The feasible region doesn't depend in any way on the choice of objective, and since this particular feasible region is non-empty, no choice of objective is going to give a linear program that has no feasible solutions. I don't see how to make any sense of the third question. LARGE TIME BEHAVIOR OF UNBOUNDED SOLUTIONS OF FIRST-ORDER HAMILTON-JACOBI EQUATIONS IN THE WHOLE SPACE GUY BARLES, OLIVIER LEY, THI-TUYEN NGUYEN, AND THANH VIET PHAN Abstract. In this paper, we establish the existence of an unbounded solution to a second order boundary value problem on the half line. Since the feasible region is bounded, there is no linear function which could be unbounded on it.Any x (x 1, x n) that satisfies all the constraints. It does not violate even a single constraint. The reactions that can be involved in loops are unbounded and show the default maximal or minimal value set in the COBRA Toolbox (1000 or 1000). For objectives that would have multiple extrema on this feasible set, you could take that of minimising $\ z=y-3x\ $, as you note in a comment, minimising $\ z=5y+x\ $, minimising $\ z=-3y+4x\ $, or maximising $\ z=4y-x\ $. A feasible solution for a linear program is a solution that satisfies all constraints that the program is subjected.Ignoring the redundant fifth constraint, and plotting the feasible region in the $x$- $y$ plane, shows that it's bounded by a quadrilateral whose sides are the segments of the lines $\ -3y+4x=5\ $ between the points $\ (2,1)\ $ and $\ (5,5)\ $, $\ 4y-5x=15\ $ between the points $\ (5,5)\ $ and $\ (9,6)\ $, $\ y-3x=-21\ $ between the points $\ (9,6)\ $ and $\ (7,0)\ $, and $\ 5y+x=7\ $ between the points $\ (7,0)\ $ and $\ (2,1)\ $ (see the diagram below). More generally, any continuous function from a compact space into a metric space is bounded.
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