4.3 Single & Systems of Linear Inequalities on the Coordinate Plane

4.3 Single & Systems of Linear Inequalities on the Coordinate Plane

Linear inequalities extend linear equation concepts into regions of numerical order and multi-dimensional coordinate spaces. On the FTCE Mathematics 6–12 examination, educators must demonstrate conceptual and algebraic mastery of order axioms, boundary line conventions, half-plane test point methods, convex polygonal feasible regions, and linear programming optimization protocols.

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1. Single Linear Inequalities & Order Inversion Axioms

A linear inequality in one variable over the real numbers $\mathbb{R}$ involves expressions of degree one separated by inequality symbols ($<, \le, >, \ge$). The algebraic properties governing inequalities mirror those of equations, with one crucial, foundational distinction: multiplying or dividing by a negative real number reverses the direction of the inequality.

The algebraic justification stems directly from the axioms of an ordered field. By definition, $a < b$ means that the difference $b - a$ is strictly positive ($b - a > 0$). If $c$ is a negative real number ($c < 0$), then by definition $-c$ is strictly positive ($-c > 0$). Multiplying the positive quantity $(b - a)$ by the positive quantity $(-c)$ must yield a positive product by field closure under multiplication:

$(-c)(b - a) > 0 \implies -bc + ac > 0 \implies ac - bc > 0 \implies ac > bc$

Hence, $a < b$ and $c < 0$ rigorously imply $ac > bc$. Failing to reverse the inequality symbol when dividing or multiplying by a negative scalar is one of the most widespread student errors diagnosed on the FTCE Mathematics 6–12 examination.

Compound inequalities combine two conditions:

• Conjunction ("And"): $a \le x \le b$ denotes the intersection of two conditions: $x \ge a$ and $x \le b$. The solution set is a bounded interval $[a, b]$.
• Disjunction ("Or"): $x \le a \text{ or } x \ge b$ (with $a < b$) denotes the union of two disjoint unbounded intervals: $(-\infty, a] \cup [b, \infty)$.

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2. Two-Variable Linear Inequalities & Half-Plane Geometry

A linear inequality in two variables can be written in standard form as $Ax + By \le C$ (or with $<, >, \ge$). In the Cartesian plane $\mathbb{R}^2$, the corresponding linear equation $Ax + By = C$ defines the boundary line.

By the Jordan Curve Theorem for lines, any straight line partitions the Euclidean plane into three mutually disjoint subsets: the boundary line itself, and two open half-planes lying on opposite sides of the line.

• Boundary Line Style:
  • Solid Line ($\le, \ge$): Indicates an inclusive inequality where the boundary points satisfy the relation.
  • Dashed Line ($<, >$): Indicates a strict inequality where points on the boundary line are excluded from the solution set.
• The Test-Point Protocol: To identify which half-plane contains the solution set:
  1. Graph the boundary line $Ax + By = C$.
  2. Select a test point $(x_0, y_0)$ that does not lie on the boundary line. The origin $(0, 0)$ is the standard algebraic choice whenever $C \neq 0$.
  3. Substitute $(x_0, y_0)$ into the inequality. If the resulting numerical statement is true, shade the entire half-plane containing $(x_0, y_0)$. If false, shade the opposite half-plane.
• Solving for $y$ Warning: When converting $Ax + By > C$ to slope-intercept form $y > mx + b$, if $B < 0$, dividing by $B$ reverses the inequality to $y < mx + b$. Thus, "greater than" in standard form with $B < 0$ corresponds to shading below the boundary line.

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3. Systems of Linear Inequalities & Convex Feasible Regions

A system of linear inequalities consists of two or more simultaneous inequalities:

$\begin{cases} A_1 x + B_1 y \le C_1 \\ A_2 x + B_2 y \le C_2 \\ \quad \vdots \\ A_m x + B_m y \le C_m \end{cases}$

The overall solution set $\mathcal{F}$ is the simultaneous intersection of all individual closed (or open) half-planes:

$\mathcal{F} = \bigcap_{i=1}^m H_i$

A fundamental geometric property of the feasible region $\mathcal{F}$ is convexity. A set $\mathcal{S} \subseteq \mathbb{R}^2$ is convex if for every pair of points $P, Q \in \mathcal{S}$, the line segment $\overline{PQ}$ is entirely contained within $\mathcal{S}$. Because every half-plane is convex and the intersection of any collection of convex sets is itself convex, every linear inequality feasible region is a convex polygonal set.

Feasible regions are classified into two categories:

• Bounded (Compact): The region can be enclosed entirely within a circle of finite radius. It forms a closed convex polygon with finite area and a finite number of vertices (corner points).
• Unbounded: The region extends infinitely in one or more directions.

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4. Linear Programming & The Fundamental Extreme Value Theorem

Linear Programming is a mathematical optimization technique used to maximize or minimize a linear objective function:

$z = f(x, y) = ax + by$

subject to a set of linear inequality constraints defining a feasible region $\mathcal{F}$.

• Fundamental Theorem of Linear Programming: If an optimal value (maximum or minimum) of the objective function $z = ax + by$ exists over a non-empty closed and bounded convex feasible region $\mathcal{F}$, it must occur at one or more vertices (corner points) of $\mathcal{F}$.
• Geometric Rationale: Level sets of the objective function $ax + by = k$ represent a family of parallel lines with constant slope $m = -\frac{a}{b}$. Increasing (or decreasing) $k$ translates this line across the coordinate plane. The extreme values of $k$ where the line still makes contact with the closed convex polygon $\mathcal{F}$ occur at the boundary extremes—which are inevitably corner vertices (or an entire boundary edge connecting two vertices if the level line is parallel to an edge).

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5. Worked Exemplar: Linear Programming Optimization

Problem: A small craft furniture shop manufactures custom desks ($x$) and bookcases ($y$). The shop seeks to maximize weekly profit $P(x, y) = 40x + 30y$ subject to the constraints:

$\begin{cases} x + 2y \le 16 & \text{(Carpentry hours)} \\ x + y \le 9 & \text{(Finishing hours)} \\ x \ge 0, \; y \ge 0 & \text{(Non-negativity constraints)} \end{cases}$

Step 1: Determine boundary lines and intercepts

1. $x + 2y = 16$: Intercepts are $(16, 0)$ and $(0, 8)$.
2. $x + y = 9$: Intercepts are $(9, 0)$ and $(0, 9)$.
3. Non-negativity $x \ge 0, y \ge 0$ restricts the region to Quadrant I.

Step 2: Identify the vertices of the feasible region The feasible region is a convex polygon bounded by the coordinate axes and constraint lines:

• Vertex 1: The origin $(0, 0)$.
• Vertex 2: The $y$-intercept bounded by carpentry: $(0, 8)$ (since $(0, 9)$ violates $x + 2y \le 16$, as $0 + 2(9) = 18 > 16$).
• Vertex 3: The $x$-intercept bounded by finishing: $(9, 0)$ (since $(16, 0)$ violates $x + y \le 9$).
• Vertex 4: The intersection of $x + 2y = 16$ and $x + y = 9$: Subtract the second equation from the first: $(x + 2y) - (x + y) = 16 - 9 \implies y = 7$ Substitute $y = 7$ into $x + y = 9 \implies x = 2$. Intersection vertex: $(2, 7)$.

Step 3: Evaluate the objective function $P(x, y) = 40x + 30y$ at each vertex

Conclusion: The maximum weekly profit is $$360$, achieved by manufacturing $9$ desks and $0$ bookcases.

[!IMPORTANT] A common student error is assuming the maximum must always occur at the interior intersection vertex $(2, 7)$. As demonstrated above, the slope of the objective function ($m = -\frac{40}{30} = -\frac{4}{3}$) relative to the constraint boundaries ($m_1 = -\frac{1}{2}, m_2 = -1$) dictates which vertex is struck last by the expanding level line. Rigorous corner-point evaluation is mandatory.