I may not understand your question fully but if you are asking what are the conditions for a system of linear inequalities to form a closed shape with area in the first quadrant I have an answer that may help. In order to create a closed shape in any quadrant we would need at least 3 inequalities (a triangle is the smallest closed in figure). I am also assuming our equations are order 1 and that you want an area only indicated by the intersection of the solution sets of your inequalities. So we know we need at least 3. Before we talk about the first quadrant lets just try to get a closed shape. Lets start with lines to make a triangle then move to inequalities. We would need 3 lines f1, f2, and f3 such that (f1 intersects f2 and f3) and (f2 and f3 intersect) to make a triangle. So we need inequalities such that the "lines of equality" generated by them intersect in the same manner. That is to say, on the graph of your inequalities, the dotted or solid lines generated intersect in the same manner, whether or not the point of intersection is a solution to the ineqaulities. So that is a condition. And the only other condition you need is that a point in the interior of the figure generated is a solution to all three inequalities. With that you achieve a closed in figure (in this case a triangle) that is representative of the intersection of the solution sets of the three inequalities. Now you just need to make sure that this closed in figure is an elment of the first quadrant which would mean that for all inequalities in question, the intersection of the "lines of equality" are all located in the first quadrant (i.e. that the points of intersection are limited to x>0 and y>0). With all of these conditions I believe your inequalities will result in a closed in figure in the first quadrant indicative of area. I hope this helps and that I didn't waste your time. Sorry if I did. :)