SOLUTION: A school is preparing a trip for 400 students. The company who is providing the transportation has 10 buses of 50 seats each and 8 buses of 40 seats, but only has 9 drivers availab
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Question 1196212: A school is preparing a trip for 400 students. The company who is providing the transportation has 10 buses of 50 seats each and 8 buses of 40 seats, but only has 9 drivers available. The rent cost for a large bus is #800 and #600 for the small bus. Calculate how many buses of each type should be used for the trip for the least possible cost.
Found 2 solutions by math_tutor2020, ikleyn:
Answer by math_tutor2020(3817) (Show Source): You can put this solution on YOUR website!
Let
A = larger bus (50 seats)
B = smaller bus (40 seats)
Bus A costs $800
Bus B costs $600
x = number of buses of type A
y = number of buses of type B
Bus A can hold 50 students, so x of them contribute 50x seats
Bus B can hold 40 students, so y of them contribute 40y seats
In total there are 50x+40y seats available
We want that seat count to be 400 or larger.
We may or may not have extra/empty seats.
which is one of the restrictions.
Another restriction is that since x+y is the total number of buses used (of both types combined).
We only have 9 drivers, so this total must be 9 or fewer.
Now either graph by hand, or use a graphing app such as Desmos or GeoGebra.
In the screenshot below, I'm using GeoGebra
The shaded triangle consists of all possible solutions to make both inequalities true, and also to keep x > 0 and y > 0
It makes no sense to have x or y be negative.
Take note that the green shaded region is below the boundary x+y = 9, and above the boundary 5x+4y = 40, and also above the x axis.
When it comes to finding the min cost, we'll be looking at the vertices. This is standard for linear programming problems.
The shaded triangle has vertices of:
(4,5)
(8,0)
(9,0)
Plug each of those x,y coordinates into the cost function
C(x,y) = 800x + 600y
Plug in (x,y) = (4,5)
C(x,y) = 800x + 600y
C(4,5) = 800*4 + 600*5
C(4,5) = 3200 + 3000
C(4,5) = 6200
Repeat for (x,y) = (8,0)
C(x,y) = 800x + 600y
C(8,0) = 800*8 + 600*0
C(8,0) = 6400 + 0
C(8,0) = 6400
And lastly do so for (x,y) = (9,0)
C(x,y) = 800x + 600y
C(9,0) = 800*9 + 600*0
C(9,0) = 7200 + 0
C(9,0) = 7200
The cost outputs were: 6200, 6400, and 7200
The smallest of which is 6200 and this corresponds to when (x,y) = (4,5)
Answer: They should use 4 buses of type A (the 50 seater buses) and 5 buses of type B (the 40 seater buses) to achieve the lowest cost of $6200
Answer by ikleyn(52788) (Show Source): You can put this solution on YOUR website!
.
See the solution under this link
http://moodle.hollandchristian.org/pluginfile.php/51815/mod_resource/content/1/1182_001.pdf
Since it is just solved, solved correctly and placed in the Internet for public viewing,
I do not see much sense to re-write it again.
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