I have a problem from an economics course. I worked out a utility function to be given by $$ U (F, C_1) = \ln C_1 + \beta \ln F + \frac{\ln ((1+r)\left( V_1 + w(1-\tau)(1-F) - C_1\right))}{1+\phi}$$ and I need to maximize this function w.r.t. $F$ and $C_1$
The final answer is supposed to be
$C_1 = \tilde\theta (V_1 + w(1-\tau))\quad $ where $\displaystyle \tilde \theta = \frac {1+\phi}{1+(1+\phi)(1+\beta)}$
$ F = \displaystyle\frac{\beta C_1}{w(1-\tau)}$
But my solution seems to have
$\displaystyle \tilde \theta = \frac {(1+\phi)^2}{1+(1+\phi)(1+\beta(1+\phi))}$
$\ $
My working is as follows:
$\displaystyle 0=U_{C_1}=\frac 1{C_1} - \frac 1{( V_1 + w(1-\tau)(1-F) - C_1)(1+\phi)}$
$\implies\ C_1 = ( V_1 + w(1-\tau)(1-F) - C_1)(1+\phi)$
$\implies\ C_1 = \displaystyle\frac {V_1 + w(1-\tau)(1-F)}{2+\phi}$
$\ $
$0=U_F = \beta \displaystyle \frac1F -\frac{w(1-\tau)}{( V_1 + w(1-\tau)(1-F) - C_1)(1+\phi)}$
$\implies\ Fw(1-\tau) = \beta ( V_1 + w(1-\tau)(1-F) - C_1)(1+\phi)$
Substituting in:
$Fw(1-\tau) = \beta \left( V_1 + w(1-\tau)(1-F) - \displaystyle\frac {V_1 + w(1-\tau)(1-F)}{2+\phi}\right)(1+\phi)$
$\hspace{4em} =\beta\displaystyle\frac{(1+\phi)(1+\phi)\left(V_1 + w(1-\tau)(1-F)\right)}{2+\phi}$
$\hspace{4em} =\beta\displaystyle\frac{(1+\phi)(1+\phi)(V_1 + w(1-\tau))-Fw(1-\tau)(1+\phi)^2}{2+\phi}$
$Fw(1-\tau)\left(1 + \displaystyle \frac{\beta (1+\phi)^2}{2+\phi}\right)=\displaystyle\frac{\beta(1+\phi)^2\left(V_1 + w(1-\tau)\right)}{2+\phi}$
$\ $
Do you see where the mistake lies?
My error lies in the following step:
$C_1 = (V_1 + w(1-\tau)(1-F) - C_1)(1+\phi)$
$\implies\ C_1 = \displaystyle\frac {(V_1 + w(1-\tau)(1-F))(1+\phi)}{2+\phi}$
$\ $
In any case, a quicker method would be to realise that $$0 = U_F = \frac{\beta}F - \frac {w(1-\tau)}{C_1}$$ and then move on from there.