# Profit Function: Just revenue maximization subject to constraint? if so where is $\lambda$?

Disclaimer: I dont know i'm getting confused with basic microeconomic prodoucer theory (note: Ive been watching some Hak Choi Videos 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?

• How is $C(x) = wx$ a constraint? Apr 27, 2018 at 8:45
• 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. Apr 27, 2018 at 9:14
• @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
Apr 27, 2018 at 17:36
• I believe denesp's approach embodies what I said. If that is unclear and you insist, however, I can elaborate with an answer. Apr 27, 2018 at 19:07

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 profits. 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 then and there are diminishing returns then dollars should be spent exactly if they earn at least one dollar.

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.

• so I have to be careful with what assumptions i'm making with $\lambda=1$
– EconJohn
Apr 27, 2018 at 17:40
• @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. Apr 27, 2018 at 19:37
• I was saying "C(x)-bar" implying a budget of sorts to the producer.
– EconJohn
Apr 27, 2018 at 19:56
• @EconJohn Including the variable $x$ in the budget limit was very confusing to me. Apr 27, 2018 at 20:10
• Ah I see. This answer is very comprehensive. Thank you!
– EconJohn
Apr 27, 2018 at 20:14