A Linear Programming ProblemA funfair car park is being designed.

It can hold cars and minibuses.

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 minibusesThere must be spaces for at least three times as many cars as for minibusesThe cost of parking a car for an hour is $20, while it is $40 for a minibus

.

Here is some information regarding the parking bays:

.

What combination of bays for cars and bays for buses maximises income?

Step 1 – express the constraints as inequalities

Let x be the number of bays for cars, y the number for minbuses

Each car needs 15m2, so x cars need 15x m2

Similarly y minibuses need 24y m2So 15x + 24y < 10 000

“Reserve space for a minimum of 100 minibuses” So y > 100

“Need spaces for at least three times as many cars as for minibuses”

So x > 3y

“The car park has a total available area of 10000m2 ”

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 and x = 3y

This line has intercepts:(0, 416.7) and (666.7, 0)

This is a horizontal line through (0, 100)

This line goes through:(0, 0) and (600, 200)

The lines

Here are the lines

y

xO

y=1/3x

416.715x + 24y = 10000

y=100100

666.7

The feasible region left unshaded

y

xO

416.7

100

666.7

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.

The Profit line is added, them moved

y

x

O

416.7

100

666.7

$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 solve the problemy

x

O

416.7

100

666.7

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)

=$14440