Maximum ratio between diameter of shape and diameter of enclosing circle For a 2-dimensional closed convex shape $C$, define:


*

*$d(C)$ = the diameter of $C$ (the largest distance between two points in $C$).

*$D(C)$ = the diameter of the smallest circle containing $C$.


Often $d(C)=D(C)$, for example in a unit square they are both $\sqrt 2$.
Sometimes $d(C)<D(C)$, for example in a unit equilateral triangle $d(C)=1$ and $D(C)=2/\sqrt 3$.
MY QUESTION: What is the largest value of $D(C)/d(C)$? (Can it be larger than $2/\sqrt 3$?)
 A: Exercise 6.1, with a hint from Convex Figures,  Yaglom and Boltyanskii,Holt, Rinehart and Winston,1961. 
Prove that a plane figure $\Phi$ cannot have two distinct circumcircles. Moreover prove that of necessity the circumcircle of a plane figure $\Phi$ contains two boundary points of $\Phi$ which are the ends of a diameter of the circle, or else contains three boundary points of $\Phi$ which form an acute-angled triangle. Deduce from this that if $\Phi$ has diameter $1$, then the radius $R$ of the circumcircle of $\Phi$ satisfies the inequalities $$0.5\le R\le 1/\sqrt3=0.577...$$ 
Clearly this answers your question, you mean plane convex figures (to be able to talk about circumcircle),
and the above inequality shows that the diameter of the circumcircle is at most $\dfrac2{\sqrt3}$.
One might want to generalize for $\Bbb R^n$, $n\ge 3$? 
A: I think $2/\sqrt 3 $ is the largest possible value: For simplicity only consider $C$ to be a convex polygon. If $C$ is a line, then the ratio is $1$. 
Now think about triangles: If we fix the diameter of the enclosing circle, then the triangle that produces the largest ratio (i.e. the smallest $d(C)$) is equilateral. 
If we still fix $D(C)$ and try to construct a polygon $C$ in this circle that has a very small diameter, we can always embed a triangle into $C$, which has the same diameter as $C$. However, this diameter is greater or equal to the diameter of an equilateral triangle with the same enclosing circle. Hence, the ratio $D(C)/d(C)$ is smaller or equal to the ratio of an equilateral triangle. 
Since arbitrary convex shapes can be approximated by polygons, the claim follows. 
