I am rather new to economics in general and to the Neoclassical Growth Model in particular and I was wondering if there was a way to get the Euler equation for consumption without using the Lagrangian multiplier? Assuming no corner solutions, let's say the agent solves $$\max_{c_t, k_t} \sum_{t=0}^{\infty} \beta^t U(c_t)$$ subject to $$c_t + k_{t+1} \leq f(k_t) \\ c_t \geq 0 \quad \forall t \\ k_{t+1} \geq 0 \quad \forall t$$

The Lagrangian is given by $$L = \beta^t [U(c_t) + \lambda_t ( f(k_t)- c_t - k_{t+1})]$$ which can then be solved for the Euler equation $U'(c_t) = \beta U'(c_{t+1})f'(k_{t+1})$.

Why is that if I try to use the tangency condition of the gradient of $U$ and the resource constraint I don't get the same EE? Taking partials with respect to consumption and investment ($\delta =0$ so $k_{t+1}=i_t$)

$\nabla U = \langle \beta^tU'(c_t), 0\rangle$ and

$\nabla RC = \langle -\beta^t, -\beta^t+\beta^{t+1}f'(k_{t+1})\rangle$.

Using tangency I should get $\frac{U'(c_t)}{-1} = \frac{0}{\beta f'(k_{t+1})-1} \Rightarrow U'(c_t) = \beta U'(c_{t})f'(k_{t+1})$. Clearly my time subscript is off. Can someone tell me if my approach is wrong? or if I made a mistake in the algebra? I think the optimizing condition should hold with tangency so its not obvious to me why this approach gives a result different from the Lagrangian.

  • $\begingroup$ $\frac{0}{\beta f'(k_{t+1})-1} = 0$, don't you think? $\endgroup$ Apr 23, 2018 at 19:50
  • $\begingroup$ @AlecosPapadopoulos I agree, which is why I thought it was strange and posted here $\endgroup$
    – ptr64
    Apr 24, 2018 at 13:01

1 Answer 1


In answer to your first question (can you get the Euler equation without a Lagrangean?), the answer is 'yes'. At least, there are less formal ways of deriving it. In general, assuming that a consumer chooses to consume strictly positive quantities of two perfectly divisible goods $x$ and $y$, then it must be that:

$$\frac{MU_x}{MU_y} = \frac{p_x}{p_y}$$

(If this were not true, then the consumer could increase their utility by shifting spending a little more on one good and a little less on the other.)

The so-called 'Euler equation' is nothing more than an application of this result, viewing the 'goods' and $C_t$ and $C_{t+1}$. If you delay your consumption by one period, you put your cash in the bank for one period, earning real interest at rate $r_{t+1}$. Hence, if we normalise $P_t = 1$, then effectively $P_{t+1}=1/(1+r_{t+1})$. Consumption tomorrow is cheaper than consumption today (assuming $r_{t+1} > 0$) since delaying your consumption allows you to earn some interest.

We assume that utility is constant, time-separable and exponentially discounted (as in your set-up). Thus, we need to remember to put a $\beta$ before $MU_{t+1}$. We now write $u'(c_t)$ for $MU_x$, $\beta u'(c_{t+1})$ for $MU_y$ and $(1+r_{t+1})$ for the price ratio.

Finally, observe that if firms maximise profits taking the cost of capital $r_{t+1}$ as given with total depreciation (as you assume), they hire capital until $f'(k_{t+1})=1+r_{t+1}$. Plugging in yields the Euler equation.

In answer to your second question (how can you derive it with a Lagrangean?), I would recommend that you try the following steps:

  • First of all, you should really have a summation before your Lagrangean (summing over an infinite number of periods). This is a dynamic problem.

  • Differentiate the objective function with respect to $c_t$.

  • Roll forward by one period and combine the equations. You should get:

$u'(c_t)/\beta u'(c_{t+1}) = \lambda _t/\lambda _{t+1}$

  • Then differentiate the objective function with respect to $k_{t+1}$. Don't forget the $f(k_t)$ term! This may look irrelevant, but the Lagrangean should be an infinite sum so it shows up when you 'roll forward' by one period. This should give you:

$\lambda _t = \lambda _{t+1}f'(k_{t+1})$

Plug that in and you have it! The equation on which so much pointless macroeconomics built.

  • $\begingroup$ Thank you, you answered my first question. But my second question was whether we can use the gradient of the utility function and the gradient of the resource constraint to use the tangency conditions. I just can't seem to find where my algebra is wrong. $\endgroup$
    – ptr64
    Apr 23, 2018 at 12:57
  • $\begingroup$ @afreelunch even if I might (or might not) agree with you I think your last remark about macroeconomics is uncalled for. $\endgroup$ Apr 23, 2018 at 14:53
  • $\begingroup$ @ptr. I am not completely clear on how you tried to solve the problem. However, the approach is not correct: the equation you get actually implies that $U'(C_t)=0$, which is incorrect. (Though in a very trivial way would actually imply the Euler equation!) Instead, I would suggest you approach the problem using either of the two methods I outlined. $\endgroup$
    – user17900
    Apr 23, 2018 at 14:53
  • $\begingroup$ I never implied $U'(c_t) = 0$. I just argued that at equilibrium the resource constraint should bind, meaning we should be able to use the gradient to get equilibrium conditions (one of which is the Euler equation). The gradient is given by the partials with respect to the controls right (ie $c_t, k_{t+1}$)?$\nabla U'(c_t) = \langle \beta^t u'(c_t), 0 \rangle$. The second should be zero since $c$ and $k$ are independent. My issue was that I kept getting the wrong subscripts using gradients. $\endgroup$
    – ptr64
    Apr 23, 2018 at 21:19
  • $\begingroup$ My point was that $\frac{0}{\beta f'(k+1)-1} = 0$, as another commenter has now pointed out, which is why your equation implies that $U'(c_t) = 0$ (which is not true at the optimum). $\endgroup$
    – user17900
    Apr 24, 2018 at 10:32

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