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Question 1153923: The officers of a high school senior class are planning to rent buses and vans
for a class trip. Each bus can transport 81 students, requires 5 chaperones, and
costs $1,100 to rent. Each van can transport 9 students, requires 1 chaperone,
and costs $100 to rent. Since there are 729 students in the senior class that
may be eligible to go on the trip, the officers must plan to accommodate at
least 729 students. Since only 65 parents have volunteered to serve as
chaperones, the officers must plan to use at most 65 chaperones. How many
vehicles of each type should the officers rent in order to minimize the
transportation costs? What are the minimal transportation costs?
Found 2 solutions by ikleyn, Edwin McCravy:
Answer by ikleyn(52803)   (Show Source):
Answer by Edwin McCravy(20060)  (Show Source):
You can put this solution on YOUR website! The officers of a high school senior class are planning to rent buses and vans
for a class trip. Each bus can transport 81 students, requires 5 chaperones, and
costs $1,100 to rent. Each van can transport 9 students, requires 1 chaperone,
and costs $100 to rent. Since there are 729 students in the senior class that
may be eligible to go on the trip, the officers must plan to accommodate at
least 729 students. Since only 65 parents have volunteered to serve as
chaperones, the officers must plan to use at most 65 chaperones. How many
vehicles of each type should the officers rent in order to minimize the
transportation costs? What are the minimal transportation costs?
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
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