Least area and least perimeter triangles that contain a convex planar region - how different can they be? 
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*Is there a planar convex region whose enclosing triangles of least perimeter and least area have different areas and different perimeters? And if so, which region maximizes the difference between the two enclosing triangles?
Note: The difference can be quantified as the ratio between areas (or perimeters) of the least area and least perimeter triangular containers.


*An analogous question can be asked about maximum area and maximum perimeter triangles contained within the convex planar region.


*A Claim: For any convex planar region C and specified integer n, the n-gon with least perimeter that contains C and the n-gon with least diameter that contains C are the same. I have no proof or counter to this.
Remarks:
In questions 1 and 2, one can replace triangles with quadrilaterals or in general, n-gons and  to derive many further questions.
A specific example where a convex polygon has quite different least area and least perimeter containing quadrilaterals is given in Oriented Convex Containers of Polygons (see figure 2; the containing quads are rectangles but there seem to be no smaller general quadrilateral containers). This example does not claim to maximize the difference between the least area and least perimeter quadrilateral containers.
A related earlier question: On convex polygons contained in convex polygons
Note added on 5th November 2021: The same set of questions - difference between min area and min perimeter containers - can also be asked with restricted classes of triangles (right, isosceles, ...) or quadrilaterals (say, kites) as containers of a given convex planar region (Indeed, in the examples given in the answer below the containers are right triangles); and about largest 'contained' triangles in regions etc..
Ref: http://nandacumar.blogspot.com/2021/11/still-more-on-oriented-containers-and.html
 A: The quadrilateral $((0,10),(0,0),(6,0),(3,6))$ is an example where the minimum-area triangle (in blue) is distinct from the minimum-perimeter triangle (in red). The analysis for it shows how to answer the full title question numerically.

We use the following results.
Lemma for Area: If triangle $ABC$ has minimal area among triangles enclosing a convex polygon, then either side $CA$ overlaps with a side of the polygon, or $CA$ includes a vertex $V$ of the polygon with
$$\cot(BVA) = \frac12(\cot(VBC) - \cot(ABV))$$
Lemma for Perimeter: If triangle $ABC$ has minimal perimeter among triangles enclosing a convex polygon, then either side $CA$ overlaps with a side of the polygon, or $CA$ includes a vertex $V$ of the polygon with
$$\cot(BVA) = \frac{\pm \sin(\frac12(ABV-VBC))}{ \sqrt{\sin(ABV)\sin(VBC)}}$$
Proof of the Lemmas: Since $ABC$ is of minimal area or perimeter, we can assume wlog that $CA$ touches some vertex $V$ of the polygon; otherwise we could replace $ABC$ with a smaller $A’BC’$, where $C’A’ \parallel CA$ and $C’A’$ does touch a vertex.
Now we can use coordinates with $V = (0,0)$, $B = (-1,0)$, $AB = (y=ax-1)$, $BC = (y=bx-1)$, $CA = (y=mx)$, and take derivatives to find $m\in(a,b)$ which minimizes the area or perimeter of $ABC$. Translating those minimums using $m = \cot(BVA)$, $a = -\cot(ABV)$, $b = \cot(VBC)$ gives the equations in the lemmas; the alternative that $m$ is minimized at an extreme of $a$ or $b$ leads to $CA$ overlapping with one of the sides. $\square$
So for any side of the triangle, and for any side or vertex of the convex polygon, we get one case of the lemma for area. Each case provides two constraints on the vertices of the triangle. There are 512 ways of getting cases for all three sides of the triangle, and each way has six constraints. Any minimal-area triangle will be determined uniquely in such a way, and there are therefore finitely many possibilities for a minimal-area triangle.
In fact, after eliminating degeneracies, there are only four possible minimum-area triangles:
\begin{matrix}
(0,0),    & (6,0),          & (0,12): & \text{area } 36, \text{ perimeter }31.4\\
(0,0),    & (\frac{15}2,0), & (0,10): & \text{area } 37.5, \text{ perimeter }30\\
(-6,10),  & (6,-10),        & (6,10): & \text{area } 120, \text{ perimeter }55.3\\
(-33,54), & (33,-54),       & \ (3,6)\ :  & \text{area } 360, \text{ perimeter }253.7\\
\end{matrix}
The first is therefore the unique minimal-area triangle, and since it is not a minimal-perimeter triangle, any minimal-perimeter triangle must be distinct.
So to answer the full title question, we would compose the following functions:

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*$f: \{(0,0)\} \times \{(1,0)\} \times \mathbb{R}^{8} \to \text{convex hexagons}$ (as here)

*$g: \text{convex hexagons} \to \text{finite lists of possible minimal-area and minimal-perimeter triangles}$

*$h: \text{lists }a\text{ and }b\text{ of triangles} \to 
\text{(minimal area in }a\text{) / (area of minimal-perimeter element of }b\text{)}$
We can reasonably optimize $h\circ g\circ f$, and the maximum value would answer the title question. The advantage of using the lemmas is that we can find the external triangles from a finite list, without needing to examine a six-parameter family.
