# 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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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.**

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**