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A Linear Programming Problem

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

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  1. 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?

  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 So 15x + 24y < 10 000 Similarly y minibuses need 24y m2 • “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

  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 line goes through: (0, 0) and (600, 200) This is a horizontal line through (0, 14)

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

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

  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(this is actually a family of lines) y 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 666.7 O

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

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