Question 215574: Well, I didn't see linear programming anywhere, so I suppose this is the place to send in my question.
The available parking area of a parking lot is 600 square meters. A car requires 6 square meters of space, and a bus requires 30 square meters of space. The attendant can handle no more than 60 vehicles. If the parking fees are $2.50 for cars and $7.50 for buses, how many o feach type of vehicle should the attendant accept to maximize product? What would that profit be? Let X= the number of cars, let Y= The number of buses.
Note: My teacher used what he called "natural restrictions" when doing this on the board, and graphed the equations he got out of them. There was 4. (The >/= stnads for greater than or equal to, since I can't put in the actual symbol.)
1. x >/= 0
2. y >/= 0
3. x + y
4. 6x + 30y
Then he shaded in the area inbetween all the lines, and found the points which made up the shaded area. If it wasn't an already given exact point, he had to find it throught the two line equations which made up that particular point.
He went incredibally fast, and thus I wasn't able to understand much of any of it...
Answer by stanbon(75887)   (Show Source):
You can put this solution on YOUR website! Well, I didn't see linear programming anywhere, so I suppose this is the place to send in my question.
The available parking area of a parking lot is 600 square meters. A car requires 6 square meters of space, and a bus requires 30 square meters of space. The attendant can handle no more than 60 vehicles. If the parking fees are $2.50 for cars and $7.50 for buses, how many of each type of vehicle should the attendant accept to maximize profit? What would that profit be?
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Let X= the number of cars, let Y= The number of buses.
Note: My teacher used what he called "natural restrictions" when doing this on the board, and graphed the equations he got out of them. There was 4.
(The >/= stanads for greater than or equal to, since I can't put in the actual symbol.)
1. x >/= 0 (This means the number of cars is not negative)
2. y >/= 0 (This means the number of busses is not negative)
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3. x + y <= 60 (Because the number of vehicles is limited to 60)
4. 6x + 30y <= 600 (Because the square footage is limited to 600)
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At this point he probably graphed the boundary lines for all
these inequalities:
x = 0
y = 0
y = -x+60
y = (-1/5)x+20
-------------------------
Then he aded in the area inbetween all the lines, and found the points
which made up the shaded area. If it wasn't an already given exact point, he had to find it throught the two line equations which made up that particular point.
He was shading in the half-plane areas corresponding to the INEQUALITIES.
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At some point he developed the "Objective Function":
Profit = 2.5x + 7.5y
-----------------------------
The intersection of the shaded areas are all the points that satisfy
the four inequality statements.
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He then found the coordinates of the corners of the shaded area
and substituted those into the Objective Function to see which
pair gave the maximum profit.
That pair is the solution for the problem.
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Using Google you can find other examples of linear programming
on the Web.
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Cheers,
Stan H.
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