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Topos Uranos
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Inequalities in the plane and linear programming

Summary

This lesson carries the study of inequalities into the plane. It proves that a line divides the plane into two half-planes, on each of which the linear expression defining it keeps its sign, and from this it derives the test point method. It describes the solution region of a system of linear inequalities as an intersection of half-planes, with its vertices, and proves that it is convex; it then studies regions bounded by conics. It ends with linear programming: the search for the maximum or the minimum of a linear function over a polygonal region, whose fundamental theorem ensures that on a bounded convex polygon both are attained at vertices and that, on an unbounded region with some vertex, every optimum that exists is also attained at one of them, and with production, diet and transport problems.

Learning objectives

  1. Represent the solution set of a linear inequality in two variables as a half-plane, open or closed, by means of a test point.
  2. Determine the solution region of a system of linear inequalities, its vertices and whether it is bounded or unbounded, and justify its convexity.
  3. Describe by inequalities regions bounded by lines and conics, and represent the regions so described.
  4. Set up and solve linear programming problems in two variables with the vertex theorem and level lines, distinguishing a unique optimum, multiple optima and the absence of an optimum.

Half-planes

In the lesson on the equation of a line it was seen that every line is the set of points satisfying an equation Ax+By+C=0Ax + By + C = 0, with AA and BB not both zero: if B≠0B \neq 0, one solves to get y=−ABx−CBy = -\frac{A}{B}x - \frac{C}{B}, a line with slope −AB-\frac{A}{B}; if B=0B = 0, the equation reduces to x=−CAx = -\frac{C}{A}, a vertical line. If an inequality is required instead of the equality, the solution set is no longer a curve and becomes a region. In the lesson on first-degree equations and inequalities, a linear inequality in one unknown had an interval of the real line as its solution; with two unknowns, the solution is a half-plane.

Let rr be a non-vertical line and let P=(x0,y0)P = (x_0, y_0) be a point. We say that PP lies above rr if y0y_0 is greater than the ordinate of the point of rr with abscissa x0x_0, and below if it is less. If rr is vertical, one speaks in the same way of the points lying to its right and to its left. Each of these two sets is an open half-plane with boundary rr; if the line is added to it, it is a closed half-plane.

TheoremHalf-planes

Let rr be the line Ax+By+C=0Ax + By + C = 0, with AA and BB not both zero, and let E(x,y)=Ax+By+CE(x, y) = Ax + By + C. The expression EE has constant sign on each side of rr: it is positive at all points of one of the two open half-planes and negative at all points of the other. More precisely, if B>0B > 0, it is positive above rr and negative below; if B<0B < 0, the other way round; and if B=0B = 0, it is positive to the right of rr when A>0A > 0, and to its left when A<0A < 0.

Proof

  1. Ax0+By1+C=0\dato{q}{A x_0 + B y_1 + C = 0}

    Let P=(x0,y0)P = (x_0, y_0) be a point not on rr, and suppose first that B≠0B \neq 0. The vertical line through PP meets rr at a single point Q=(x0,y1)Q = (x_0, y_1), whose ordinate is obtained by solving for yy in the equation of rr; the point QQ satisfies that equation.

  2. E(x0,y0)=(Ax0+By0+C)−(Ax0+By1+C)=B⋅(y0−y1)E(x_0, y_0) = \resaltar{(A x_0 + B y_0 + C) - (A x_0 + B y_1 + C)} = \dato{dif}{\resaltar{B \cdot (y_0 - y_1)}}

    We subtract this zero from the value of EE at PP: the terms in x0x_0 and the constants cancel, and a product remains.

  3. B>0→(E(x0,y0)>0↔y0>y1)B > 0 \rightarrow \left( E(x_0, y_0) > 0 \leftrightarrow y_0 > y_1 \right)

    If B>0B > 0, by the rule of signs studied in the lesson on sign charts and inequalities, the product has the sign of y0−y1y_0 - y_1: it is positive when PP lies above rr and negative when it lies below.

  4. B<0→(E(x0,y0)>0↔y0<y1)B < 0 \rightarrow \left( E(x_0, y_0) > 0 \leftrightarrow y_0 < y_1 \right)

    If B<0B < 0, the product has the sign opposite to that of y0−y1y_0 - y_1: the expression is negative above rr and positive below.

  5. E(x0,y0)=(Ax0+C)−(Ax1+C)=A⋅(x0−x1)E(x_0, y_0) = \resaltar{(A x_0 + C) - (A x_1 + C)} = \resaltar{A \cdot (x_0 - x_1)}

    If B=0B = 0, then A≠0A \neq 0 and rr is the vertical line x=x1x = x_1, with x1=−CAx_1 = -\frac{C}{A}. The computation is repeated with the point of rr that has the ordinate of PP.

  6. A>0→(E(x0,y0)>0↔x0>x1)A > 0 \rightarrow \left( E(x_0, y_0) > 0 \leftrightarrow x_0 > x_1 \right)

    If A>0A > 0, the expression is positive exactly when x0>x1x_0 > x_1, that is, to the right of rr; if A<0A < 0, to its left. In all cases the sign of EE at a point depends only on the side of rr on which the point lies.

The theorem says more than is used in practice: there is no need to remember which sign corresponds to each side, because it is enough to know the sign of EE at a single point of one side to know it at all of them. This is the test point method.

To represent an inequality Ax+By+C>0Ax + By + C > 0, or one with <<, ≥\ge or ≤\le, one draws the line Ax+By+C=0Ax + By + C = 0 and computes the value of the expression at a test point not on it, usually the origin. If the point satisfies the inequality, the solution set is the half-plane containing it; if not, it is the other one. If the inequality is strict, the line does not belong to the solution set and is drawn dashed; if it is not, the line belongs to it and is drawn solid.
ExampleA half-plane by the test point

Represent the solution set of the inequality 2x−3y+6>02x - 3y + 6 > 0.

Proof

  1. 2x−3y+6=0,(0,2),(−3,0)2x - 3y + 6 = 0, \quad (0, 2), \quad (-3, 0)

    We draw the boundary line through the points where it meets the axes: with x=0x = 0 we get y=2y = 2, and with y=0y = 0, x=−3x = -3.

  2. 2⋅0−3⋅0+6=6>02 \cdot 0 - 3 \cdot 0 + 6 = \resaltar{6} > 0

    The origin is not on the line. We compute the value of the expression there: it is positive, so the origin satisfies the inequality.

  3. 2x−3y+6>0↔y<23x+22x - 3y + 6 > 0 \leftrightarrow \resaltar{y < \frac{2}{3}x + 2}

    The solution set is the half-plane containing the origin. The theorem confirms it: since B=−3<0B = -3 < 0, the expression is positive below the line.

  4. 2⋅(−3)−3⋅0+6=02 \cdot (-3) - 3 \cdot 0 + 6 = \resaltar{0}

    The inequality is strict: the points of the line, such as (−3,0)(-3, 0), make the expression zero and do not satisfy it. The boundary is excluded and is drawn dashed.