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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?. A.$1,400. B. $1,600. C.$1,700. D.$1,800

Respuesta :

Answer : $1700

From the given information we make constraints

Let x be the number of large buses

and y be the number of small buses

C be the total cost for the trip

Cost C  = 300 * number of large buses + 200 * number of small buses

So C = 300 x + 200 y

Constraints are :

at least 360 students, So [tex]60 x + 45 y >= 360[/tex]

at most 7 buses , So x + y < = 7

Number of large buses x>=0

Number of large buses y > =0

Now we graph the constraints and find all the endpoints

Graph is attached below

End points are (3,4) , (6,0) and (7,0)

Now we find cost at each end points

C = 300 x + 200 y

(3,4) ----> 300(3) + 200(4) = 1700

(6,0) ----> 300(6) + 200(0) = 1800

(7,0) ----> 300(7) + 200(0) = 2100

The minimum cost is 1700



Ver imagen lisboa

Answer:

$1700

Step-by-step explanation:

Let the number of large buses be x

Let number of small buses be y

Let C be the total cost for the trip

Since w eare given that cost of 1 large bus is 300 and 1 small bus is 200

So, Cost C  = 300 * number of large buses + 200 * number of small buses

So  [tex]C = 300 x + 200 y[/tex]

Constraints are :

Since we are given that large buses can hold 60 students each

So, total students in x buses = 60 x

We are also given that small buses can hold 45 students each.

So, total students in y buses = 45y

Since we are also given that A school district needs to arrange transportation for at least 360 students

So, [tex]60x+45y\geq 360[/tex]

Number of large buses[tex]x\geq 0[/tex]

Number of small buses [tex]y\geq 0[/tex]

We are also given that The district only has chaperones for, at most, seven buses.

So, [tex]x+y\leq 7[/tex]

Now we graph the constraints and find all the endpoints

Refer the attached Graph

So,End points are (3,4) , (6,0) and (7,0)

Now substitute these points in Cost function to find minimum cost.

with(3,4)

[tex]C = 300(3) + 200(4)[/tex]

[tex]C = 1700[/tex]

with (6,0)

[tex]C = 300(6) + 200(0)[/tex]

[tex]C = 1800[/tex]

With (7,0)

[tex]C = 300(7) + 200(0)[/tex]

[tex]C = 2100[/tex]

Hence the minimum cost is $1700.

Ver imagen wifilethbridge