# 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.

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:
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 $$x =10 - 2 \cdot 0 = 10$$.