I'm sure your teacher expects you to use the simplex method, changing it to the 
maximization of the dual problem.  I did one like that recently, here: 

http://www.algebra.com/tutors/students/your-answer.mpl?question=1153399

and tutor Ikleyn gave you some links.  So I thought I'd see if I could solve
it another way: 
 
Let x = the number of busses
Let y = the number of vans



Divide the first one through by 9



Solve both for y





Combine those into a 3-sided inequality:



So we must have:



or





So the least number of busses we could have is 4. Since busses are so much more 
expensive than vans, that may give us the minimum cost.  Let's see:

4 busses can carry 4∙81=324 students. So that leaves 729-324=405 students to go
by van.  That would require 405/9=45 vans.

The cost would be $1100∙4=$4400 for the busses plus $100∙45=$4500 for the vans.
The total cost would be $8900.

Suppose we add n busses. Then we'd have 4+n busses which would carry
(4+n)(81)=324+81n students. That would leave 729-(324+81n) = 729-324-81n =
405-81n students to go by van.  That would require (405-81n)/9 = 45-9n vans.

The cost would be $1100∙(n+4) for the busses and $100∙(45-9n) for the
vans, or 

1100∙(n+4)+100∙(45-9n) dollars
1100n+4400+4500-900n dollars
200n+8900 dollars

So it would cost $200 more to even add 1 bus.  So we will use just 4 busses.

So the minimum cost would be as I suspected, when 4 busses and 45 vans are used.
The minimum cost would then be $8900.

Now go do it by the simplex method, changing it to the maximization of the dual
problem. You should get that same answer.

Edwin