<p>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 $$c_t + k_{t+1} \leq f(k_t) \\ c_t \geq 0 \quad \forall t \\ k_{t+1} \geq 0 \quad \forall t$$</p>

<p>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 $U'(c_t) = \beta U'(c_{t+1})f'(k_{t+1})$. </p>

<p>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 $k_{t+1}=i_t$)</p>

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

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

<p>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. </p>

CC BY-SA 3.0
2018-04-23T19:50:41.197Z
$\frac{0}{\beta f'(k_{t+1})-1} = 0$, don't you think?
Alecos Papadopoulos
CC BY-SA 3.0
2018-04-24T13:01:05.713Z
@AlecosPapadopoulos I agree, which is why I thought it was strange and posted here
ptr64
<macroeconomics><mathematical-economics>
Euler equation through tangency conditions
ptr64
# Euler equation through tangency conditions

(Asked by: username_0 on 2018-04-23)

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.

**Tags:** macroeconomics, mathematical-economics

> **Comments:**
> **username_1**: $\frac{0}{\beta f'(k_{t+1})-1} = 0$, don't you think?  
> **username_0**: @AlecosPapadopoulos I agree, which is why I thought it was strange and posted here  

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## 0 answers

