Question

A school district needs to arrange transportation for at least 360 students for a field trip. The district can use large buses or small buses. The large buses hold 60 students each and have an operating expense of $300 for the day. The small buses hold 45 students each and have an operating expense of $200 for the day. The district only has chaperones for, at most, seven buses. What is the minimum cost for the transportation for the field trip if each bus has chaperones? $1,400 $1,600 $1,700 $1,800

Answer 1

Answer:

The minimum cost for the transportation for the field trip is $1700

Step-by-step explanation:

Let x be the no. of large buses and y be the no. of small buses

Number of students in large bus = 60

Number of students in x large buses = 60x

Cost of 1 large bus =$300

Cost of x large buses = 300x

Number of students in small bus = 45

Number of students in y small buses = 45y

Cost of 1 small bus =$200

Cost of y small buses = 200y

Cost = 300x+200y

We are given that A school district needs to arrange transportation for at least 360 students

$60x+45y \geq 360$

The district only has chaperones for, at most, seven buses.

$x+y \leq 7$

Plot the lines on graph

$60x+45y \geq 360$ --- Black region

$x+y \leq 7$ --- Red region

Feasible points :

At(3,4)

Cost  = 300x+200y=300(3)+200(4)=1700

At(7,0)

Cost  = 300x+200y=300(7)+200(0)=2100

At(6,0)

Cost  = 300x+200y=300(6)+200(0)=1800

So, The minimum cost for the transportation for the field trip is $1700

Answer 2

Answer:

1700

Step-by-step explanation:

2020 edge