Lagrangian multiplier is zero I am trying to maximize the utility function $u = x + y$ according to the budget constraint $x + 2y = 10$. After taking first-order conditions I get the following:
$$L = u + \lambda(10-x-2y) = 0$$
$$\frac{dL}{dx} = 1 + \lambda(-1) = 0$$
$$\frac{dL}{dy} = 1 + \lambda(-2) = 0$$
At this point, I am stuck because I get $\lambda = 0$. How would I proceed from this point? Any help would be appreciated. Thanks. 
 A: you do not use Lagrange 
 but put x=10-2y in u=x+y , you have u=10-y
 for y>=0 you have a max for y=0 and x=10
A: Actually, you have the two incompatible solutions for lambda: from $\mathrm{d}L/\mathrm{d}x$: $1 = \lambda$ and from $\mathrm{d}L/\mathrm{d}y$: $1/2 = \lambda$.  This behaviour is very common with the objective function and the constraint are linear and the level sets of the objective are not parallel to the constraint.  Here's a picture for the current problem:

The constraint is the green line -- any feasible solution is on that line.  The level sets of the objective function are in orange and the objective increases going up and to the right.
The Lagrange multiplier method tries to find a place where the constraint is tangent to the objective level sets -- that is a place where to first order a change in the independent variables does not change the value of the objective (which is a way to characterize any point of tangency).  I can't explain this better than the Lagrange multiplier article at the English Wikipedia.
Notice that your constraint is nowhere tangent to the objective level sets : the Lagrange multiplier method has no solution, which is what the incompatible solutions for $\lambda$ are telling you.
But ...  Linear systems are "easy".  There are much less theoretically complex ways to optimize a linear objective subject to linear constraints.  In this case, solve the constraint for $x$:
    $$  x = 10 - 2y  $$
and plug into the objective
    $$  u(y) = (10 - 2y) + y = 10 - y  $$
and maximize.  I assume you intend $x,y \geq 0$, so this is maximized when $y = 0$, so when $x =10 - 2 \cdot 0 = 10$.
