TITLE: Max area: Quadrilateral with fixed perimeter and interior angle
QUESTION [10 upvotes]: Edit (May 1, 2018): There appears to be an implicit assumption related to convexity in several of the answers posted thus far; see the answer and image posted here. 
So: Let us restrict to the case in which the quadrilateral is convex.

My post is based on a tweet linking to a blog post where the following problem is posed:

A [convex] quadrilateral has perimeter $60$ and a $30^{\circ}$ angle. What is the maximum possible area?

In the comments, there is a link to a GeoGebra applet for exploring this problem. Tinkering with the applet led me to believe that the maximization occurs when the other three angles are each $110^{\circ}$, but I neither possess a proof that this is optimal, nor have computed the exact area (in closed form) for this case.

I would appreciate an answer that broaches the above special case, although I think it is reasonable to ask about a natural generalization for quadrilaterals; in either scenario, a proof or a link to an already existing one would be great.
Generalization: 

A [convex] quadrilateral has perimeter $N$ and a $d^{\circ}$ angle. What is the maximum possible area?

(So, as a well-known example, for the case in which $d = 90$, the maximum possible area occurs for a square with side-length $N/4$; therefore, the max area is $N^2/16$.)

REPLY [2 votes]: There's no requirement given that the quadrilateral must be convex.  People are assuming that the result must be convex, but that's not correct.  If the quadrilateral is concave at $D$, that cannot be switched to a convex quadrilateral without violating the constraint that the angle $A$ is $30$ degrees.
A quadrilateral with an arbitrarily short side $AD$, and with the other three sides equal in length, will have an area approaching the area of an equilateral triangle: $\frac{\sqrt{3}}{4} \cdot 20^2$ or about $173.21$.  This is significantly larger than any other solution proposed so far.
Below is an image from the linked GeoGebra app in which $D$ is dragged towards $A$; by zooming in, one can see that the $30^{\circ}$ angle is maintained, but the result is much higher than the previously claimed maximum for convex figures.