 # Application of Linear Inequalities

Welcome to class!

In today’s class, we will be talking about the application of linear inequalities. Enjoy the class!

### Application of Linear Inequalities #### APPLICATION OF LINEAR INEQUALITIES IN REAL LIFE

##### Greatest and Least Values

Example

Draw a diagram to show the region which satisfies the following inequalities.

5x + y  – 4, x + y  4, y  x + 2, y – 2x  – 4

Find the greatest and the least value of the linear function F = x + 2y within the region.

Solution

For the inequality 5x + y  – 4, first draw the line 5x + y = – 4.

When x = 0, y = 4, when x = -1, y = 1

Add a third point on your own and then draw line 5x + y = -4. You may need to extend the axes to do this:

Now use a test point such as x = 0, y = 0

When x = 0, y = 0, then 0  – 4 is true, so shade the region below the line 5x + y = -4.

For the inequality x + y  4, first draw the line x + y = 4.

When x = 0, y = 4 and when y = 0, x = 4.

So draw a line that passes through (0, 4) and (4, 0).

Test point: (0, 0), so 0  4 is true. Shade the region above the line.

Similarly, for y  x + 2 and y – 2x  – 4, shade the unwanted regions.

The required region is labelled as R as shown. R is also called the feasible region (i.e. the region that satisfies a set of inequalities).

The greatest (maximum) and the least (minimum) of any linear function such as F = x + 2y occurs at the vertices (corner points) of the region which satisfies the given set of the inequalities.

At        A(-1, 1)                      F = x + 2y

F = -1 + 2 = 1

At        B(1, 3)                                    F = x + 2y

F = 1 + 6 = 7

At        C(2.67, 1.33)                        F = x + 2y

F = 2.67 + 2.66 = 5.33

At        D(0, -4)                      F = x + 2y

F = 0 – 8 = -8

F = x + 2y is least at the point D(0, 4).

F = x + 2y is greatest at the point B(1, 3).

Note: The coordinates at point C can also be found by solving the simultaneous equations x + y = 4 and y – 2x = -4,    Example 2

To start a new transport company, a businessman needs at least 5 buses and 10 minibuses. He is not able to run more than 30 vehicles altogether. A bus takes up 3 units of parking space, a minibus takes up to 1 unit of parking space and there are only 54 units available.

If x and y are the numbers of buses and minibuses respectively,

1. Write down four inequalities which represent the restrictions on the businessman
2. Draw a graph that shows a region representing possible values x and y.   In our next class, we will be talking about Algebraic Fractions.  We hope you enjoyed the class.

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