I have $z=\sqrt{x^2+y^2}$ and need to find the point on that surface that is closest to $(3,4,0)$.
Absolutely no clue about where to start. This one has me puzzled. I was suggested that I should use Lagrange multipliers but I don't see how finding a max or min value would help me here, plus I don't have a constraint function...
I do know that distance is $\sqrt{(x1-x0)^2 + (y1-y0)^2 + (z1-z0)^2}$ so I'd plug in my point as $\sqrt{(x-3)^2 + (y-4)^2 + (z-0)^2}$ but that's as far as I got.
hint: the distance can be modeled by the function $f(x,y) = \sqrt{(x-3)^2+(y-4)^2+(\sqrt{x^2+y^2} - 0)^2}= \sqrt{x^2-6x+9+y^2-8y+16+x^2+y^2}=\sqrt{2x^2+2y^2-6x-8y+25}$. From this you can take the partial derivatives $f_x = 0, f_y= 0$ and solve for the critical points. From this the answer is around the corner...