Question 1081751: The officers of a high school senior class are planning to rent buses and vans for a class trip. Each bus can transport 72 students, requires 3 chaperones, and costs $1,400 to rent. Each van can transport 9 students, requires chaperone, and costs $100 to rent. Since there are 720 students in the senior class that may be eligible to go on the trip, the officers must plan to accommodate at least 720 students. Since only 45 parents have volunteered to serve as chaperones, the officers must plan to use at most 45 chaperones. How many vehicles of each type should the officers rent in order to minimize the transportation costs? What are the minimal transportation costs?
Answer by jim_thompson5910(35256)   (Show Source):
You can put this solution on YOUR website! Lets make
x = number of buses
y = number of vans
Because we can't have a negative number of buses or vans, let's restrict x and y such that  and 
If you rent x buses, and each bus costs $1400 to rent, then you'll pay 1400x dollars for just the buses (ignore the vans for now).
If you rent y vans, and each van costs $100 to rent, then you'll 100y dollars for just the vans (ignore the buses for now).
In total, the overall cost would be C(x,y) = 1400x+100y
The goal is to minimize the cost function C(x,y). This is known as the objective function. We'll use Linear Programming to achieve this goal.
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Each bus can hold 72 students. If we have x buses, then we can transport 72x students.
Each van can hold 9 students. If we have y vans, then we can transport 9y students.
In total, we can transport S = 72x+9y students where S is the total number of students combined for both sets of vehicles.
"The officers must accommodate for "at least 720 students" meaning that  which is the same as saying  after doing a substitution.
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Let P = total number of parents or chaperones on the trip.
"The officers plan to use at most 45 chaperones" tells us that 
Each bus can hold 3 chaperones. If we have x buses, then we can transport 3x chaperones.
Each van can hold 1 chaperone. If we have y vans, then we can transport 1y chaperones.
In total, we have P = 3x+1y chaperones
Since and P = 3x+1y, we can say 
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If you graph the system
then you'll get this
Image generated by GeoGebra (free graphing software).
Take note how the feasible region has the vertices A = (7, 24), B = (10, 0), C = (15, 0).
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Those vertices are important: the min cost will happen at one of the vertices. Specifically the min cost will happen at the (x,y) pairing that makes C(x,y) the smallest output.
Let's plug the coordinates of point A into the C(x,y) function.
C(x,y) = 1400x+100y
C(7,24) = 1400(7)+100(24)
C(7,24) = 12,200
If we have 7 buses and 24 vans, then the total cost is $12,200.
Repeat for point B. Plug in (x,y) = (10,0).
C(x,y) = 1400x+100y
C(10,0) = 1400(10)+100(0)
C(10,0) = 14000
If we have 10 buses, and 0 vans, then the total cost is $14,000. This cost is higher than the last cost so we'll ignore point B. So far, point A is the winner in terms of keeping the costs down to a minimum.
Finally, plug in the coordinates of point C
C(x,y) = 1400x+100y
C(15,0) = 1400(15)+100(0)
C(15,0) = 21,000
Like before, this cost of $21,000 is larger than the result we got for point A.
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Point A = (7,24) leads to the lowest cost for C(x,y) = 1400x+100y
Let's check to make sure that there is enough for both the students and chaperones
We have x = 7 buses, so 72*x = 72*7 = 504 students have been accounted for. The y = 24 vans provide transportation for 9*y = 9*24 = 216 additional students. In total, 504+216 = 720 students have a ride in one vehicle or another. So that takes care of the students.
For the chaperones, we have 3*x = 3*7 = 21 for the buses and 1*y = 1*24 = 24 for the vans. Adding these results up gives 21+24 = 45 which also checks out.
So we meet our goal of getting 720 students and 45 chaperones to the destination.
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Final Answers:
Min Cost = $12,200
Number of buses needed = 7
Number of vans needed = 24
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