<p>You have to look at this from the derivation of the profit equation.</p>

<p>From the equation <span class="math-container">$\mathbb{E}(\pi_n) = e - c(e)$</span>, you can see that the <em>marginal benefit</em> of increasing <span class="math-container">$e$</span> is equal to 1. That is, for each extra <span class="math-container">$e$</span> you put in, you get that exact amount back in terms of expected profits. The <span class="math-container">$e$</span> term is the benefits, and <span class="math-container">$\frac{de}{de} = 1$</span> is then the marginal benefits. The same goes for marginal cost with the <span class="math-container">$c(e)$</span> term.</p>

<hr>

<h2>Follow up</h2>

<p>In any objective function, labeling things "benefits" or "cost" could be tricky. Here, one could assume, from the functional form, that the function looks like <span class="math-container">$\pi = TR - TC$</span>. It's then quite natural to label <span class="math-container">$TR = e$</span> and <span class="math-container">$TC = c(e)$</span>. The benefit is the revenue you get, and cost is, well, cost.</p>

<p>It could very well be that, actually, <span class="math-container">$TR = e + f(e)$</span> while <span class="math-container">$TC = c(e) + f(e)$</span>, where <span class="math-container">$f(e)$</span> could be any well-behaved function. The "correct" marginal benefits would then be <span class="math-container">$1 + f'(e)$</span>, while marginal cost would be <span class="math-container">$c'(e) + f'(e)$</span>. At the end of the day, though, you'd still have</p>

<p><span class="math-container">$$1 + f'(e) = c'(e) + f'(e) \qquad\Leftrightarrow\qquad 1 = c'(e)$$</span></p>

CC BY-SA 4.0
2020-01-06T04:58:24.187Z
Sorry but for added clarity, if the e term is 'cost' I'm confused as to how it'd also be benefit. Isn't $ \pi_n $ the benefit? Since the profit function includes e twice, how is adding extra e = 1? 

So for example if we assume e* = 10:
$ \pi $ = (10) - c(10)


This way of thinking about though may be wrong?
aisync
CC BY-SA 4.0
2020-01-06T05:16:51.930Z
It makes sense from the standpoint of differentiating the profit function. Thanks
aisync
CC BY-SA 4.0
2020-01-06T03:30:40.247Z
2020-01-06T05:37:45.700Z
2020-01-06T05:37:45.700Z
Art
<p>The function below is a utility function simplified after subject to an implied participation constraint. 
<span class="math-container">$$
E\left(\pi_{n}\right)=e^{*}-E\left(s^{*}\right)=e^{*}-c\left(e^{*}\right)
$$</span> where <span class="math-container">$ \pi_n $</span> is net profits, s is salary, <span class="math-container">$ e^*$</span> is optimal effort and <span class="math-container">$ c(e^*) $</span> is the personal cost (disutility) of that effort</p>

<p>The function is maximized for <span class="math-container">$ e^* $</span> satisfying the first-order condition
<span class="math-container">$$
c^{\prime}\left(e^{*}\right)=1
$$</span>
The text goes on to state that at the optimum, the marginal cost of effort, <span class="math-container">$c^{\prime}\left(e^{*}\right)$</span>, equals the marginal benefit, 1. </p>

<p>Q: Where did marginal benefit just come from? I could be just forgetting some basics, but I don't recall MC = 1 as an implication that MC  =  MB? </p>

<microeconomics><expected-utility>
If the marginal cost is equal to 1, how does that imply marginal cost is equal to marginal benefit?
aisync
# If the marginal cost is equal to 1, how does that imply marginal cost is equal to marginal benefit?

(Asked by: username_0 on 2020-01-06)

The function below is a utility function simplified after subject to an implied participation constraint.
$$
E\left(\pi\_{n}\right)=e^{\*}-E\left(s^{\*}\right)=e^{\*}-c\left(e^{\*}\right)
$$ where $ \pi\_n $ is net profits, s is salary, $ e^\*$ is optimal effort and $ c(e^\*) $ is the personal cost (disutility) of that effort

The function is maximized for $ e^\* $ satisfying the first-order condition
$$
c^{\prime}\left(e^{\*}\right)=1
$$
The text goes on to state that at the optimum, the marginal cost of effort, $c^{\prime}\left(e^{\*}\right)$, equals the marginal benefit, 1.

Q: Where did marginal benefit just come from? I could be just forgetting some basics, but I don't recall MC = 1 as an implication that MC = MB?

**Tags:** microeconomics, expected-utility


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## 1 answers

### Answer by username_1 on 2020-01-06. Score: 2 | Accepted answer

You have to look at this from the derivation of the profit equation.

From the equation $\mathbb{E}(\pi\_n) = e - c(e)$, you can see that the *marginal benefit* of increasing $e$ is equal to 1. That is, for each extra $e$ you put in, you get that exact amount back in terms of expected profits. The $e$ term is the benefits, and $\frac{de}{de} = 1$ is then the marginal benefits. The same goes for marginal cost with the $c(e)$ term.

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## Follow up

In any objective function, labeling things "benefits" or "cost" could be tricky. Here, one could assume, from the functional form, that the function looks like $\pi = TR - TC$. It's then quite natural to label $TR = e$ and $TC = c(e)$. The benefit is the revenue you get, and cost is, well, cost.

It could very well be that, actually, $TR = e + f(e)$ while $TC = c(e) + f(e)$, where $f(e)$ could be any well-behaved function. The "correct" marginal benefits would then be $1 + f'(e)$, while marginal cost would be $c'(e) + f'(e)$. At the end of the day, though, you'd still have

$$1 + f'(e) = c'(e) + f'(e) \qquad\Leftrightarrow\qquad 1 = c'(e)$$

> **Comments:**
> **username_0**: Sorry but for added clarity, if the e term is 'cost' I'm confused as to how it'd also be benefit. Isn't $ \pi_n $ the benefit? Since the profit function includes e twice, how is adding extra e = 1?   So for example if we assume e* = 10: $ \pi $ = (10) - c(10)   This way of thinking about though may be wrong?  
> **username_0**: It makes sense from the standpoint of differentiating the profit function. Thanks  

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