Planar convex region maximizing the difference in 'orientation' between its smallest containing rectangle and largest contained rectangle

Question

We say a rectangle has orientation $\theta$ if the vector from its center to the middle of its shortest side (parallel to the longest side) has some angle $\theta$ with X axis.

Consider a planar convex region $C$ fixed in $\mathbb{R}^2$. Let us draw $R_{1}$, the smallest rectangle containing $C$ and call the orientation of $R_{1}$, $\theta_1$. Let us also draw the rectangle $R_{2}$, the largest rectangle contained within $C$ and let its orientation be $\theta_2$.

Question: Which planar convex region $R$ maximizes $|\theta_1-\theta_2|$ ?

Note 1: If either rectangle is a square, the $\theta$ values won't be unique and we take the smallest value of $|\theta_1-\theta_2|$ as the orientation difference.

Note 2: The same question can be asked with area replaced with perimeter.

Related: bounds on largest internal rectangle area and considering a simple case: right triangles

Answer 1

Consider the convex hull of the union of a unit circle centered at the origin and an axis-aligned rectangle centered at the origin with dimensions of $2\times \sqrt{2}$. Now trim this shape slightly by cutting away the left and right edges by a width of $\epsilon$. This shape has a horizontal maximal inscribed rectangle and a vertical minimal containing rectangle.