lagrange multiplier slope I was reading the link given in the thread's last comment. I understood initial part. I understand that in case of the hill, if we take any point on that hill, the gradient of the original function will always point towards the peak of the mountain. But I am a bit confused about the gradient of the constraint. Why would gradient of the constraint at constraint's topmost point will point in the direction of hill's peak? I thought that it will be 0 as we have reached the topmost point of the constraint. any explanation? 
 A: You are moving on a level set of the constraint -- $g(x) = 0$ -- and since the gradient of $g$ is always perpendicular to its level sets, $\nabla g$ is always perpendicular to $g(x)=0$.
When you are at the constrained maximum of $f$, $\nabla f$ is also perpendicular to $g(x)$. Therefore $\nabla f$ and $\nabla g$ are parallel, i.e. $\nabla f = \lambda \nabla g$ and since $\nabla f$ points uphill, so does $\nabla g$. But that is a consequence of the fact that the two gradients are parallel, and only true at the top of the constrained hill.
EDIT: No, the gradient is never zero. (In fact, the method of Lagrange multipliers requires zero to be a regular value of $g$, i.e., $\nabla g \neq 0$ whenever $g=0$.)
Here's a concrete example:
$$\max\, x \quad \textrm{s.t.}\quad x^2+y^2=1$$
In other words, you are trying to maximize $x$ over the unit circle. This has obvious solution $x=1, y=0$.
Now the constraint function is $g(x,y) = x^2+y^2-1.$ The gradient is
$$\nabla g = (2x, 2y).$$
At the top of the constrained hill, $(1,0)$, this gradient points further uphill -- in the direction $(2,0)$ -- but is nowhere near zero. In fact it is easy to check that when $g=0$, the gradient of $g$ is never zero: 
$$\|\nabla g\| = \sqrt{4x^2+4y^2} = 2\sqrt{g+1} = 2.$$
