a school graduation class wants to hire buses and vans for a trip to jasper national park. each bus holds 40…

a school graduation class wants to hire buses and vans for a trip to jasper national park. each bus holds 40 students and 3 teachers and costs $1200 to rent. each van holds 8 students and 1 teacher and costs $100 to rent. the school has at least 400 students who want to go on the trip, but at most 36 teachers. determine the minimum transportation cost by graphing a system of linear inequalities (let x = the number of vans).

Answer

Explanation:

Step1: Set up the student - capacity inequality

Let $y$ be the number of buses. The number of students is at least 400. Each bus holds 40 students and each van holds 8 students. So the inequality for student - capacity is $40y + 8x\geq400$. Simplify it to get $5y + x\geq50$, or $y\geq\frac{50 - x}{5}$.

Step2: Set up the teacher - number inequality

The number of teachers is at most 36. Each bus has 3 teachers and each van has 1 teacher. So the inequality for teachers is $3y+x\leq36$, or $y\leq\frac{36 - x}{3}$.

Step3: Consider non - negativity

Also, $x\geq0$ and $y\geq0$ since the number of vans and buses cannot be negative.

Step4: Analyze cost function (not required for finding minimum transportation but for context)

The cost function $C = 1200y+100x$. We want to minimize the number of vehicles (a proxy for cost in a sense of resource - usage). We can find the intersection points of the boundary lines of the inequalities. First, set $5y + x=50$ and $3y + x=36$. Subtract the second equation from the first: $(5y + x)-(3y + x)=50 - 36$. $2y=14$, so $y = 7$. Substitute $y = 7$ into $3y + x=36$, we get $3\times7+x=36$, $21+x=36$, $x = 15$. We also consider the $x$ and $y$ - intercepts of the boundary lines. For $5y + x=50$, when $y = 0$, $x = 50$; when $x = 0$, $y = 10$. For $3y + x=36$, when $y = 0$, $x = 36$; when $x = 0$, $y = 12$. By checking the corner - points of the feasible region (the region defined by the inequalities $5y + x\geq50$, $3y + x\leq36$, $x\geq0$, $y\geq0$):

Intersection of $5y + x=50$ and $x = 0$ gives $(0,10)$
Intersection of $3y + x=36$ and $x = 0$ gives $(0,12)$
Intersection of $5y + x=50$ and $3y + x=36$ gives $(15,7)$
Intersection of $3y + x=36$ and $y = 0$ gives $(36,0)$
Intersection of $5y + x=50$ and $y = 0$ gives $(50,0)$

We want to minimize the total number of vehicles $N=x + y$. For $(0,10)$, $N=10$ For $(0,12)$, $N=12$ For $(15,7)$, $N=15 + 7=22$ For $(36,0)$, $N=36$ For $(50,0)$, $N=50$

The minimum value of $x$ (number of vans) occurs when we use as many buses as possible while still satisfying the constraints. From $5y + x=50$ and $3y + x=36$, we find that when we try to meet the student - capacity and teacher - number constraints, the minimum value of $x$ for which the system of inequalities is satisfied is $x = 0$.