A linear programming problem
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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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A linear programming problem

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?


Step 1 express the constraints as inequalities

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


Step 2 graph the inequalities

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, 100)


The lines

The lines

Here are the lines

y

15x + 24y = 10000

416.7

y=1/3x

y=100

100

x

666.7

O


The feasible region left unshaded

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


The profit line is added them moved

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


We solve the problem

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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