A Linear Programming Problem A funfair car park is being designed. It can hold cars and minibuses. 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 m 2 ) 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. 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 15m 2, so x cars need 15x m 2 Similarly y minibuses need 24y m 2 So 15x + 24y < “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 10000m 2 ”

Step 2 – Graph the inequalities Remember – we start by drawing lines. And in this problem the lines we shall draw will be: 15x + 24y = 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 x O y=1/3x x + 24y = y=

The feasible region left unshaded y x O 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 $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 problem y x O 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