lagrangian multipliers proof I am not able to understand the theory behind the Lagrangian multipliers .
$f(x,y)$ is the function $g(x,y)$ is the constraint , then let 
$F(x,y)= f(x)- \lambda g(x,y)$
Now can someone please tell me what is the reasoning for assuming that  of $F(x,y)$ will have the same maxima and minima as the $f(x)$. Can anyone tell in simple terms the reasoning for this , I couldn't understand the explanation in the text book. 
 A: I find it easier to understand the method of Lagrange multipliers by looking at $f$ and $g$ separately.  
The constraint is expressed as $g(x,y)=\text{const}$, i.e., as a level curve of $g$. If we look at how $f$ changes along one of these curves, we see that at a stationary point, the directional derivative along a tangent to this curve vanishes, just as the derivative vanishes at a stationary point in the elementary one-dimensional case. This directional derivative is $\nabla f\cdot u$, where $u$ is tangent to the curve, so we must have that $\nabla f$ is orthogonal to the curve at this point.  
On the other hand, the gradient of any differentiable function is orthogonal to its level curves, so at a stationary point $P$ of $f$ along the curve $g=\text{const}$, it must be the case that the two gradients are parallel, i.e., that $\nabla f(P)=\lambda\nabla g(P)$ for some constant $\lambda$. Note that at a different stationary point, $\lambda$ might also be different. Rearranging this condition gives the equation $$\nabla(f-\lambda g)=0\tag{1}$$ for the stationary points of the constrained function.  
Sometimes the constraint is instead given as $g(x,y)=0$, with the constant absorbed into $g$. The above reasoning still holds, but now the constraint  can be folded into equation $(1)$ in the following way: Consider $F=f+\lambda g$ as a function of $\lambda$ as well as $x$ and $y$. Then $F_\lambda(x,y,\lambda) = g(x,y)$, so equation $(1)$ also encompasses the constraint $g=0$.
