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: LaTeX: C=300x+45yC = 300 x + 45 y

Constraints:

LaTeX: 200+60y\ge360200 + 60 y ≥ 360

LaTeX: x+y\le7x + y ≤ 7

LaTeX: x\ge0x ≥ 0

LaTeX: y\ge0y ≥ 0

Dontavious decided the following represented this situation:

Let x = number of large buses

Let y = number of small buses

Objective Fx: LaTeX: C=300x+200yC = 300 x + 200 y

Constraints:

LaTeX: 60x+45y\ge36060 x + 45 y ≥ 360

LaTeX: x+y\le7x + y ≤ 7

LaTeX: x\ge0x ≥ 0

LaTeX: y\ge0y ≥ 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 feasible region?

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

Step-by-step explanation:

The constraints are on the number of students (≥360) and the number of buses (≤7). The numbers in each constraint inequality need to relate to that constraint. The numbers in the cost function need to relate to costs. Misha doesn't seem to understand that.

Dontavious wrote appropriate inequalities. Except for the constraints that the values cannot be negative, they are graphed in the attached. The feasible region is where the solution spaces overlap: a small triangular region with vertices at (6, 0), (7, 0) and (3, 4).

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The recommended method of finding the minimum cost is to evaluate the cost function at the vertices of the feasible region. The purpose is to choose the vertex that places the cost function line as close to the origin as possible. Once the slope of the cost function line is found* in comparison to the slopes of the edges of the feasible region, it is clear that the point (3, 4) will minimize the cost.

If you choose to evaluate C at the vertices, you find that ...

3 large buses and 4 small buses will achieve the minimum cost of $1700.

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You can also look at the relative costs of the buses. The large bus costs $300 for 60 students, or $5 per student. The small bus costs $200 for 45 students, or $4.44 per student. If chaperones were not an issue, we would prefer small buses in order to minimize cost. However, small buses use our chaperone budget faster than large buses do, so we want to use as many small buses as we can within the constraint imposed by chaperones. That is what the solution (3, 4) does.

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* Since we're using a graphing calculator to draw the solution space for the inequalities, we can use it also to draw a line representing the const function. We can arbitrarily pick a cost, including 0, to see what the slope of the line looks like relative to the solution space. It is usually a simple matter to visually locate the point that minimizes cost.

idk

step-by-step explanation:

this is false ^.^ lol