Why Constraint Extrema(The Method of Lagrange Multipliers) gives the max point of function f in a given path? I am studying partial derivatives.  Now we use Method of Lagrange Multipliers to find the max of function of the given path or way. The thing that I am curious about it why this equation gives this condition, what makes this happen ?
 A: Suppose $\nabla f$ is not parallel to $\nabla g$ at a point $x$ that is proposed to be a critical point. Then there is a vector $y$ perpendicular to $\nabla g(x)$ which is not perpendicular to $\nabla f(x)$; for definiteness suppose $\nabla f(x) \cdot y>0$. Since $y$ is perpendicular to $\nabla g(x)$, one can move in the surface $g=0$ in the direction of $y$ to increase $f$, or in the direction of $-x$ to decrease $f$. (More rigorously, there is a curve in the surface $g=0$ whose tangent vector is $y$ at $x$.) Thus $x$ can't be an extreme point. 
Note that this is actually the same as why the first derivative test in 1 variable works: if the derivative were not zero, then you could go in one direction to increase $f$, so the point can't be a max, and you could go in the opposite direction to decrease $f$, so the point can't be a min. The only catch is that in the presence of a constraint, not every direction is accessible, so we don't need $\nabla f$ to be zero, we just need it to be perpendicular to all the directions that we are permitted to go.
Note that I've only argued that the Lagrange condition is necessary, but this is all that we can get. Indeed the Lagrange condition can give solutions that aren't actually extreme points, analogous to the situation in 1D with $f(x)=x^3$ at $x=0$.
