# If the marginal cost is equal to 1, how does that imply marginal cost is equal to marginal benefit?

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.

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
$$1 + f'(e) = c'(e) + f'(e) \qquad\Leftrightarrow\qquad 1 = c'(e)$$
• 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? Jan 6 '20 at 4:58