# Why equilibrium efficiency wage maximizes worker effort per dollar wage

In the Keynesian model,

to make as much profit as possible, firms will choose the level of the real wage that gets the most effort from workers for each dollar of real wages paid

• Source: Macroeconomics, 10e Andrew B. Abel, Ben S. Bernanke, Dean Croushore Copyright © 2020 Book ID: 1FR37TIEM0I

I think this is assuming that worker effort is equivalent to output, which is equivalent to revenue.

But why does maximizing revenue/cost maximize profit? I get why maximizing revenue-cost would maximize profit (because revenue-cost is the definition of profit).

Assuming revenue is r(w) and wage is w, we are setting $$\frac{\partial}{\partial w}\frac{r(w)}{w}=0$$, which is $$\frac{r'(w)w-r(w)}{w^2}=0$$, so $$r'(w)w-r(w)=0$$. But how is that the same as $$\frac{\partial}{\partial w}(r(w)-w)=0$$ or $$r'(w)-1=0$$ which is profit maximization?

• Thanks, but we'll probably need a lot more specific details about this specific model, as most of us will not have access to this textbook. You've asked questions about $w$ and $r(w)$, but based on the screenshot of the textbook you've posted, most of us won't have any idea of how these actually work in this specific model of this specific textbook. If there is a Google Books link, you could share that. – user18 Jun 27 '19 at 2:10
• I agree, it is hard to know if you are interpreting correctly their statement without having the full model. As you interpret it, it is clear these cannot be the same. for example let $r(w)=log(w)$, then $log(w)/w$ is maximized at $w=e$, while $r(w)-w$ is maximized at $w=1$. – Regio Jun 27 '19 at 5:25