
2022-11-03T08:56:27.933Z
Imagine all input prices are equal to $1$ and you have a discrete production technology such that you can produce $6$ widgets by using input vectors $(6, 6)$ or $(7, 2)$, but you can produce $5$ widgets *only* by using input vector $(5, 5)$. Is it cost-minimizing to use $(5, 5)$ to produce $q^*=5$? With your definition, trivially yes, since it's the only way to do it. With Varian's definition, no. The definitions are equivalent under free disposal (where you can "produce" $5$ widgets also by producing $6$ and throwing away $1$), which is usually assumed, but need not be.
VARulle
CC BY-SA 4.0
2022-11-03T09:46:39.943Z
@VARulle I don't understand that. If $x^{*}$ does not minimize cost of $f(x^{*})$, it means that some other $x$ minimizes the cost of $f(x^{*})$. Written mathematically, $\exists x^{**} \neq x^{*} : x^{**} = \text{arg}\min \text{Cost}(f(x^*))$. Isn't it so? I don't get the use of $\geq$. Can you write it in words instead of an example for me please? I hope I will get it that way.
Rick_Morty
CC BY-SA 4.0
2022-11-03T10:38:11.167Z
It's just two slightly different definitions of what exactly it means to "*minimize cost for the output $f(x^*)$*". You say it means producing exactly $f(x^*)$ at minimal cost, Varian says it means producing at least $f(x^*)$ at minimal costs. However, it doesn't really matter, since under both definitions, profit maximization implies cost minimization.
VARulle
CC BY-SA 4.0
2022-11-03T10:39:23.987Z
@VARulle Ah! Now that makes sense. Thanks a lot.
Rick_Morty
CC BY-SA 4.0
2022-11-03T10:41:23.853Z
@VARulle Also, your example makes sense now. It might indeed be the case that we can't produce exactly the earlier quantity of output with the new inputs (but if we produce more, that should work as long as the cost doesn't go up).
Rick_Morty
<microeconomics><monopoly><profit-maximization><perfect-competition>
Does profit maximization imply cost minimization in both pure competition and monopoly?
Rick_Morty
# Does profit maximization imply cost minimization in both pure competition and monopoly?

(Asked by: username_0 on 2022-10-31)

1. **How do I show that profit maximization implies cost minimization (in pure competition)?**

   Suppose we only consider inputs $l,k$ whose prices are $w,r$ and output price $p$. Profit is $\pi = pf(k,l) - wl - rk$ where $f$ is the production function. Let's assume further that $q^\* = f(k^\*, l^\*)$ maximizes profit.

   **To show that this minimizes cost, we need to show that $(k^\*, l^\*) = \text{arg}\min (wl+rk)$ subject to $f(k,l) = q^\*$.**

   Is this all we have to show? It seems too trivial, so I am confused.

   But in case that's what we are to show: Assume on the contrary that $(k', l') = \text{arg}\min (wl+rk)$ subject to $f(k,l) = q^\*$.

   Then $wl' + rk' < wl^\* + rk^\* \implies \pi(l',k',q^\*) = pq^\* - (wl' + rk') > pq^\* - (wl' + rk') = \pi(l^\*, k^\*, q^\*)$ contradicting the fact that $(k^\*, l^\*)$ maximizes profit in the long run.

   **Is the proof correct? The fixing of $q^{\*}$ is what I am concerned about.**
2. Is it true for monopoly? Does profit maximization imply cost minimization in monopoly?

**Tags:** microeconomics, monopoly, profit-maximization, perfect-competition

> **Comments:**
> **username_1**: The argument doesn't use the assumption of perfect competition in the output market, so it holds for any market structure. Just replace $p$ by $p(q^*)$.  
> **username_0**: @username_1 Yes! (1) My main doubt is if the assumption of $q^{*} = f(k,l)$ is correct. [Varian's original solution seems a bit different](https://i.imgur.com/FOFk9Fc.png). Can you tell me why he used $\geq$ instead of $=$? (2) Does *"profit maximization imply cost minimization"* here refer to "cost minimization **at the same** ${q^{*}}$ at which profit is maximized"? Thank you!  
> **username_1**: Imagine all input prices are equal to $1$ and you have a discrete production technology such that you can produce $6$ widgets by using input vectors $(6, 6)$ or $(7, 2)$, but you can produce $5$ widgets *only* by using input vector $(5, 5)$. Is it cost-minimizing to use $(5, 5)$ to produce $q^*=5$? With your definition, trivially yes, since it's the only way to do it. With Varian's definition, no. The definitions are equivalent under free disposal (where you can "produce" $5$ widgets also by producing $6$ and throwing away $1$), which is usually assumed, but need not be.  
> **username_0**: @username_1 I don't understand that. If $x^{*}$ does not minimize cost of $f(x^{*})$, it means that some other $x$ minimizes the cost of $f(x^{*})$. Written mathematically, $\exists x^{**} \neq x^{*} : x^{**} = \text{arg}\min \text{Cost}(f(x^*))$. Isn't it so? I don't get the use of $\geq$. Can you write it in words instead of an example for me please? I hope I will get it that way.  
> **username_1**: It's just two slightly different definitions of what exactly it means to "*minimize cost for the output $f(x^*)$*". You say it means producing exactly $f(x^*)$ at minimal cost, Varian says it means producing at least $f(x^*)$ at minimal costs. However, it doesn't really matter, since under both definitions, profit maximization implies cost minimization.  
> **username_0**: @username_1 Ah! Now that makes sense. Thanks a lot.  
> **username_0**: @username_1 Also, your example makes sense now. It might indeed be the case that we can't produce exactly the earlier quantity of output with the new inputs (but if we produce more, that should work as long as the cost doesn't go up).  

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## 2 answers

### Answer by username_2 on 2022-10-31. Score: 2

Is the current production level reached at minimum cost? If not, reach it by lower cost. Revenue is unchanged, hence profit increases, thus it was not maximized.

> **Comments:**
> **username_0**: Yes, so the proof I wrote is correct? I actually saw Varian's solution manual saying, in the contradiction part: $\exists \ x^{**} : f(x^{**}) \geq f(x^*)$ and $w \cdot x^{*} > w \cdot x^{**}$. I got confused with the $f(x^{**}) \geq f(x^*)$ part as I thought $q^{*} = f(x^*)$ is fixed.  
> **username_2**: I am unwilling to comment beyond my proof.  

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### Answer by username_3 on 2023-12-15. Score: 0

Hey hope this helps you out. The maximization problem is:

$$\max\_{x\_1,x\_2,y} py - w\_1x\_1 - w\_2x\_2$$

$s.t. y=f(x\_1,x\_2)$

In order to maximize, we could first choose to maximize with respect to $x\_1$ then to $x\_2$ and later to $y$. So we are going to use this in our advantage. First, we are going to maximize with respect to $x\_1,x\_2$ and later on with respect to $y$.

First Step:

$$\max\_{x\_1,x\_2} py - w\_1x\_1 - w\_2x\_2$$

$s.t. y=f(x\_1,x\_2)$

Because we are not choosing $y$, its like having a constant value. Then $py$ is constant we could take it off for a moment, and our problem would be:

$$\max\_{x\_1,x\_2} - w\_1x\_1 - w\_2x\_2= -(w\_1x\_1 + w\_2x\_2)$$

$s.t. y=f(x\_1,x\_2)$

Because we are maximizing $-(w\_1x\_1 + w\_2x\_2)$ we could write it like minimizing $w\_1x\_1 + w\_2x\_2$. We have arrived to our minimizing problem. And we would get $C^\*(w\_1,w\_2,y)$.

Second Step:

$$\max\_{y} py - C^\*(w\_1,w\_2,y)$$

$$\max\_{y} py- C^\*(w\_1,w\_2,y)\equiv \max\_{x\_1,x\_2,y} pf(x\_1,x\_2) - w\_1x\_1 - w\_2x\_2$$

And these problems are equivalent. We have just altered the order of maximizing. That's why the maximization of profit problem implies minimizing the cost!!


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