Minimize squared distance to origin from a paraboloid I have to use Lagrange multilpiers to find the minimum distance from the paraboloid with equation
$z = \left({x-1/}{\sqrt{2}}\right)^2 + \left({y-1/}{\sqrt{2}}\right)^2$   to the origin, and from this I managed to solve that $y = x$. After this, I can't figure out what the $z$ value would be. 
I minimized the function $d^2 = x^2+y^2+z^2$ with the paraboloid being the constraint.
Any pointers would be appreciated.
 A: The general form of a lagrange multipliers problem is that you want to maximize/minimize an objective function $f(x,y,z)$ with some constraint $g(x,y,z) = 0$. To do this, you have to solve the equations $\nabla f(x,y,z) = \lambda \nabla g(x,y,z)$ and $g(x,y,z) = 0$ to get the possible points $(x,y,z)$ at which a maximum/minimum is obtained. 
In this problem, you want to minimize $f(x,y,z) = x^2+y^2+z^2$ (the distance squared) with the constraint $g(x,y,z) = (x-\frac{1}{\sqrt{2}})^2+(y-\frac{1}{\sqrt{2}})^2-z = 0$ (the point must be on the paraboloid). 

Then, $\nabla f(x,y,z) = \langle2x,2y,2z\rangle$ and $\nabla g(x,y,z) = \langle2x-\sqrt{2},2y-\sqrt{2},-1\rangle$. 
So, setting $\nabla f(x,y,z) = \lambda \nabla g(x,y,z)$ yields 3 equations which we can solve: 
$2x = \lambda(2x-\sqrt{2}) \leadsto x = \dfrac{\lambda \sqrt{2}}{2-2\lambda}$
$2y = \lambda(2y-\sqrt{2})\leadsto y = \dfrac{\lambda \sqrt{2}}{2-2\lambda}$
$2z = -\lambda \leadsto z = -\dfrac{\lambda}{2}$
Now, plug these back into the constraint equation  $g(x,y,z) = (x-\frac{1}{\sqrt{2}})^2+(y-\frac{1}{\sqrt{2}})^2-z = 0$ and solve for $\lambda$. Then you get the possible values of $(x,y,z)$ at which $f(x,y,z)$ is minimized. 
