Computing the Lagrange multiplier in constrained optimization problem As I know the value of the Lagrange multiplier does not have a valuable meaning in most cases. Our goal is to compute the values of the parameters.
Here is an example for constrained optimization: https://www.svm-tutorial.com/2016/09/duality-lagrange-multipliers/#comment-11576
$$\text{minimize }\; f(x,y)=x^2+y^2 ~\text{ s.t. }~  g(x,y)= x+y-1=0$$ 
He constructs the Lagrange multipliers and computes the value of $x$ and $y$. But at the end and after having the values of $x$ and $y$, he also computes the value of $\lambda$, which I don't understand why?
What it is the purpose to compute the value of $\lambda$? Why do we need its value at the very end?
You can simplify your problem writing

$$f(x,1-x)=x^2+(1-x)^2$$
If we denote

$$g(x)=2x^2-2x+1$$
and with
$$g'(x)=4x-2$$ you will find your Minimum.
You can also consider

$$f(x,y)+\lambda(x-y-1)=F(x,y,\lambda)$$
$$\frac{\partial F(x,y,\lambda)}{\partial x}=2x+\lambda$$
$$\frac{\partial F(x,y,\lambda)}{\partial y}=2y+\lambda$$
$$\frac{\partial F(x,y,\lambda)}{\partial \lambda}=x+y-1$$
You can simplify your problem writing

$$f(x,1-x)=x^2+(1-x)^2$$
If we denote

$$g(x)=2x^2-2x+1$$
and with
$$g'(x)=4x-2$$ you will find your Minimum.
You can also consider

$$f(x,y)+\lambda(x-y-1)=F(x,y,\lambda)$$
$$\frac{\partial F(x,y,\lambda)}{\partial x}=2x+\lambda$$
$$\frac{\partial F(x,y,\lambda)}{\partial y}=2y+\lambda$$
$$\frac{\partial F(x,y,\lambda)}{\partial \lambda}=x+y-1$$