<p>The problem
<span class="math-container">$$\max py(x) $$</span>
<span class="math-container">$$s.t. wx \leq \bar{C} $$</span>
could be interpreted as revenue maximization subject to an operational budget contraint. However the solution of this can differ from the solution of the profit maximization problem, as costs do not appear in the goal function. <span class="math-container">$\lambda$</span> here would show by how much an additional dollar spent on operational costs would increase revenue. If <span class="math-container">$y$</span> is strictly increasing in <span class="math-container">$x$</span> the entire budget would be spent even when <span class="math-container">$\lambda &lt; 1$</span>. However with such a low <span class="math-container">$\lambda$</span> a dollar spent earns less than a dollar, so it should not be spent if the goal is to maximize profits.</p>
<p>In fact, if the goal is profit maximization nd there are diminishing returns then the dollars in the budget should be spent exactly if they earn at least one additional dollar in revenue.</p>
<p>If <span class="math-container">$y(x)$</span> is strictly increasing and strictly concave in <span class="math-container">$x$</span> and you set <span class="math-container">$\bar{C}$</span> at such a level that the Lagrangian's solution yields <span class="math-container">$\lambda = 1$</span>, then the solution of the above problem would also solve the profit maximization problem.</p>
<p>The Lagrangian is</p>
<p><span class="math-container">$$L=py(x) - \lambda(wx - \bar{C})$$</span></p>
<p>The equation system solving the Lagrangian is</p>
<p><span class="math-container">$$
\begin{align*}
pMP(x) -\lambda w &amp; = 0 \\
wx - \bar{C} &amp; = 0 
\end{align*}
$$</span></p>
<p>Getting <span class="math-container">$x = \bar{C}/w$</span> from the second equation one and plugging it into the second equation one has</p>
<p><span class="math-container">$$
\frac{p}{w}MP\left( \bar{C}/w \right) = \lambda.
$$</span></p>
<p>So <span class="math-container">$\bar{C}$</span> determines <span class="math-container">$\lambda$</span>. If <span class="math-container">$\bar{C}$</span> is set at such a level that <span class="math-container">$\lambda = 1$</span> then the first order condition</p>
<p><span class="math-container">$$
pMP(x) -\lambda w = 0
$$</span></p>
<p>becomes</p>
<p><span class="math-container">$$pMP(x) - w = 0$$</span></p>
<p>which is also what you would get by solving the unconstrained profit maximization problem.</p>

CC BY-SA 3.0
2018-04-27T17:40:45.173Z
so I have to be careful with what assumptions i'm making with $\lambda=1$
EconJohn
CC BY-SA 3.0
2018-04-27T19:37:56.950Z
@EconJohn unfortunatelly this comment is unclear to me. As is what exactly you mean by $C(x) = wx$ as a constraint. This seems like the definition of a function to me.
Giskard
CC BY-SA 3.0
2018-04-27T19:56:16.817Z
I was saying "C(x)-bar" implying a budget of sorts to the producer.
EconJohn
CC BY-SA 3.0
2018-04-27T20:10:19.253Z
@EconJohn Including the variable $x$ in the budget limit was very confusing to me.
Giskard
CC BY-SA 3.0
2018-04-27T20:14:22.347Z
Ah I see. This answer is very comprehensive. Thank you!
EconJohn
CC BY-SA 4.0
2018-04-27T09:06:44.370Z
2023-03-20T06:38:08.247Z
2023-03-20T06:38:08.247Z
Giskard
<p><strong>Disclaimer:</strong> I dont know i'm getting confused with basic microeconomic prodoucer theory (note: Ive been watching some <a href="https://www.youtube.com/watch?v=Zj165wWC5Wc&amp;list=PLa71hPrarxLghw22PziNQbA7oGBtxLO1w" rel="nofollow noreferrer">Hak Choi Videos</a> on youtube),but this is my overall thought process.</p>

<p>We know that the standard profit maximization problem can be described as:
$$\max\pi=py(x)-wx$$</p>

<p>let $y(x)=x^\alpha$</p>

<p>it follows that the optimal $x^*$ from such a formula is:</p>

<p>$$\frac{\partial\pi}{\partial x}=p\alpha x^{\alpha-1}-w=0$$
$$x^*=\left(\frac{w}{p\alpha}\right)^{\frac{1}{\alpha-1}}$$</p>

<p>it follows that the optimal $y^*$</p>

<p>$$y^*=\left(\frac{w}{p\alpha}\right)^{\frac{\alpha}{\alpha-1}}$$</p>

<hr>

<p><strong>however, isn't the profit maximization problem also just a revenue maximization problem subject to a constraint?</strong>
$$\max py(x) $$
$$s.t.\bar{C}(x)=wx $$</p>

<p>let 
$y(x)=x^\alpha$</p>

<p>It follows that our Lagrangian for this problem is:</p>

<p>$$L=py(x)+\lambda(\bar{C(x)}-wx)$$</p>

<p>taking the derivative of our Lagrangian With respect to $x$ we find
$$\frac{\partial L}{\partial x}=p\alpha x^{\alpha-1}-\lambda w=0$$</p>

<p>$$x^*=\left(\frac{\lambda w}{p\alpha}\right)^{\frac{1}{\alpha-1}}$$</p>

<p>it follows:
$$y^*=\left(\frac{\lambda w}{p\alpha}\right)^{\frac{\alpha}{\alpha-1}}$$</p>

<p>I find that there is a great deal of similarity between these two problems and their solutions are also very similar. based on this reason I ask, is this an appropriate way to solve for the input and out of a profit maximizing firm? if so what happens to $\lambda$ in the profit function?</p>

CC BY-SA 3.0
2018-04-27T08:45:33.087Z
How is $C(x) = wx$ a constraint?
Giskard
CC BY-SA 3.0
2018-04-27T09:14:52.167Z
The answer to "however, isn't the profit maximization problem also just a revenue maximization problem subject to a constraint?" is no. In your second approach you are putting an upper bound on costs, which is not what profit maximization does. Furthermore, in your example you do not need Lagrange since you only have one unknown.
BB King
CC BY-SA 3.0
2018-04-27T17:36:02.623Z
@BBKing can you elaborate on "In your second approach you are putting an upper bound on costs, which is not what profit maximization does" as an answer?
EconJohn
CC BY-SA 3.0
2018-04-27T19:07:00.750Z
I believe denesp's approach embodies what I said. If that is unclear and you insist, however, I can elaborate with an answer.
BB King
<microeconomics><profit-maximization>
Profit Function: Just revenue maximization subject to constraint? if so where is $\lambda$?
EconJohn
# Profit Function: Just revenue maximization subject to constraint? if so where is $\lambda$?

(Asked by: username_0 on 2018-04-27)

**Disclaimer:** I dont know i'm getting confused with basic microeconomic prodoucer theory (note: Ive been watching some [Hak Choi Videos](https://www.youtube.com/watch?v=Zj165wWC5Wc&list=PLa71hPrarxLghw22PziNQbA7oGBtxLO1w) on youtube),but this is my overall thought process.

We know that the standard profit maximization problem can be described as:
$$\max\pi=py(x)-wx$$

let $y(x)=x^\alpha$

it follows that the optimal $x^\*$ from such a formula is:

$$\frac{\partial\pi}{\partial x}=p\alpha x^{\alpha-1}-w=0$$
$$x^\*=\left(\frac{w}{p\alpha}\right)^{\frac{1}{\alpha-1}}$$

it follows that the optimal $y^\*$

$$y^\*=\left(\frac{w}{p\alpha}\right)^{\frac{\alpha}{\alpha-1}}$$

---

**however, isn't the profit maximization problem also just a revenue maximization problem subject to a constraint?**
$$\max py(x) $$
$$s.t.\bar{C}(x)=wx $$

let
$y(x)=x^\alpha$

It follows that our Lagrangian for this problem is:

$$L=py(x)+\lambda(\bar{C(x)}-wx)$$

taking the derivative of our Lagrangian With respect to $x$ we find
$$\frac{\partial L}{\partial x}=p\alpha x^{\alpha-1}-\lambda w=0$$

$$x^\*=\left(\frac{\lambda w}{p\alpha}\right)^{\frac{1}{\alpha-1}}$$

it follows:
$$y^\*=\left(\frac{\lambda w}{p\alpha}\right)^{\frac{\alpha}{\alpha-1}}$$

I find that there is a great deal of similarity between these two problems and their solutions are also very similar. based on this reason I ask, is this an appropriate way to solve for the input and out of a profit maximizing firm? if so what happens to $\lambda$ in the profit function?

**Tags:** microeconomics, profit-maximization

> **Comments:**
> **username_1**: How is $C(x) = wx$ a constraint?  
> **username_2**: The answer to "however, isn't the profit maximization problem also just a revenue maximization problem subject to a constraint?" is no. In your second approach you are putting an upper bound on costs, which is not what profit maximization does. Furthermore, in your example you do not need Lagrange since you only have one unknown.  
> **username_0**: @BBKing can you elaborate on "In your second approach you are putting an upper bound on costs, which is not what profit maximization does" as an answer?  
> **username_2**: I believe denesp's approach embodies what I said. If that is unclear and you insist, however, I can elaborate with an answer.  

---

## 1 answers

### Answer by username_1 on 2018-04-27. Score: 7 | Accepted answer

The problem
$$\max py(x) $$
$$s.t. wx \leq \bar{C} $$
could be interpreted as revenue maximization subject to an operational budget contraint. However the solution of this can differ from the solution of the profit maximization problem, as costs do not appear in the goal function. $\lambda$ here would show by how much an additional dollar spent on operational costs would increase revenue. If $y$ is strictly increasing in $x$ the entire budget would be spent even when $\lambda < 1$. However with such a low $\lambda$ a dollar spent earns less than a dollar, so it should not be spent if the goal is to maximize profits.

In fact, if the goal is profit maximization nd there are diminishing returns then the dollars in the budget should be spent exactly if they earn at least one additional dollar in revenue.

If $y(x)$ is strictly increasing and strictly concave in $x$ and you set $\bar{C}$ at such a level that the Lagrangian's solution yields $\lambda = 1$, then the solution of the above problem would also solve the profit maximization problem.

The Lagrangian is

$$L=py(x) - \lambda(wx - \bar{C})$$

The equation system solving the Lagrangian is

$$
\begin{align\*}
pMP(x) -\lambda w & = 0 \\
wx - \bar{C} & = 0
\end{align\*}
$$

Getting $x = \bar{C}/w$ from the second equation one and plugging it into the second equation one has

$$
\frac{p}{w}MP\left( \bar{C}/w \right) = \lambda.
$$

So $\bar{C}$ determines $\lambda$. If $\bar{C}$ is set at such a level that $\lambda = 1$ then the first order condition

$$
pMP(x) -\lambda w = 0
$$

becomes

$$pMP(x) - w = 0$$

which is also what you would get by solving the unconstrained profit maximization problem.

> **Comments:**
> **username_0**: so I have to be careful with what assumptions i'm making with $\lambda=1$  
> **username_1**: @username_0 unfortunatelly this comment is unclear to me. As is what exactly you mean by $C(x) = wx$ as a constraint. This seems like the definition of a function to me.  
> **username_0**: I was saying "C(x)-bar" implying a budget of sorts to the producer.  
> **username_1**: @username_0 Including the variable $x$ in the budget limit was very confusing to me.  
> **username_0**: Ah I see. This answer is very comprehensive. Thank you!  

---

