![]() As a result, the objective value may increase (maximization case) or decrease (minimization case) indefinitely. You are employed by Acme Corporation, a company that makes two products: widgets and bobbles. In a Linear Programming problem, both the objective function and the constraints are linear functions of the variables. Each of these constraints is specied as an inequality. 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. In some LP models, the values of the variables may be increased indefinitely without violating any of the constraints-meaning that the solution space is unbounded in at least one variable. For a maximization problem, an optimal solution to an LP is a point in the feasible region with the largest objective function value. A set of constraints that limits the feasible solution space. I don't see how to make any sense of the third question. ![]() Since the feasible region is bounded, there is no linear function which could be unbounded on it. In some cases, the solution of a linear programming problem may exist despite an unbounded feasible region.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\ $. ![]() 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). ![]()
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