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?


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.

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

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

|improve this answer|||||
  • $\begingroup$ 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? $\endgroup$ – aisync Jan 6 at 4:58
  • $\begingroup$ It makes sense from the standpoint of differentiating the profit function. Thanks $\endgroup$ – aisync Jan 6 at 5:16

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.