x< or equal to 1; y > or equal to 0; 3x+y< or equal to 5
a. infeasible
b. unbounded
c. an optimal solution
d. alternate optimal solutions
2007-06-11 11:31:17 · 4 answers · asked by HOPING 1 in Science & Mathematics ➔ Mathematics
The feasible set is unbounded (and closed, because the constraints are not strict inequalities). 3x+y<= 5 implies x <= (5 - y)/3. Letting y-> oo and putting x = (5 - y)/3, we satisfy all constraints , while x -> -oo and y -> oo. The feasible set being unbounded, the problem is automatically feasible, no matter what the objectve function is.
Since you didn't give the objective function and the feasible set is unbounded , it's impossible to say anything about the optimal solution Depending on this function, the problem may be unbounded, may have only one optimal solution or may have infinitely many optimal solutions. If a linear programming problem has more than 1 optimal solution, then it has infinitely (actually uncountably) many optimal solutions.
2007-06-11 11:52:22 · answer #1 · answered by Steiner 7 · 2⤊ 0⤋
It is unbounded. x Can take on any value < = 1 and y can take on the corresponding value 5-3x.
2007-06-11 18:45:14 · answer #2 · answered by ironduke8159 7 · 2⤊ 0⤋
b. unbounded
As the left side of the solution goes on indefinitely.
2007-06-11 18:46:25 · answer #3 · answered by Kemmy 6 · 3⤊ 0⤋
...what MsMath said....
2007-06-11 18:41:45 · answer #4 · answered by mbucket 3 · 1⤊ 1⤋