In the plane $\mathbb{R}^2$, the intersection of two wedge-shaped regions should still be wedge-shaped.
However I don't see where to go from here. I searched for wedge-shaped and couldn't find anything on SE. If I write a wedge-shaped region as $y > ax+b$ and $y> cx + d$ where $a,c$ are both non-zero, how do I write the intersection of two such regions in the same form? Is this a linear algebra question?
On
If a cone-shaped region is one of the four "pieces" (connected components) of the complement of two lines in the plane, i.e., the convex region defined by two non-parallel rays, then:
The intersection of two cone-shaped regions is ("usually") not cone-shaped. Even if "both regions open in the same direction", the rays defining one region may intersect both rays defining the other. (This is the case in the diagram provided: The left-hand ray of the blue cone hits both rays of the orange cone if you continue the picture far enough up and to the left.)
If each ray of one cone-shaped region is parallel to a ray of the other, then yes, the intersection is a cone-shaped region.
It depends on how restrictive your definition of cone is, but in almost any case I can think of the answer is generally that "cone-ness" is not preserved under intersection. Indeed, imagine in your example if the orange and blue cones were very narrow and wide (resp.). It's then possible that their intersection doesn't emanate from a single vertex. I made a (bad) illustration of this here. (Sorry, I'm new to Geogebra.)
Even if the intersection is a new cone in the form you want, you can (referring to the cones in your picture) solve for the equation of the line bounding the orange cone with positive slope and that bounding the blue cone with negative slope. If those lines are $y = ax+b$ and $y = cx+d$, then the desired cone is of the form $y > a x + b$ and $y > c x + d$.