Primal: Minimize cx subject to Ax >= b, x>=0 Dual: Maximize yb subject to yA <=c, y>=0
In the primal, c in cost function and b was in the constraint. In the dual, reversed.
Inequality sign is changed and minimization turns to maximization.
Example: minimize 2x + 3y subject to
x+2y>=4, 2x + 5y >= 1, x - 3y >= 2, x>=0, y>=0
Dual problem: maximize 4p +q + 2r subject to
p+2q + r <= 2, 2p+5q -3r <= 3, p,q,r >=0