A school is preparing a trip for 400 students. the company who is providing the transportation has 10 buses of 50 seats each and 8 buses of 40 seats, but only has 9 drivers available. The rental...

A school is preparing a trip for 400 students. the company who is providing the transportation has 10 buses of 50 seats each and 8 buses of 40 seats, but only has 9 drivers available. The rental cost for a large bus is $800 and $600 for the small bus. Calculate how many buses of each type should be used fro the trip for the least possible cost.

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justaguide's profile pic

justaguide | College Teacher | (Level 2) Distinguished Educator

Posted on

A school is preparing a trip for 400 students. The company that is providing the transportation has 10 buses of 50 seats each and 8 buses of 40 seats, but only has 9 drivers are available. The rental cost for a large bus is $800 and $600 for the small bus.

The cost of taking a large bus on rent is $800 and it has 50 seats. The cost per student is 800/50 = 16. The cost of taking a small bus on rent is $600 and it has 40 seats. The cost per student is 600/40 = 15. To minimize the cost, as many small buses should be used as possible. All 8 small buses and one large bus can only accommodate 370 students. This leaves 30 students. Three of the smaller buses should be replaced with the larger buses.

The school should use 4 large buses and 5 small buses.

aruv's profile pic

aruv | High School Teacher | (Level 2) Valedictorian

Posted on

Let formulate the problem mathematically.

Rent of hired busses =R

No. of busses capacity 50  =x

No. of busses capacity 40   =y

Find

Minimum R= 800x+ 600y

subject to conditions

`x<=10 ............(i)`

`y<=9...................(ii)`

`x+y<=9`     ( condition for drivers)     (iii)         

`50x+40y>=400`    ( students restriction)   (iv)

`x,y>=0`                               (v)

Soving these inequalitie (i) to (v)

We have following solutions

1. x=8  ,y=0  and     R=$6400

2.  x=9  , y=0  and   R= $7200

3.  x= 4   ,y= 5  and  R=$ 6200

Minimum R= ( 6400 ,7200,6200)=6200

Thus solution for this problem is

Minimum rent for hired busses =$6200

No. of busses with capacity 50  = 4

No. of busses with capacity 40   = 5

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