Solve across both K-Maps to minimize cost and select the best answer from below-(Assume that the cost is proportional to inputs. An inverter cost is 1x, a 2-input AND gate's

cost is2x, etc.) O K1 = A'D'C + A'BC + B'CD + AC (Cost 17x), K2= C'D'+A'BC+BCD+A'C'D' (Cost 18x), Total ImplementationCost (K1+K2) is 25xO K1 = A'D'C + A'BC + BCD + AC (Cost 18x), K2= C'D'+A'BC+BCD+A'CD' (Cost 18x), Total ImplementationCost (K1+K2) is 36xK1 = A'D'C + A'BC + BCD + A'C (Cost 14x), K2= C'D+A'BC+BCD+A'CD' (Cost 17x), Total ImplementationCost (K1+K2) is 29xO K1 = A'D'C + A'BC + BCD + AC (Cost 18x), K2= C'D'+A'BC+BCD+A'CD' (Cost 18x), Total ImplementationCost (K1+K2) is 18x