Euler equation through tangency conditions 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.
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 $r_{t+1} &gt; 0$) 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.