SOLUTION: A company has a vehicle parking area of 1200m^2 with space for x cars and y trucks. Each car requires 20m^2 of space and each truck 100m^2 of space. a)show that x + 5y is equal

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Question 549747: A company has a vehicle parking area of 1200m^2 with space for x cars and y trucks. Each car requires 20m^2 of space and each truck 100m^2 of space.
a)show that x + 5y is equal or less than 60.
b) there must be space for at least
i) 40 vehicles
ii) 2 trucks
write down inequalities to show this information.
Thank you.
Answer by KMST(5328) About Me  (Show Source):
You can put this solution on YOUR website!
x cars take up 20x of space (in m%5E2)
y trucks take up 100y of space (in m%5E2)
If x cars and y trucks will fit into 1200m%5E2
20x%2B100y%3C=1200 <--> x%2B5y%3C=60 dividing everything by 20.
At least 40 vehicles translates into x%2By%3E=40
At least 2 trucks translates into y%3E=2
The inequalities asked for are
x%2B5y%3C=60, y%3E=2 and y%3E=2
EXTRA THOUGHTS
It's only logical to ask that x%3E=0 too, but that will happen anyway, as a consequence of the other restrictions above.
The inequalities determine a feasibility region. The boundaries for that feasibility region are:
x%2B5y=60, x%2By=40, and y=2.
Those boundaries can be graphed as:
graph%28300%2C300%2C-10%2C90%2C-5%2C45%2C2%2C40-x%2C12-x%2F5%29 The feasibility region is the tiny triangle between the 3 lines, sides included.