Profit Function: Just revenue maximization subject to constraint? if so where is $\lambda$?
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 $\lambda &lt; 1$. 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

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

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.