I am thankful to any hints:

What I have:

Simple log-utility form: $u = \log c_1 + \beta \log c_2$

Budget constraints:

$c_1 + s \leq w$

$c_2 \leq R\; s$


For utility maximization: $s = \frac{\beta}{1+\beta} \cdot w \; \; \; \; \;$ I am not getting this!

I have tried this:

Since the utility is monotonic, we use equality and then I substitue $c_1 \; and \; c_2$ so I get:

$ max \; \; \log (w-s) + \log Rs = 0$

Thus, $\frac{1}{w-s} + \frac{\beta}{s} = 0$

$ \implies s = - \beta w - \beta s \therefore$

$s = \frac{- \beta}{1 + \beta} w$ !!

Any ideas are appreciated.


The cost is an increasing function of $c_1$, so we can take $c_1 = w-s$.

If $\beta<0$, then we can choose $c_2$ close to zero to get arbitrarily large cost, so presumably we have $\beta \ge 0$. In this case, the cost is a non-decreasing function of $c_2$, so we should take $c_2 = Rs$. The utility now reduces to the unconstrained $f(s)= \log (w-s) + \beta \log (Rs)$.

Differentiating gives $-\frac{1}{w-s} + \beta \frac{1}{s} = 0$ (note the first minus sign!), which yields $s = \frac{\beta}{1+\beta} w$.

  • $\begingroup$ Oh God - I forgot to multiply by -1, thanks heaps Copper :) If I do such a mistake in real life, I am screwed :) $\endgroup$ – Ali NasserEddine Jun 24 '13 at 6:40
  • 1
    $\begingroup$ Everyone makes mistakess. $\endgroup$ – copper.hat Jun 24 '13 at 6:41

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