Question 1196212
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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.


{{{total_seats_available >= 400}}}


{{{50x+40y >= 400}}}


{{{10(5x+4y) >= 400}}}


{{{5x+4y >= 400/10}}}


{{{5x+4y >= 40}}}
which is one of the restrictions.


Another restriction is that {{{x+y <= 9}}} 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
<img src = "https://i.imgur.com/56Il4Ns.png">
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
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