Maxima and Minima

Let's focus on the upper-half cone ( $z \ge 0$ ) for the purpose of this minimization problem. Taking a point $P(x_0, y_0, \sqrt{x_0^{2} + y_0^{2}})$ on the cone's surface, let the distance from $P$ to $(1,2,0)$ be the function:

$l(x_0,y_0) = \sqrt{(x_0-1)^2 + (y_0-2)^2 + (x_0^{2}+y_0^{2})} = \sqrt{2x_0^{2} - 2x_0 + 2y_0^{2} - 4y_0 + 5}$ (i).

Taking $\nabla l(x_0,y_0) =0$ yields:

$\frac{\partial l}{\partial x_0} = \frac{4x_0-2}{2 \sqrt{2x_0^{2} - 2x_0 + 2y_0^{2} - 4y_0 + 5}} = 0 \Rightarrow x_0 = \frac{1}{2},$

$\frac{\partial l}{\partial y_0} = \frac{4y_0-4}{2 \sqrt{2x_0^{2} - 2x_0 + 2y_0^{2} - 4y_0 + 5}} = 0 \Rightarrow y_0 = 1$

Finally, calculating the Hessian matrix of $l$ at the critical point $Q(\frac{1}{2},1)$ yields:

$L(\frac{1}{2},1) = \begin{bmatrix} l_{x_{0}x_{0}} & l_{x_{0}y_{0}} \\ l_{y_{0}x_{0}} & l_{y_{0}y_{0}} \end{bmatrix} = \begin{bmatrix} \frac{4}{5} & 0 \\ 0 & \frac{4}{5} \end{bmatrix} = \frac{4}{5} \cdot I_{2x2}$

which is positive-definite at $Q$ . Hence, the minimum distance from $P$ to $(1,2,0)$ computes to $l(\frac{1}{2},1) = \boxed{\sqrt{\frac{5}{2}}}.$