Represent the Maximal Domain of the Given Laws (Multivariable Inequalities) So I have to represent the Maximal Domain of this law
$$ A = \{ x \in R^2 : x_1 < x_2 \} $$
My Idea was to take $x_1 = x$ and $x_2 = y$. I then found $ y > x$ , so I made it an equation as $ y = x$ and sketched the line in a graph. I thought that the domain would have been the upper right corner; yet plugging some random numbers proves my idea wrong. 
I am completely unsure that a graph is what the professor wants. The question for the exercise is exactly the title of this post. 
What would your take be? 
I have also do to the same for
$$B = \{ x \in R^2 : x^2_1 + x_2^2 < 4 \}$$
Thanks for your help! It would be very appreciated, I have been banging my head on these exercises for two days now.
(The subject is "Mathematical Methods for Economics" and the topic is optimization problems)
 A: The professor is using the common notation $(x_1,x_2)$ instead of $(x,y)$. It is often used since you can speak of the $k^{\textrm th}$ coordinate $x_k$ using the same notation for all coordinates. But I will address the question using $(x,y)$ since you seem more comfortable with that.
Your first shot was correct. It is the half-plane above the line $y=x$ (it doesn't include the bounding line since the inequality is strict).
You can often answer such questions by considering the corresponding equality, which usually describes the boundary. The boundary breaks the plane into various connected regions, and you then can test points within these regions to determine which regions satisfy the given inequality.
Note that if the inequality is not strict ("$\leq$" instead of "$<$"), then the boundary is included.
So for the second problem, the boundary is the curve $x^2+y^2=4$, a circle. The complementary regions are the interior of the circle (an open disk) and the exterior of the circle (plane with a hole in it). Points in the inner region satisfy the original inequality, as you can see that the test point $(0,0)$ does.
