10.01 Linear Inequalities

Feasible regions

The graph of several inequalities describes the set of all points that satisfy the inequalities. As in the case of one inequality, the region that satisfies all the inequalities is called the required region. Alternatively, and especially in the context of linear programming, which we'll later see, this region is called the feasible region.

Exploration

Consider these two inequalities $y\le x+1$y≤x+1 and $y\ge0$y≥0. If we construct each inequality individually, we see an "overlap" of the two inequalities between the lines $y=x+1$y=x+1 and $y=0$y=0. All the points in this "overlap" are guaranteed to satisfy both inequalities as shown below.

We say the "overlap" region that satisfies both of these two inequalities is the feasible region or required region. When constructing the graph of these two inequalities, we can either state that the feasible region is the "darker" region, or we can just shade the feasible region and accompany it with a key.

In the above example, the inequalities were not strict ($\ge$≥ and $\le$≤) so the boundaries are included in the required region, and indicated by a solid line. For inequalities that are strict ($>$> or $<$<), the boundaries are not included, and indicated by a dashed line. Sometimes there is a combination of both, where the required region is bordered by solid and dashed lines.

Worked example

Construct the feasible region satisfying the inequalities $x\ge0$x≥0, $y\ge0$y≥0, $2x-5y>-10$2x−5y>−10 and $x+y\le5$x+y≤5.

Think: We first want to draw each of the inequalities as if they were equations. We use dashed lines for strict inequalities, and solid lines for inequalities that aren't strict. Then we can shade the region where they intersect.

Do: Below is a graph of all the inequalities as if they were equations.

The inequalities $x\ge0$x≥0 and $y\ge0$y≥0 describe the first quadrant (the top-right quarter of the $xy$xy-plane)–so we know the feasible region must lie there. It is easier to graph the other two inequalities by making $y$y the subject:

$2x-5y$2x−5y	$>$>	$-10$−10
$-5y$−5y	$>$>	$-2x-10$−2x−10
$y$y	$<$<	$\frac{2x}{5}+2$2x5​+2

 

(The inequality sign changes direction because both sides were divided by $-5$−5 to make $y$y the subject.)

$x+y$x+y	$\le$≤	$5$5
$y$y	$\le$≤	$-x+5$−x+5

 

To ensure we shade the correct region, we can choose a set of coordinates from the predicted feasible region, and substitute into each of the inequalities. Let's use $(2,1)$(2,1).

Inequality $1$1	Inequality $2$2	Inequality $3$3	Inequality $4$4
$x\ge0$x≥0	$y\ge0$y≥0	$y<\frac{2x}{5}+2$y<2x5​+2	$y\le-x+5$y≤−x+5
$2\ge0$2≥0	$1\ge0$1≥0	$1<\frac{2\times2}{5}+2$1<2×25​+2	$1\le-2+5$1≤−2+5
		$1<\frac{14}{5}$1<145​	$1\le3$1≤3

As all of these inequalities hold for the coordinates $(2,1)$(2,1), the area enclosed by the lines which includes this set of coordinates is the feasible region. By using these equations, we know to shade below the two lines.

It's good practice to indicate the corner points of the feasible region if you're drawing them by hand. That way, it's clear where the boundaries of the feasible region lie.

Reflect: The feasible region can be either bounded or unbounded. In the worked example, the feasible region was bounded. In the exploration, the feasible region was unbounded. Also note that the feasible region in the worked example has a combination of dashed and solid boundary lines.

Practice questions

Question 1

Question 2

 

Several word problems can be "disguised" as systems of linear inequalities. Typically, these problems are to do with the allocation of resources subject to constraints.

The way to approach these types of problems is to first choose variables to represent the amounts of each resource. Then, we want to construct inequalities using these variables that reflect the constraints.

Worked example

example 1

There are $200$200 employees in an organisation. Some of the employees help produce product $A$A and some help produce product $B$B. Some employees may work in other areas of the organisation that aren't involved in either product. In order to produce any meaningful quantity of either product, at least $50$50 workers must be assigned to each product.

Express the above problem as a system of linear inequalities.

Think: We first want to choose some variables. Let $x$x be the number of workers producing product $A$A, and $y$y be the number of workers producing product $B$B. Then, using only $x$x and $y$y, we want to define a set of inequalities that match the above constraints.

Do: The sum of the workers on product $A$A and product $B$B must be at most the total number of workers in the organisation. This translates to:

$x+y\le200$x+y≤200 (1)

There must be at least $50$50 workers on each product, which translates to:

$x\ge50$x≥50 (2)

$y\ge50$y≥50 (3)

All together, we have the following system of linear inequalities:

$x+y\le200$x+y≤200, $x\ge50$x≥50 and $y\le50$y≤50

Graphically, the set of points satisfying these inequalities is given below.

Reflect: Sometimes we can express the same constraints, but using different units. For instance, suppose each unit of product $A$A requires $2.4$2.4 hours worth of work and each unit of product $B$B requires $3$3 hours worth of work. If each worker can work $36$36 hours per week, then there must be at most $200\times36=7200$200×36=7200 hours available.

Let $\alpha$α be the number of units of product $A$A produced in a week, and $\beta$β be the number of units of product $B$B produced in a week.

Then the time spent on product $A$A in a week is $2.4\alpha$2.4α hours, and the time spent on product $B$B in a week is $3\beta$3β hours. Since the total amount of time available in a week is $7200$7200 hours, then we have the inequality:

$2.4\alpha+3\beta\le7200$2.4α+3β≤7200

This is equivalent to inequality (1), but rephrased in terms of hours and not workers. What are the equivalent inequalities for (2) and (3)?

Practice questions

Question 3

Question 4