AP Precalculus: Systems of Inequalities & Linear Programming

Master graphing inequalities, finding feasible regions, and optimization techniques

📊 Graphing 🎯 Feasible Regions 📍 Vertices 📈 Optimization

📚 Understanding Systems of Inequalities

A system of inequalities consists of two or more inequalities with the same variables. Unlike equations, the solution is a region containing infinitely many points. This guide covers graphing techniques, finding vertices, and using linear programming to optimize real-world problems.

1 Graphing Linear Inequalities

A linear inequality in two variables creates a half-plane of solutions. The boundary line divides the coordinate plane, and shading indicates the solution region.

Boundary Line Types

≤ or ≥

Solid Line

Points ON the line ARE included in the solution

< or >

Dashed Line

Points ON the line are NOT included

Shading Direction

$y > mx + b$ or $y \geq mx + b$

Shade ABOVE the line

$y < mx + b$ or $y \leq mx + b$

Shade BELOW the line

Steps to Graph

Solve for $y$ to get slope-intercept form (if needed)
Graph the boundary line (solid or dashed based on symbol)
Choose a test point NOT on the line (usually $(0, 0)$)
Substitute the test point into the inequality
If TRUE: shade the side containing the test point. If FALSE: shade the opposite side

📌 Example

Graph: $2x + y < 4$

Solve for y: $y < -2x + 4$

Boundary: $y = -2x + 4$ (dashed line, since $<$)< /p>

Test $(0, 0)$: $0 < 4$ ✓ TRUE

Shade: Below the line (side containing origin)

2 Solving Systems of Linear Inequalities

The solution to a system of inequalities is the intersection (overlap) of all individual solution regions. This overlapping area is called the feasible region.

Steps to Solve

Graph each inequality on the same coordinate plane
Shade the solution region for each inequality
Identify the overlapping region (feasible region)
The solution is all points in the overlapping region

📌 Example

System: $\begin{cases} y \geq x - 2 \\ y \leq -x + 4 \\ x \geq 0 \end{cases}$

Graph each:

• $y \geq x - 2$: solid line, shade above

• $y \leq -x + 4$: solid line, shade below

• $x \geq 0$: solid vertical line, shade right

Solution: Triangular region where all three shadings overlap

💡 No Solution Case

If the shaded regions don't overlap at all, the system has no solution. This happens when constraints are contradictory.

3 Absolute Value Inequalities

Absolute value inequalities create V-shaped boundaries. To solve, split into two separate linear inequalities.

"Less Than" Case

$|ax + by + c| \leq d$

Splits into:
$-d \leq ax + by + c \leq d$

Solution: region BETWEEN two parallel lines

"Greater Than" Case

$|ax + by + c| \geq d$

Splits into:
$ax + by + c \leq -d$ OR $ax + by + c \geq d$

Solution: region OUTSIDE two parallel lines

📌 Example

Solve: $|x + 2y| \leq 6$

Split: $-6 \leq x + 2y \leq 6$

Two inequalities:

• $x + 2y \leq 6$ → shade below

• $x + 2y \geq -6$ → shade above

Solution: Strip between the two parallel lines

4 Finding Vertices of the Feasible Region

Vertices (corner points) are where boundary lines intersect. These points are crucial for optimization problems.

How to Find Vertices

Identify all boundary lines (convert inequalities to equations)
Find where each pair of boundary lines intersects
Solve the system of two equations for each pair
Keep only vertices that lie within the feasible region

📌 Example

System: $\begin{cases} x + y \leq 6 \\ x \geq 0 \\ y \geq 0 \end{cases}$

Boundaries: $x + y = 6$, $x = 0$, $y = 0$

Find intersections:

• $x = 0$ and $y = 0$ → $(0, 0)$

• $x = 0$ and $x + y = 6$ → $(0, 6)$

• $y = 0$ and $x + y = 6$ → $(6, 0)$

Vertices: $(0, 0)$, $(0, 6)$, $(6, 0)$

⚠️ Check Each Vertex

Not every intersection point is a vertex of the feasible region! Verify that each candidate point satisfies ALL inequalities in the system.

5 Introduction to Linear Programming

Linear programming is a method to find the maximum or minimum value of a linear function subject to constraints (inequalities). It's used in business, economics, and operations research.

Objective Function

$Z = ax + by$

The linear function you want to maximize or minimize (profit, cost, etc.)

Constraints

System of linear inequalities

Limitations or requirements that define the feasible region

Key Theorem If the feasible region is bounded, the maximum and minimum values of the objective function occur AT VERTICES of the feasible region.

6 Solving Linear Programming Problems

To solve a linear programming problem, graph the constraints, identify vertices, and evaluate the objective function at each vertex.

The Linear Programming Process

Step 1

Graph all constraints

Step 2

Identify feasible region

Step 3

Find all vertices

Step 4

Evaluate $Z$ at each vertex

Step 5

Choose max or min value

📌 Complete Example

Problem: Maximize $Z = 3x + 4y$ subject to:

$\begin{cases} x + y \leq 8 \\ 2x + y \leq 10 \\ x \geq 0, \ y \geq 0 \end{cases}$

Step 1-2: Graph constraints and find feasible region

Step 3: Find vertices:

• $(0, 0)$ — intersection of $x = 0$ and $y = 0$

• $(5, 0)$ — intersection of $y = 0$ and $2x + y = 10$

• $(2, 6)$ — intersection of $x + y = 8$ and $2x + y = 10$

• $(0, 8)$ — intersection of $x = 0$ and $x + y = 8$

Step 4: Evaluate $Z = 3x + 4y$:

• $Z(0, 0) = 0$

• $Z(5, 0) = 15$

• $Z(2, 6) = 6 + 24 = 30$

• $Z(0, 8) = 32$

Step 5: Maximum is $Z = 32$ at $(0, 8)$

7 Bounded vs Unbounded Feasible Regions

The shape of the feasible region affects whether maximum and minimum values exist.

Bounded Region

Feasible region is enclosed (polygon shape)

✅ Both maximum AND minimum exist
✅ Both occur at vertices

Unbounded Region

Feasible region extends infinitely in some direction

⚠️ May have only max OR only min
⚠️ Or may have neither (unbounded objective)

💡 Real-World Constraints

Most real-world problems include $x \geq 0$ and $y \geq 0$ (non-negativity constraints) because you can't have negative quantities of resources, products, etc.

8 Real-World Applications

Linear programming solves optimization problems in business, manufacturing, logistics, and resource allocation.

Common Application Types

Profit Maximization: Find product mix that maximizes profit given limited resources
Cost Minimization: Meet requirements at minimum cost
Resource Allocation: Distribute limited resources optimally
Diet Problems: Meet nutritional needs at minimum cost
Transportation: Minimize shipping costs between locations

📌 Word Problem Example

Problem: A company makes chairs and tables. Each chair requires 2 hours of labor and yields $40 profit. Each table requires 5 hours and yields $80 profit. With 40 labor hours available, maximize profit.

Variables: $x$ = chairs, $y$ = tables

Objective: Maximize $Z = 40x + 80y$

Constraints:

• $2x + 5y \leq 40$ (labor hours)

• $x \geq 0, \ y \geq 0$ (non-negative)

Vertices: $(0, 0)$, $(20, 0)$, $(0, 8)$

Evaluate: $Z(0, 8) = 640$ ← Maximum

Answer: Make 0 chairs & 8 tables for $640 profit

📋 Quick Reference

Solid Line

$\leq$ or $\geq$ — boundary included

Dashed Line

$<$ or $>$ — boundary excluded

Feasible Region

Intersection of all shaded regions

Vertices

Solve pairs of boundary equations

Objective Function

$Z = ax + by$ (maximize or minimize)

Optimal Value

Evaluate $Z$ at all vertices, choose best

📚 Understanding Systems of Inequalities

1 Graphing Linear Inequalities

Boundary Line Types

Shading Direction

Steps to Graph

2 Solving Systems of Linear Inequalities

Steps to Solve

3 Absolute Value Inequalities

"Less Than" Case

"Greater Than" Case

4 Finding Vertices of the Feasible Region

How to Find Vertices

5 Introduction to Linear Programming

Objective Function

Constraints

6 Solving Linear Programming Problems

The Linear Programming Process

7 Bounded vs Unbounded Feasible Regions

Bounded Region

Unbounded Region

8 Real-World Applications

Common Application Types

📋 Quick Reference

Solid Line

Dashed Line

Feasible Region

Vertices

Objective Function

Optimal Value

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