Let the school hire x buses and y vans.
A bus can hold 40 students and 3 teachers.
A van can hold 8 students and 1 teacher.
The number of students riding in  buses and vans is at least 400, therefore
40x + 8y ≥ 400          (1)
The number of teachers riding in buses and vans is at most 36, therefore
3x + y ≤ 36                (2)
Write (1) and (2) as
y ≥ 50 - 5x                 (3)
y ≤ 36 - 3x                 (4)
The equality portion of the solution of (3) and (4) is
36 - 3x = 50 - 5x
2x = 14
x = 7   =>  y = 36 - 3*7 = 15
A graph of the inequalities indicates the acceptable solution in shaded color, as shown below.
The minimum cost of renting buses and vans is
7*$1200 + 15*$100 =  $9900
Answer: The minimum cost is $9,900