Consider a consumer who is trying to maximize:
max $\sum_{t=1}^{M}\left(\frac{1}{2}\right)^{t} (x_{t})^{1/2}$
s.t. $\sum_{t=1}^{M} x_{t} \leq 1$
($\forall x_{t} \geq 0, t=1, \ldots, M$ $(t < \infty)$)
I do not know how to incorporate $x_{t} \leq 1 $ and $x_{t} \geq 0 $ as constraints. Any ideas?
The objective function can be rewritten:
max $(\frac{1}{2})(x_1)^{1/2}+(\frac{1}{2})^2(x_2)^{1/2}+(\frac{1}{2})^3(x_3)^{1/2}...$
but I am still confused as to how to set up the lagrangian.
Thanks!
By the Cauchy-Schwarz inequality, $\sum_{t=1}^M (\sqrt{x_t})^2\sum_{t=1}^M (\dfrac1{2^t})^2\ge \sum_{t=1}^M (\dfrac12)^t(x_t)^{1/2}$
$\sum_{t=1}^M (\dfrac1{2^t})^2=\sum_{t=1}^M (\dfrac1{4^t})=\dfrac{1-(1/4)^M}{1-1/4}=\dfrac{4-(1/4)^{M-1}}{3}$
Thus $\sum_{t=1}^M (\dfrac12)^t(x_t)^{1/2}\le\sum_{t=1}^M (\sqrt{x_t})^2\dfrac{4-(1/4)^{M-1}}{3}=\sum_{t=1}^M (x_t)\dfrac{4-(1/4)^{M-1}}{3}\le \dfrac{4-(1/4)^{M-1}}{3}$
Assuming M is a given fixed integer, the desired maximum would be $\dfrac{4-(1/4)^{M-1}}{3}$.