A Linear Programming Problem. . Here is some information regarding the parking bays:. A parking bay for a car takes up 15m 2 , while that for a minibus takes up 24m 2 The car park has a total available area of 1ha (i.e. 10000m 2 )
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A Linear Programming Problem
Here is some information regarding the parking bays:
A parking bay for a car takes up 15m2, while that for a minibus takes up 24m2
The car park has a total available area of 1ha (i.e. 10000m2)
The car park must reserve space for a minimum of 100 minibuses
There must be spaces for at least three times as many cars as for minibuses
The cost of parking a car for an hour is $20, while it is $40 for a minibus
A funfair car park is being designed. It can hold cars and minibuses.
What combination of bays for cars and bays for buses maximises income?
Let x be the number of bays for cars, y the number for minbuses
Each car needs 15m2, so x cars need 15x m2
So 15x + 24y < 10 000
Similarly y minibuses need 24y m2
So y > 100
So x > 3y
Remember – we start by drawing lines.
And in this problem the lines we shall draw will be:
15x + 24y = 10 000
y = 100
and x = 3y
This line has intercepts:
(0, 416.7) and (666.7, 0)
This line goes through:
(0, 0) and (600, 200)
This is a horizontal line through (0, 100)
Here are the lines
15x + 24y = 10000
Remember – we take “test” points off the line. If the coordinates of these points don’t satisfy the inequality, we shade all of that region in.
$20 per car, $40 per bus, so the income is 20x + 40y
Draw the line P = 20x + 40y(this is actually a family of lines)
The line 4000 = 20x + 40y is drawn
Now the line 8000 = 20x + 40y
The greater we make P, the further the Profit line moves
We make P as big as possible – until it is just about to leave the feasible region
We find the x and y coordinates of the final lattice point the profit line hits, just before it leaves the feasible region
In this case it’s (434, 144)
So the maximum income is:
$(434x20) + $(144 x 40)