Goes through a problem step by step - changing constraints into inequalities, graphing them, finding a feasible region then moving a profit line until it just leaves the region.

1. A Linear Programming Problem
A funfair car park is being designed.
It can hold cars and minibuses.
.
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
.
What combination of bays for cars and bays for buses maximises income?

2. Step 1 – express the constraints as inequalities
Let x be the number of bays for cars, y the number for minbuses
“The car park has a total available area of 10000m2 ”
Each car needs 15m2, so x cars need 15x m2
Similarly y minibuses need 24y
m2
“Reserve space for a minimum of 100 minibuses”
So 15x + 24y < 10 000
So y > 100
“Need spaces for at least three times as many cars as for minibuses”
So x > 3y

3. Step 2 – Graph the inequalities
Remember – we start by drawing lines.
And in this problem the lines we shall draw will be:
15x + 24y = 10 000
y = 100
This line has intercepts:
(0, 416.7) and (666.7, 0)
and x = 3y
This line goes through:
(0, 0) and (600, 200)
This is a horizontal
line through (0, 100)

4. The lines
Here are the lines
y
416.7
15x + 24y = 10000
y=1/3x
y=100
100
O
666.7
x

5. The feasible region left unshaded
y
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.
416.7
100
O
666.7
x

6. The Profit line is added, them moved
$20 per car, $40 per bus, so the income is 20x + 40y
Draw the line P = 20x + 40y
y
(this is actually a family of lines)
The line 4000 = 20x + 40y is drawn
416.7
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
100
x
O
666.7

7. We solve the problem
y
We find the x and y
coordinates of the final lattice
point the profit line hits, just
before it leaves the feasible
region
416.7
In this case it’s (434, 144)
So the maximum income is:
100
x
O
666.7
$(434x20) + $(144 x 40)
=$14440