How to compute the diameter of a truncated cone?

Question

Let us consider the truncated cone given by $$2x^2 + 2y^2 \leq z^2, \quad z \in [1,2].$$ How can I prove that its diameter is $2\sqrt{2}.$ By choosing the points $(\sqrt{2},0,2)$ and $(-\sqrt{2},0,2)$, it is clear that the diameter is bigger than $2\sqrt{2}.$ Now, by choosing two arbitrary points $(x,y,z)$ and $(a,b,c)$ in the cone, I have to show that $$dist((x,y,z),(a,b,c)) \leq 2\sqrt{2}.$$ By using Cauchy-Schwartz, I can get a bound of $2\sqrt{3+\sqrt{6}}$ but not less. Any help would be much appreciated.

EDIT : As precised in the comments, by diameter I mean $\sup\limits_{c_1,c_2 \in C} d(c_1,c_2)$, the biggest distance in the cone.

Answer 1

Use cylindrical coordinates. Then $x^2+y^2=r^2$, so you can write your equation as $$2r^2\le z^2$$ or $$r\le\frac z{\sqrt 2}$$ The diameter of the base is $d=2r\le\sqrt 2 z$. The maximum value is then $d_{max}=2\sqrt 2$

Answer 2

Diameter in the sens of $D=\sup\limits_{c_1,c_2 \in C} d(c_1,c_2)$:

$P_1 = (\frac{1}{\sqrt 2}, 0, 1)$ and $P_2 = (- \sqrt 2, 0, 2)$ both belong to the truncated cone.

And $$d(P_1,P2) = \sqrt{\left(\frac{1}{\sqrt 2} + \sqrt 2 \right)^2 + 1} = \sqrt{\frac{11}{2}}$$

By geometrical arguments, you can prove that $D=\sqrt{\frac{11}{2}} < 2 \sqrt 2$.

Diameter in the sens of the largest face:

The largest face is the one obtained for $z=2$ and has a diameter equal to $2 \sqrt 2$.

Answer 3

Here is the graph to better visualize:

$$AB=3\sqrt{\frac32}\approx 3.67, \quad BC=2\sqrt2\approx 2.83.$$ Note: For the interested, the graph was produced in Geogebra with the commands:

1) if $(1<=2x^2+2y^2<=4,2x^2+2y^2)$

2) $A=(0,1/\sqrt{2},1)$

3) $B=(0,-\sqrt{2},4)$

4) $C=(0,\sqrt{2},4)$

5) $f=\text{segment}(A,B)$

6) $g=\text{segment}(B,C)$.