Help understanding Lagrangian multipliers? I am trying to understand Lagrangian multipliers and using an example problem I found online. 

Problem Set Up:

Consider a consumer with utility function $u(x,y) = x^{\alpha} y^{1-\alpha}$, where $\alpha \in (0,1)$. Suppose this consumer has wealth $w$ and the prices $p =(p_x,p_y)$. That's all we were given.

Work I Did:

I then defined a budget constraint equation: $w = xp_x + yp_y$. I also then defined an associated Lagrangian for the consumer's maximization problem: 
$\Lambda(x,y,\lambda) = x^{\alpha} y^{1-\alpha} + \lambda ((xp_x+yp_y)-w)$. 

My question:

What does this equation allow me to do? Although I set it up given the formula on Wikipedia's page on Lagrangian multipliers, I really have no idea what the purpose of this equation is. Like I don't understand how the equation as given allows me to determine how to maximize my utility function. 

Note: I am familiar with multivariable calculus and Lagrangians ($L = T -V$) in physics, but this method is new to me.
A constrained optimization function maximizes or minimizes an objective subject to one or more constraints. As I understand it, the Lagrangian multiplier approach transforms a constrained optimization problem (I) into an unconstrained optimization problem (II) where the optimal control values to problem II are also the optimal control values to problem I. Additionally, the the objective functions in problems I and II take the same optimal values. The trick is a clever way of putting the constraints into the objective function directly rather than using them separately. 

I agree with your presentation of the consumer's maximization problem: 
$\Lambda(x,y,\lambda) = x^{\alpha} y^{1-\alpha} + \lambda ((xp_x+yp_y)-w)$. 

Now we take the partial derivatives with respect to x an y, set them equal to zero, and then solve for x* and y*.

$0=\partial\Lambda / \partial x  = \alpha x^{\alpha -1 } y^{1-\alpha} + \lambda p_x = (\alpha / x ) x^{\alpha  } y^{1-\alpha} + \lambda p_x$ 

$\Rightarrow -\lambda = (\alpha / (x p_x)) x^{\alpha  } y^{1-\alpha}$ 

$0 =\partial\Lambda / \partial y  = (1 - \alpha) x^{\alpha} y^{-\alpha} + \lambda p_y = ((1 - \alpha) / y ) x^{\alpha  } y^{1-\alpha} + \lambda p_y$ 

$\Rightarrow -\lambda = ((1- \alpha) / (y p_y)) x^{\alpha  } y^{1-\alpha}$

$\Rightarrow (\alpha / (x p_x)) x^{\alpha  } y^{1-\alpha} = -\lambda = ((1- \alpha) / (y p_y)) x^{\alpha  } y^{1-\alpha}$

$\Rightarrow (\alpha / (x p_x)) = ((1- \alpha) / (y p_y))$

$\Rightarrow ( y p_y ) / (1- \alpha) = (x p_x) / \alpha$ (eqn 1)

Recover the budget constraint equation by taking the partial derivative $\partial\Lambda / \partial \lambda = 0$.

$0 = \partial\Lambda / \partial \lambda = xp_x + yp_y - w \Rightarrow xp_x /  w + yp_y /w = 1$ (eqn 2)

We now have two equations and two unknowns (x,y) and can solve for x* and y*.

$\Rightarrow yp_y / w  = xp_x/w \cdot (1 / \alpha - 1) = xp_x/w / \alpha - xp_x/w$

$\Rightarrow 1 = yp_y / w + xp_x/w = xp_x/w / \alpha$

$\rightarrow \alpha = xp_x/w$ (result 1)

$\Rightarrow \alpha = xp_x/w = 1 - yp_y /w$

$\rightarrow 1-\alpha =yp_y/w$ (result 2) 

Results 1 and 2 form the famous constant expenditure shares result for the Cobb-Douglas utility and production functions. Which can also be explicitly solved for x* and y*: $x^* = \alpha w /p_x$ and $y^* = (1-\alpha) w /p_y$ which are the optimal values for both the Lagrangian and the original problems.