2.7 Linear Inequalities in Two Variables & Feasible Regions

Linear Inequalities in Two Variables & Feasible Regions

A linear inequality in two variables (such as $y \le 2x + 4$ or $3x + 5y > 15$) represents an entire half-plane of coordinate pairs $(x, y)$ in the coordinate plane. When multiple inequalities are combined into a system, their overlapping intersection forms a feasible region that models resource limits, production constraints, and financial allocations.

---

1. Graphing Boundary Lines & Determining Line Styles

Every two-variable linear inequality divides the coordinate plane into two regions separated by a boundary line.

+-----------------------------------------------------------------------------+
|                        BOUNDARY LINE STYLE RULES                            |
|                                                                             |
|   INEQUALITY SYMBOL     BOUNDARY LINE STYLE    POINTS ON LINE INCLUDED?     |
|   -----------------     -------------------    ------------------------     |
|   <  or  >              Dashed / Broken Line   NO  (Boundary not in set)    |
|   <= or  >=             Solid / Continuous     YES (Boundary is in set)     |
+-----------------------------------------------------------------------------+

   DASHED LINE (< or >)                          SOLID LINE (<= or >=)
           y                                             y
           |   - - /                                     |     /
           |    - /                                      |    / 
           |   - /                                       |   /  
   --------+----/--- x                           --------+--/----- x
           | - /                                         | /    
   Boundary points EXCLUDED                      Boundary points INCLUDED

---

2. Determining Shading Direction: Two Reliable Methods

Once the boundary line is graphed, you must shade the correct half-plane representing the inequality's solution set.

Method 1: The Test Point Method (The Universal Approach)

1. Choose any test point $(x_0, y_0)$ that does not lie directly on the boundary line. The origin $(0, 0)$ is the fastest and easiest test point unless the boundary line passes through $(0, 0)$.
2. Substitute $(x_0, y_0)$ into the original inequality.
3. If the statement is TRUE: Shade the entire half-plane containing $(x_0, y_0)$.
4. If the statement is FALSE: Shade the opposite half-plane away from $(x_0, y_0)$.

Method 2: The Slope-Intercept Vertical Position Rule

When the inequality is solved explicitly for $y$ in slope-intercept form:

• $y > mx + b$ or $y \ge mx + b$ : Shade ABOVE the boundary line (greater $y$-values).
• $y < mx + b$ or $y \le mx + b$ : Shade BELOW the boundary line (lesser $y$-values).

[!CAUTION] Standard Form Negative Coefficient Trap: In the inequality $3x - 2y \le 6$, do not assume "$\le$" means shade below. When converting to slope-intercept form, dividing by $-2$ reverses the inequality: $-2y \le -3x + 6 \implies y \ge \frac{3}{2}x - 3$. The "$\ge$" indicates that the solution set is actually ABOVE the boundary line!

---

3. Systems of Linear Inequalities & Feasible Regions

A system of linear inequalities consists of two or more inequalities in the same variables. The solution set of the system is the overlapping intersection of the shaded regions of all individual inequalities.

+-----------------------------------------------------------------------------+
|                   SYSTEM OF INEQUALITIES FEASIBLE REGION                    |
|                                                                             |
|   Inequality 1: y >= -x + 4                                                 |
|   Inequality 2: y <= 2x + 1                                                 |
|                                                                             |
|                        y                                                    |
|                        |          Feasible Region                           |
|                        |            (Overlap)                               |
|                        |             /  /                                   |
|                        |            /  /                                    |
|                        |   (1, 3)  *  /   y = 2x + 1                        |
|                        |          / \/                                      |
|                        |         /  /\                                      |
|                        |        /  /  \                                     |
|                        |       /  /    \  y = -x + 4                        |
|   ---------------------+------+---------+----- x                            |
|                        |                                                    |
+-----------------------------------------------------------------------------+

Calculating Feasible Region Vertices

The vertices (corner points) of a feasible region are the points where the boundary lines intersect. To find a vertex, solve the system of linear equations formed by the intersecting boundary lines using substitution or elimination.

Worked Example: Multi-Constraint Feasible Region

Problem: Find all four vertices of the feasible region defined by the following system of linear constraints:

$x \ge 0$ $y \ge 0$ $x + y \le 8$ $2x + y \le 12$

Step 1: Identify the non-negativity axes boundaries The constraints $x \ge 0$ and $y \ge 0$ restrict the feasible region strictly to the First Quadrant.

• Vertex 1: Intersection of $x = 0$ and $y = 0 \implies (0, 0)$.

Step 2: Find the $x$-axis intercept of $2x + y = 12$ Set $y = 0$: $2x + 0 = 12 \implies x = 6$.

• Vertex 2: $(6, 0)$.

Step 3: Find the intersection of the two constraint lines Solve the system:

1. $x + y = 8 \implies y = 8 - x$
2. $2x + y = 12$

Substitute $y = 8 - x$ into equation 2: $2x + (8 - x) = 12 \implies x + 8 = 12 \implies x = 4$ $y = 8 - 4 = 4$

• Vertex 3: $(4, 4)$.

Step 4: Find the $y$-axis intercept of $x + y = 8$ Set $x = 0$: $0 + y = 8 \implies y = 8$.

• Vertex 4: $(0, 8)$.

Feasible Region Summary: The bounded feasible region is a convex polygon with four vertices: $(0, 0), (6, 0), (4, 4),$ and $(0, 8)$.

---

4. Real-World Constraint Optimization (The Vertex Theorem)

In real-world business and manufacturing problems (linear programming), an objective function (such as profit $P(x, y) = ax + by$ or cost $C(x, y) = cx + dy$) is optimized subject to a set of linear inequality constraints.

[!IMPORTANT] The Vertex Principle of Linear Programming: The maximum or minimum value of any linear objective function $f(x, y) = ax + by$ over a bounded polygonal feasible region always occurs at one of the vertices (corner points) of the region.

Testing Vertices for Maximum Profit

Using the vertices found in the previous example: $(0, 0), (6, 0), (4, 4), (0, 8)$ with profit function $P(x, y) = 50x + 40y$:

The maximum possible profit is $360, achieved by producing 4 units of $x$ and 4 units of $y$.

---

5. Digital SAT Desmos Workflows for Inequalities

The built-in Desmos calculator makes solving systems of inequalities fast and visual:

1. Type each inequality on a separate line (e.g., y >= -x + 4 and y <= 2x + 1).
2. Desmos automatically renders solid or dashed boundary lines and shades each region.
3. The feasible region appears as the darkest overlapping shaded area.
4. Click directly on the intersection of boundary lines to view the exact coordinate values of vertices.