How to compute the diameter of a truncated cone? 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. 
 A: 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) $g=\text{segment}(B,C)$.
A: 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 $d_{max}=2\sqrt 2$ 
A: 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 $2 \sqrt 2$.
