Answer:
(a) The dual problem of the primal problem "Maximize Z = 7X1 + 5X2, Subject to X1 + 2X2 ≤ 6, 4X1 + 3X2 ≤ 12, X1, X2 ≥ 0" is:
Minimize Z = 6Y1 + 12Y2
Subject to:
Y1 + 4Y2 ≥ 7
2Y1 + 3Y2 ≥ 5
Y1, Y2 ≥ 0
(b) The dual problem of the primal problem "Maximize Z = 3X1 + 4X2, Subject to 5X1 + 4X2 ≤ 200, 3X1 + 5X2 ≤ 150, 8X1 + 4X2 ≥ 80, X1, X2 ≥ 0" is:
Minimize Z = 200Y1 + 150Y2
Subject to:
4Y1 + 5Y2 ≥ 3
5Y1 + 3Y2 ≥ 4
4Y1 + 8Y2 ≤ -80
Y1, Y2 ≥ 0
Step-by-step explanation: