<p>In the Keynesian model, </p>

<blockquote>
  <p>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</p>
</blockquote>

<p><a href="https://i.sstatic.net/YEI4G.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/YEI4G.png" alt="enter image description here"></a></p>

<ul>
<li>Source: Macroeconomics, 10e
Andrew B. Abel, Ben S. Bernanke, Dean Croushore
Copyright © 2020 Book ID: 1FR37TIEM0I</li>
</ul>

<p>I think this is assuming that worker effort is equivalent to output, which is equivalent to revenue.</p>

<p>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).</p>

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

<p>Not sure if I am thinking about this the right way.</p>

CC BY-SA 4.0
2019-06-27T01:56:34.793Z
Yes - I edited the post to make this clearer
Murey Tasroc
CC BY-SA 4.0
2019-06-27T05:25:56.693Z
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
CC BY-SA 4.0
2019-06-27T15:35:08.347Z
I can't post the textbook pages, but here are slides I found online that say pretty much the same thing: http://www.cameron.edu/~syeda/intermacro/ch11.pdf
Murey Tasroc
<wages><profit-maximization>
Why equilibrium efficiency wage maximizes worker effort per dollar wage
Murey Tasroc
# Why equilibrium efficiency wage maximizes worker effort per dollar wage

(Asked by: username_0 on 2019-06-25)

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

[![enter image description here](https://i.sstatic.net/YEI4G.png)](https://i.sstatic.net/YEI4G.png)

* 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?

Not sure if I am thinking about this the right way.

**Tags:** wages, profit-maximization
> **Comments:**
> **username_0**: Yes - I edited the post to make this clearer  
> **username_1**: 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 $w=1$.  
> **username_0**: I can't post the textbook pages, but here are slides I found online that say pretty much the same thing: http://www.cameron.edu/~syeda/intermacro/ch11.pdf  

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