Unit 6, lesson 73

Inequalities in Two Variables

How to graph an inequality in x and y as a region of the plane, using a boundary curve, solid or dashed lines and test points, and how to combine regions into a feasible region.

  • Theory part 145
  • Practice part 146
  • Systems of Equations and Inequalities

A farmer has 6 hectares of land and 8 days of labor to share between wheat and corn. Every possible plan is a point (x,y)(x, y), with xx hectares of wheat and yy hectares of corn, and each limit rules out part of the plane. What is left is a picture of every plan that actually works. Inequalities in two variables describe regions like this, and once the region is drawn you can search it for the best plan.

Key terms

Linear and nonlinear inequalities

Definition An inequality in two variables compares two expressions in xx and yy with one of the symbols <<, ≤\le, >> or ≥\ge. An ordered pair (a,b)(a, b) is a solution if the inequality is true for x=ax = a and y=by = b. The solution set, also called the graph of the inequality, is the set of all solutions, drawn as a region of the plane. The inequality is linear if it can be written as

Ax+By<C,Ax+By≤C,Ax+By>CorAx+By≥C\begin{gathered} Ax + By < C, \quad Ax + By \le C, \\[0.4em] Ax + By > C \quad \text{or} \quad Ax + By \ge C \end{gathered}

with real numbers AA, BB, CC and with AA and BB not both zero. Any other inequality is nonlinear, for example y≥x2y \ge x^2, x2+y2<9x^2 + y^2 < 9 or xy>1xy > 1.

In plain words A solution is now a point, not a number, and the answer is a whole region. A linear inequality is cut out by a straight line, while a nonlinear one can be cut out by a parabola, a circle or any other curve.

Example Here are four points tested in the linear inequality 3x+2y≤123x + 2y \le 12.

PointValue of 3x+2y3x + 2ySolution?
(0,0)(0, 0)00yes
(2,3)(2, 3)1212yes
(4,0)(4, 0)1212yes
(2,5)(2, 5)1616no

The point (1,2)(1, 2) is a solution of the nonlinear inequality y≥x2y \ge x^2, since 2≥12 \ge 1.

Boundary curve

Definition The boundary curve of an inequality in xx and yy is the graph of the equation obtained by replacing the inequality symbol with ==. If both sides are polynomials in xx and yy, the boundary curve splits the rest of the plane into regions, and within each region the inequality is either true at every point or false at every point. A point can switch from true to false only by crossing the boundary. If a side can be undefined, as with a fraction, the points where it is undefined must be treated as part of the boundary too, just as with the sign charts for rational inequalities.

In plain words The boundary is where the two sides are exactly equal. It is the fence between points that work and points that do not.

Example Each inequality below has a familiar boundary.

InequalityBoundary curve
2x−y<42x - y < 4the line y=2x−4y = 2x - 4
y>x2−4y > x^2 - 4the parabola y=x2−4y = x^2 - 4 with vertex (0,−4)(0, -4)
x2+y2≤25x^2 + y^2 \le 25the circle x2+y2=25x^2 + y^2 = 25 with center (0,0)(0, 0) and radius 55

The circle splits the rest of the plane into two regions, the inside and the outside.

Solid and dashed boundary

Definition For an inequality with ≤\le or ≥\ge, every point of the boundary curve makes both sides equal and so satisfies the inequality. The boundary belongs to the solution set and is drawn as a solid curve. For an inequality with << or >>, no point of the boundary satisfies it. The boundary is left out of the solution set and is drawn as a dashed curve. Points where a side is undefined are never solutions, so they are always left out.

In plain words A solid line means the fence itself counts. A dashed line means you may come as close as you like but never stand on it, just like an open endpoint on a number line.

Example The point (3,4)(3, 4) lies on the circle x2+y2=25x^2 + y^2 = 25. It satisfies x2+y2≤25x^2 + y^2 \le 25 because 25≤2525 \le 25, so that circle is drawn solid. The point (2,0)(2, 0) lies on the parabola y=x2−4y = x^2 - 4, since 22−4=02^2 - 4 = 0. It does not satisfy y>x2−4y > x^2 - 4, because 0>00 > 0 is false, so that parabola is drawn dashed.

Half-plane

Definition A line in the plane splits the points not on it into two open half-planes, one on each side. A closed half-plane is an open half-plane together with its boundary line. The solution set of a linear inequality with << or >> is an open half-plane, and with ≤\le or ≥\ge it is a closed half-plane. When B≠0B \ne 0 you can solve for yy. If B>0B > 0, then Ax+By<CAx + By < C means

y<C−AxB,y < \frac{C - Ax}{B},

the half-plane below the line. If B<0B < 0, dividing by BB reverses the inequality and gives the half-plane above the line. When B=0B = 0 the boundary is a vertical line and the half-plane lies to its left or right.

In plain words A straight boundary always leaves two sides, and the answer is one whole side. After you solve for yy, the symbol << means below the line and >> means above it. Just remember to flip the symbol if you divide by a negative number.

Example To graph 2x−3y>62x - 3y > 6, subtract 2x2x and divide by −3-3, which reverses the inequality.

−3y>6−2x⟺y<23x−2.\begin{aligned} -3y &> 6 - 2x \\[0.4em] \Longleftrightarrow \quad y &< \frac{2}{3}x - 2. \end{aligned}

The solution set is the open half-plane below the dashed line y=23x−2y = \frac{2}{3}x - 2. As a check, the origin lies above that line and fails the original inequality, since 0>60 > 6 is false.

Test point

Definition A test point is a point not on the boundary curve that you substitute into the inequality to decide whether its region belongs to the solution set. When both sides are polynomials, the inequality is true in the whole region of a test point that makes it true, and false in the whole region of one that makes it false. One test point per region is enough. A linear inequality has only two regions, but a curved boundary can create more, and then every region needs its own test point.

In plain words Pick an easy point on one side, plug it in, and shade its whole side if it works. The origin is the easiest choice whenever it is not on the boundary.

Example The boundary of xy>1xy > 1 is the hyperbola xy=1xy = 1, whose two branches split the plane into three regions. Test one point in each.

Test pointValue of xyxyIs xy>1xy > 1?
(2,2)(2, 2)44yes
(0,0)(0, 0)00no
(−2,−2)(-2, -2)44yes

So the solution set is the region beyond the branch in the first quadrant together with the region beyond the branch in the third quadrant. The middle region, which contains both axes, is left out, and the hyperbola is dashed. Testing only the origin would have missed that the answer comes in two pieces.

Intersection

Definition The intersection S∩TS \cap T of two sets of points is the set of points that belong to both SS and TT. A system of inequalities joined by “and” asks for the points that satisfy every inequality at once, so its solution set is the intersection of the individual solution sets.

In plain words Shade each inequality on its own. The answer is the part where all the shadings overlap.

Example Solve the system y≥x2y \ge x^2 and y≤x+2y \le x + 2. The first inequality gives the region on and above the parabola, the second gives the region on and below the line. The boundaries meet where x2=x+2x^2 = x + 2, so where x=−1x = -1 or x=2x = 2, at the points (−1,1)(-1, 1) and (2,4)(2, 4). The solution set is the region between the parabola and the line for −1≤x≤2-1 \le x \le 2, boundaries included. The point (0,1)(0, 1) is inside, since 1≥01 \ge 0 and 1≤21 \le 2. The point (3,5)(3, 5) satisfies the second inequality but not the first, because 5≥95 \ge 9 is false, so it is not in the intersection.

Union

Definition The union S∪TS \cup T of two sets of points is the set of points that belong to SS or to TT or to both. A compound inequality joined by “or” has as its solution set the union of the individual solution sets.

In plain words A point counts if it passes at least one of the tests. Shade every inequality, and keep everything that is shaded at least once.

Example For c>0c > 0 the condition ∣u∣≥c\lvert u \rvert \ge c means u≥cu \ge c or u≤−cu \le -c. So ∣x−y∣≥1\lvert x - y \rvert \ge 1 means

y≤x−1ory≥x+1.y \le x - 1 \quad \text{or} \quad y \ge x + 1.

The solution set is the union of two closed half-planes, everything except the open strip between the parallel lines y=x−1y = x - 1 and y=x+1y = x + 1. The origin is in that strip and fails, since ∣0∣≥1\lvert 0 \rvert \ge 1 is false. The points (3,0)(3, 0) and (0,3)(0, 3) each satisfy one of the two inequalities, so both are solutions.

Feasible region

Definition In a problem with several conditions, called constraints, the feasible region is the set of all points that satisfy every constraint, that is the intersection of their solution sets. A corner point, or vertex, of the region is a point of the region where two nonparallel boundary lines meet. If the feasible region is nonempty and bounded and is described by finitely many linear inequalities with ≤\le or ≥\ge, then every linear expression ax+byax + by reaches its largest value on the region at a corner point, and its smallest value at a corner point too.

In plain words The feasible region holds every plan that breaks no rule. When the rules are straight lines and the region is bounded, it is a polygon, and the best and worst values of a linear quantity sit at its corners.

Example Back to the farmer. Wheat needs 1 day of labor per hectare and corn needs 2, so the constraints are

x≥0,y≥0,x+y≤6,x+2y≤8.\begin{gathered} x \ge 0, \qquad y \ge 0, \\[0.4em] x + y \le 6, \qquad x + 2y \le 8. \end{gathered}

The lines x+y=6x + y = 6 and x+2y=8x + 2y = 8 meet where y=2y = 2 and x=4x = 4. The corner points are (0,0)(0, 0), (6,0)(6, 0), (4,2)(4, 2) and (0,4)(0, 4), and each satisfies all four constraints. Suppose wheat earns 3 thousand dollars per hectare and corn earns 4. The profit 3x+4y3x + 4y is 00, 1818, 2020 and 1616 at these corners, so the best plan is 4 hectares of wheat and 2 of corn. When the quantity to be made largest is not linear, checking corners is no longer enough, and calculus finds the best value with derivatives, as in optimization problems.

Check yourself

  1. Graph 3x−y≥63x - y \ge 6. Is the boundary solid or dashed, and which side is shaded?
Show the answer

The boundary is the line 3x−y=63x - y = 6, or y=3x−6y = 3x - 6, drawn solid because of ≥\ge. The origin gives 0≥60 \ge 6, which is false, so shade the side away from the origin. That is the closed half-plane below the line, y≤3x−6y \le 3x - 6.

  1. Find the corner points of the feasible region given by x≥0x \ge 0, y≥0y \ge 0, 2x+y≤102x + y \le 10 and x+y≤7x + y \le 7.
Show the answer

Subtracting x+y=7x + y = 7 from 2x+y=102x + y = 10 gives x=3x = 3, and then y=4y = 4. The corner points are (0,0)(0, 0), (5,0)(5, 0), (3,4)(3, 4) and (0,7)(0, 7). Each one satisfies all four constraints, while the other crossings (0,10)(0, 10) and (7,0)(7, 0) break one of them.

  1. Is the point (1,1)(1, 1) a solution of the system x2+y2<4x^2 + y^2 < 4 and y>x2y > x^2? Is it a solution of x2+y2<4x^2 + y^2 < 4 or y>x2y > x^2?
Show the answer

The first inequality holds, since 1+1=2<41 + 1 = 2 < 4. The second fails, since 1>11 > 1 is false, because (1,1)(1, 1) lies on the dashed parabola y=x2y = x^2. So the point is not in the intersection, but it is in the union.

How this lesson connects

Where this leads in Calculus