Given:
Objective (period-t consumer): \[ W_t=\ln c_t+\frac{1}{1+\theta}\,\ln c_{t+1}, \] Budget constraint: \[ c_t+\frac{c_{t+1}}{1+r}\le y_t+\frac{y_{t+1}}{1+r}. \] Interior optimum, \(\theta=0.05,\; r=0.08\). Find \(\dfrac{c_{t+1}}{c_t}\) (to 2 d.p.).
Step 1 β Lagrangian
\[ \mathcal{L}=\ln c_t+\frac{1}{1+\theta}\ln c_{t+1} +\lambda\!\left(y_t+\frac{y_{t+1}}{1+r}-c_t-\frac{c_{t+1}}{1+r}\right). \] FOCs:
\[ \frac{\partial\mathcal{L}}{\partial c_t}=\frac{1}{c_t}-\lambda=0 \quad\Rightarrow\quad \lambda=\frac{1}{c_t}. \] \[ \frac{\partial\mathcal{L}}{\partial c_{t+1}}=\frac{1}{1+\theta}\cdot\frac{1}{c_{t+1}} -\lambda\cdot\frac{1}{1+r}=0. \] Substitute \(\lambda=\dfrac{1}{c_t}\): \[ \frac{1}{1+\theta}\cdot\frac{1}{c_{t+1}} =\frac{1}{c_t}\cdot\frac{1}{1+r}. \] Rearrange for the consumption ratio: \[ \frac{c_{t+1}}{c_t}=(1+r)\cdot\frac{1}{1+\theta} =\frac{1+r}{1+\theta}. \] Step 2 β Substitute numbers
\[ \frac{c_{t+1}}{c_t}=\frac{1+0.08}{1+0.05}=\frac{1.08}{1.05}=1.028571\ldots \] Final answer (rounded to 2 d.p.):
\[ \boxed{\;\dfrac{c_{t+1}}{c_t}=1.03\;} \]

