Get the answers you need at Westonci.ca, where our expert community is always ready to help with accurate information. Experience the convenience of finding accurate answers to your questions from knowledgeable professionals on our platform. Discover in-depth answers to your questions from a wide network of professionals on our user-friendly Q&A platform.

Misha and Dontavious were asked to solve the following problem using linear programming.
A local school district can operate a large bus for $300 per day and a small bus for $200 per day. The district needs to arrange transportation for at least 360 students for a field trip and has enough chaperones to have one chaperone per bus for up to 7 buses. Each large bus can hold 60 students and each small bus can hold 45 students. What is the minimum cost for the transportation for the field trip?

Misha decided the following represented this situation:

Let x = number of large buses

Let y = number of small buses

Objective Fx: C=300x+45y

Constraints:

200+60y≥360

x+y≤7

x≥0

y≥0

Dontavious decided the following represented this situation:

Let x = number of large buses

Let y = number of small buses

Objective Fx: C=300x+200y

Constraints:

60x+45y≥360

x+y≤7

x≥0

y≥0



Decide which student set up the correct Objective function and Constraints and using that information answer the following:

a) What are the vertices for the correct feasible region?

b) What is the minimum cost for transportation for the field trip and how did you arrive at this cost?


Sagot :

The correct function was: Dontavious

The constraints are the number of students at least 360 and the number of buses up to 7.  

Dontavious wrote  the appropriate inequalities. cost is $300 for large buses and $200 for small buses. and the constraints are the number of students per bus.

a) The feasible region is where the solution spaces overlap: this creates a small triangular area with the vertices at (6, 0), (7, 0) and (3, 4).

(6, 0), (7, 0), (3, 4)

b) 3 large buses and  4 small buses would cost 3*300 + 4*200 = $1700  

6 large buses and  0 small buses would cost 6*300 = $1800 dollars

The cheapest solution would be  3 large buses and 4 small buses which would cost $1700

 

 

Visit us again for up-to-date and reliable answers. We're always ready to assist you with your informational needs. Thanks for stopping by. We strive to provide the best answers for all your questions. See you again soon. Keep exploring Westonci.ca for more insightful answers to your questions. We're here to help.