I need to find the points of $C=\{(x,y,z)\in \mathbb R^3:4\sqrt{x^2+y^2}\le z\le 1 \}$ with max and min distance from $(-1,0,3)$ I think the solution is:
But i don't know how to parameterize those sets (the upper surface and the external surface).
Can someone help me?
On
The inclined surface of the (single-nappe) cone is given by $ \ 4·\sqrt{x^2+y^2} \ = \ z $ $ \Rightarrow \ z^2 \ - \ 16x^2 - 16y^2 \ = \ 0 \ \ , \ $ so the boundary of this surface at the level $ \ z \ = \ 1 \ $ is the circle $ \ 1^2 \ - \ 16x^2 - 16y^2 \ = \ 0 $ $ \Rightarrow \ x^2 \ + \ y^2 \ = \ \frac{1}{16} \ \ , \ \ z \ = \ 1 \ \ . \ $ The domain $ \ C \ $ given by $ \ ( \ 0 \ \le \ ) \ 4 \sqrt{x^2+y^2} \ \le \ z \ \le \ 1 \ $ is then the inclined surface, the circular "top" surface, and all the points interior to these surfaces. For what follows, it should be kept in mind that a normal to the inclined surface of the cone points outward from its symmetry axis and "downward" toward the $ \ xy-$plane, while the normal to the "top" surface points in the positive $ \ z-$direction (and so perpendicularly to the $ \ xy-$plane) .
It is convenient to work with the "distance-squared" function related to distance from the point $ \ (-1 \ , \ 0 \ , \ 3) \ \ , \ $ which is $ \ (x + 1)^2 \ + \ y^2 \ + \ (z - 3)^2 \ \ . \ $ The Lagrange equations for this problem are then $$ 2·(x + 1) \ + \ \lambda·(-32 x) \ \ = \ \ 0 \ \ \ , \ \ \ 2·y \ + \ \lambda·(-32 y) \ \ = \ \ 0 \ \ \ , $$ $$ 2·(z - 3) \ + \ \lambda·(2 z) \ \ = \ \ 0 \ \ \ . $$
Before we work with these, let us consider what the Lagrange-multiplier method does: it tells us which particular level-surface of the function is tangent to the constraint surface and at what point, that is, it locates places on the constraint surface where the normal vector to the surface is parallel (or anti-parallel) to the normal vector on that level-surface.
The level-surfaces of this function are concentric spheres centered on $ \ (-1 \ , \ 0 \ , \ 3) \ \ . \ $ The normal vectors of these spheres are only parallel to that of the top surface of $ \ C \ $ at points $ \ (-1 \ , \ 0 \ , \ z \ ) \ , \ $ which will never be tangent to the top surface. So these Lagrange equations may not be helpful in finding the extremal distances of the top surface from the point $ \ (-1 \ , \ 0 \ , \ 3) \ \ . \ $ Since the center of the level-surfaces is "above" the top surface, the normal vectors to those spheres can only be parallel to the normal vectors of the cone's inclined surface where they point "downward" toward the $ \ xy-$plane. The tangent point on the inclined surface of $ \ C \ $ then can only be on the side of the cone "facing away" from $ \ (-1 \ , \ 0 \ , \ 3) \ \ . \ $
So saying, we can factor the second equation as $ \ y·(1 - 16·\lambda) \ = \ 0 \ \ , \ $ which gives us the possibilities $ \ y \ = \ 0 \ $ or $ \ \lambda \ = \ \frac{1}{16} \ \ ; \ $ this will guide us in our solution. Isolating $ \ \lambda \ $ in the first and third equations produces $ \ \lambda \ = \ \frac{x \ + \ 1}{16 \ · \ x} \ = \ \frac{3 \ - \ z}{z} \ \ . $
In the case for which $ \ \lambda \ = \ \frac{1}{16} \ \ , \ \ y \ \neq \ 0 \ \ , \ $ inserting this into the two ratio equations leads to $$ \frac{3 \ - \ z}{z} \ \ = \ \ \frac{1}{16} \ \ \Rightarrow \ \ 48 \ - \ 16z \ \ = \ \ z \ \ \Rightarrow \ \ z \ \ = \ \ \frac{48}{17} \ \ , $$ but $$ \frac{x \ + \ 1}{16 · x} \ \ = \ \ \frac{1}{16} \ \ \Rightarrow \ \ 16x \ + \ 16 \ \ = \ \ 16x \ \ (!?) $$ We are unable to find a consistent result for this case; in any event, the value found for $ \ z \ $ would place this putative solution outside of the domain $ \ C \ \ . $
Instead, we will examine the case for $ \ y \ = \ 0 \ \ , \ $ which is the symmetry plane for this geometrical arrangement. For the "top" surface, with $ \ z \ = \ 1 \ \ , \ $ we find $ \ 1^2 \ - \ 16·x^2 \ - \ 16·0^2 \ = \ 0 \ $ $ \Rightarrow \ x \ = \ \pm \frac14 \ \ . \ $ The points $ \ \left( -\frac14 \ , \ 0 \ , \ 1 \right) \ $ and $ \ \left( +\frac14 \ , \ 0 \ , \ 1 \right) \ $ are the points on the circular "edge" of the top surface nearest and farthest from $ \ (-1 \ , \ 0 \ , \ 3) \ \ , \ $ respectively. Their distances from that point can be calculated as $ \ \frac{\sqrt{73}}{4} \ $ and $ \ \frac{\sqrt{89}}{4} \ \ $ [the green and violet circles on the graph below].
Working with the equation of ratios, we have $$ \frac{x \ + \ 1}{16 · x} \ = \ \frac{3 \ - \ z}{z} \ \ \Rightarrow \ \ xz \ + \ z \ \ = \ \ 48·x \ - \ 16·xz \ \ \Rightarrow \ \ 48·x \ - \ (17x + 1)·z \ \ = \ \ 0 \ \ . $$ In this plane $ \ y \ = \ 0 \ \ , \ $ the cross-section of the cone is $ \ z \ = \ 4·\sqrt{x^2 + 0^2} \ = \ 4·|x| \ \ , \ $ leading to the two quadratic equations $$ 48·x \ - \ (17x + 1)·(-4x) \ \ = \ \ 0 \ \ \Rightarrow \ \ 68·x^2 \ + \ 52·x \ \ = \ \ 0 \ \ \Rightarrow \ \ x \ \ = \ \ 0 \ \ , \ \ -\frac{13}{17} $$ or $$ 48·x \ - \ (17x + 1)·(+4x) \ \ = \ \ 0 \ \ \Rightarrow \ \ 68·x^2 \ - \ 44·x \ \ = \ \ 0 \ \ \Rightarrow \ \ x \ \ = \ \ 0 \ \ , \ \ +\frac{11}{17} \ \ . $$
Two of these points are $ \ \left(-\frac{13}{17} \ , \ 0 \ , \ +\frac{52}{17} \right) \ $ and $ \ \left(+\frac{11}{17} \ , \ 0 \ , \ +\frac{44}{17} \right) \ \ , \ $ which actually are tangent points of level-surfaces of the "distance-squared" function [the red and orange circles] to the conical surface (the normal vectors of the surfaces are aligned parallel or anti-parallel). However, they both occur well "above" $ \ z \ = \ 1 \ $ and so are completely outside of the domain $ \ C \ \ . \ $ The third point is the "apex" of the cone at the origin [on the black circle] at a distance $ \ \sqrt{1^2 + 0^2 + 3^2} $ $ = \ \sqrt{10} \ \ $ from $ \ (-1 \ , \ 0 \ , \ 3 ) \ \ . $
So the Lagrange method does succeed in finding the points on the cone that are closest and farthest from the specified point, even though technically those points do not have defined normal vectors.
$$ \text{cross-section in the symmetry plane} \ \ y \ = \ 0 \ \ : $$
$$ \text{the circles are cross-sections of level-surfaces of} \ \ (x + 1)^2 \ + \ y^2 \ + \ (z - 3)^2 \ \ . $$
Hint:
What is the set $4\sqrt{x^2+y^2}\leq 1, z=1$?
What is the set $4\sqrt{x^2+y^2}=z, 0 \leq z\leq 1$?
If you need to help to visualize this geometrically, notice that $\sqrt{x^2+y^2}$ is the distance from $(x,y,z)$ to the $z$-axis, or the radius in polar $3$D coordinates.