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This is the problem I've been given. Assume the consumers’preference over current and future consumption is given by $U(c,c')=c^\frac{1}{2}*c'^\frac{1}{2}$.
Further assume there is no government.
1) solve for the optimal decisions: $c,c′$ and $s$ ?

This is as far as I have come

$\max U(c,c')=c^\frac{1}{2}c'^\frac{1}{2}$
s.t. $c+\frac{c'}{1+r}=y+\frac{y'}{1+r}$
$c, c',s \ge 0 $

$L=c^\frac{1}{2}*c'^\frac{1}{2}+λ(c+\frac{c'}{1+r}-y-\frac{y'}{1+r})$

(1)derv of $c$: $\frac{1}{2}c^{-\frac{1}{2}}*c'^\frac{1}{2}-λ=0 $
(2)derv of $c'$:$\frac{1}{2}c^\frac{1}{2} c'^{-\frac{1}{2}}-λ\frac{1}{1+r}=0$
(3)($λ$):$c+\frac{c'}{1+r}=y+\frac{y'}{1+r}$

Divide equation 1 by 2

$\frac{ \frac{1}{2}c^{-\frac{1}{2}}c'^\frac{1}{2}} {\frac{1}{2}c^\frac{1}{2}c'^{-\frac{1}{2}}}= \frac{λ}{\frac{λ}{1+r}}$ = $\frac{c'}{c}=1+r$

I would appreciate if someone could give some help on how to continue so I can find the optimal decisions for $c, c', s$

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    $\begingroup$ I recommend you edit your text to make your equations understandable! (i.e. write them in equation form) As it is it is very difficult to read $\endgroup$ – user20105 Apr 23 at 21:39
  • $\begingroup$ and please show us your efforts to answer the question $\endgroup$ – emeryville Apr 24 at 3:57
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Every lagrangian with two goods has the same two steps basically. First, find the optimal relationship between the two goods (often through derivatives). Second, plug that relationship back into the budget constraint to solve for each good.

This problem is no different and you have essentially only done step 1. So you just need to do step 2.

The idea is that now you have 2 unknowns ($c$,$c'$), so you need 2 equations. One equation is the relationship you found between consumption today and tomorrow from the first order conditions. The second equation is the the budget constraint (or equation 3 as you called it). Your equation 3 has still not been used toward solving for consumption.

Plugging in the equation for $c'$ that you found into your equation 3 you get:

$c + \frac{c(1+r)}{1+r} =y+\frac{y'}{1+r}$

So $c= 0.5 *(y+\frac{y'}{1+r})$

Now plug in this equation back into the relationship you found through the lagrangian, which was:

$c' = (1+r) c$.

Then you'll be done.

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