I am currently working on the following problem:
Let $T \geq 1$ be some finite integer. Solve the following maximization problem: $$ \max \sum_{t=1}^T \frac{1}{2^t} \sqrt{x_t} \quad \text{ s.t. } \quad \sum_{t=1}^T x_t \leq 1, x_t \geq 0 \; \forall t $$
Can I just write that constraint is binding? Somehow I get the incorrect answer.
Dynamic Programming Let $x = (x_1,\ldots,x_T)$ and define the objective by $J(x) = \sum_{t=1}^T{\sqrt{x_t}/2^t}$. We want to find the $\arg\max\{J(x) \mid \sum_t{x_t} \leq 1, x_t \geq 0\}$. As $J$ is increasing in each $x_t$, the constraint is binding such that $x_T = 1 - \sum_{t=1}^{T-1}{x_t}$. Let $J_\tau(x_1,\ldots,x_{T-1}) = \sum_{t=\tau}^T{\sqrt{x_t}/2^t}$. Now you may recursively solve from the second to final problem to the initial problem. Consider for example \begin{align} x_{T-1}^*(x_1,\ldots,x_{T-2}) &= \arg\max_{x_{T-1}}\left\{J_{T-1}(x_1,\ldots,x_{T-1}) = \frac{\sqrt{x_{T-1}}}{2^{T-1}} + \frac{\sqrt{1 - \sum_{t=1}^{T-1}{x_t}}}{2^T}\right\}\\ &=\frac{4}{5}\left(1- \sum_{t=1}^{T-2}{x_t}\right) . \end{align} Then each maximizer on each stage is a function of the previous actions. Generally \begin{align} x_\tau^*(x_1,\ldots,x_{\tau-1}) &= \arg\max_{x_\tau}J_\tau(x_1,\ldots,x_\tau,x_{\tau+1}^*(x_1,\ldots,x_\tau),x_{\tau+2}^*(x_1,\ldots,x_\tau,x_{\tau+1}^*(x_1,\ldots,x_\tau)), \ldots). \end{align} Eventually, at $t = 1$ you would only need to pick $x_1$.
Lagrange Let $L(x,\lambda) = J(x) + \lambda(1-\sum_{t=1}^{T}{x_t})$. Optimality conditions \begin{align} &\frac{\partial L(x)}{\partial x_t} = \frac{1}{2^{t+1}\sqrt{x_t}} - \lambda = 0,\\ &\frac{\partial L(x)}{\partial \lambda} = 1-\sum_{t=1}^{T}{x_t} = 0. \end{align} Solving the first equation for $x_t$ yields \begin{align} x_t = \frac{1}{2^{2(t+1)}\lambda^2}. \end{align} Summing over $t$ \begin{align} \sum_{t=1}^{T}{x_t} = \sum_{t=1}^{T}{\frac{1}{2^{2(t+1)}\lambda^2}} \quad \Longleftrightarrow \quad 1 = \frac{4^T - 1}{3 \cdot 4^{T+1} \lambda^2}. \end{align} Solving for $\lambda^2$ an substituting in the first optimality condition eventually yields \begin{align} x_t = \frac{3 \cdot 4^{T+1}}{2^{2(t+1)}(4^T - 1)}. \end{align}