I have the following Utility function: \begin{align} U = w^\prime\mu \end{align}
and Langrangian function subject to constraint: \begin{align} F (w, \lambda)= w^\prime\mu - \lambda(w^\prime i - 1) \end{align}
I would like to have an expression for w so that I can calculate the weights. Does anyone know how to solve the system and obtain the function for 'w'?
Thanks in advance!
Shouldn't the utility function be like this?
$$ L(w, \lambda) = w' \mu - \lambda\left( \sum_{i=1}^{n} w_i -1 \right) $$
In that case, let: $w'\mu $ be $\sum_{i=1}^{n} w_i \mu_i$, with the constraint that the weights must sum $1$.
$$ L(w, \lambda) = \sum_{i=1}^{n} w_i \mu_i -\lambda\left( \sum_{i=1}^{n} w_i -1 \right) $$
Then you just have to find each derivative for the problem, meaning:
$$ \frac{\partial L}{\partial w_i} = 0 \ , \forall i $$
$$ \frac{\partial L}{\partial \lambda} = 0 $$
If you had more constraints, you'd have to find also the derivative for the rest of constraints, meaning a vector $\lambda' $