Any feasible solution (e.g. x=(1/4, 1/4)) to the linear program provides a lower bound on OPT.
How can we get upper bounds? For any feasible solution x,
goal: bound x1+x2 for any feasible x.
know: x1+2x2 ≤ 1 and 3x1+x2 ≤ 1.
Linear combinations of these constraints give (all) other valid constraints:
for then (#1) above implies x1+x2 ≤ a+b.
Desired inequality (#2) will be true as long as 1 ≤ a+3b and 1 ≤ (2a+b),
in which case we know any feasible x satisfies x1+x2 ≤ a+b.
To get the best possible upper bound:
Exercise: determine optimal solution using pen and paper.
(Geometric interpretation of feasible dual solutions as valid half-spaces.)
weak duality: cost of any feasible solution to the primal ≥ cost of any feasible solution to the dual
proof: b.y ≥ (ATx)y = xAy = x(Ay) ≥ x.c
strong duality: cost of optimal solution to the primal = cost of optimal solution to the dual