A school is going to organize 290 students to carry out field investigation activities. There are 100 pieces of luggage. The school plans to rent 8 cars of two models a and B. It is understood that each car of type a can carry 40 people and 10 pieces of luggage at most, and each car of type B can carry 30 people and 10 pieces of luggage at most (1) to rent X type a cars, please help the school design the possible car rental scheme (2) if the rental fee of each car is 2000 yuan and 180 yuan, please choose the most economical one

A school is going to organize 290 students to carry out field investigation activities. There are 100 pieces of luggage. The school plans to rent 8 cars of two models a and B. It is understood that each car of type a can carry 40 people and 10 pieces of luggage at most, and each car of type B can carry 30 people and 10 pieces of luggage at most (1) to rent X type a cars, please help the school design the possible car rental scheme (2) if the rental fee of each car is 2000 yuan and 180 yuan, please choose the most economical one

Let's rent x cars of type A and (8-x) cars of type B; X cars of type a can carry 40x people and 10x pieces of luggage; and (8-x) cars of type B can carry 30 (8-x) people and 20 (8-x) pieces of luggage
40x+30(8-x)≥290 (1)
10x+20(8-x)≥100 (2)
The solution of inequality (1) is x ≥ 5
The solution of inequality (2) is x ≤ 6
The solution set of inequality system is 5 ≤ x ≤ 6
The integer solution satisfying the inequality system is x = 5 or x = 6
When x = 5, 8-x = 3
When x = 6, 8-x = 2
Therefore, there are two car rental schemes
Scheme 1: rent 5 type a cars and 3 type B cars;
Plan 2: rent 6 type a cars and 2 type B cars
The total cost of renting a car is: 2000x + 1800 (8-x) = 200X + 14400 (yuan)
When x is the minimum, the total cost is the least, so when x = 5, the total cost is the least
When x = 5, the total cost is 200 × 5 + 14400 = 15400 yuan
The most economical car rental scheme is scheme 1: 5 class a cars and 3 class B cars