1. 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. 10000m 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?

2. 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 < 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 10000m 2 ”

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 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)

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

5. The feasible region left unshaded y x O 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.

6. 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

7. We solve the problem y 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