Lesson 1.2: Linear Inequalities & Linear Programming

Master Linear Inequalities & Optimization

Explore the powerful world of linear inequalities and linear programming! Learn to solve complex inequalities, graph systems, identify feasible regions, and optimize real-world objectives using the vertex principle. Develop skills for decision-making and resource allocation.

Learning Objectives

Solve complex linear inequalities with fractions and parentheses

Graph inequality systems and identify feasible regions

Apply the vertex principle for optimization

Model real-world constraints as inequality systems

Interpret solutions in practical contexts

Use test-point method for shading regions

Core Concepts & Theoretical Foundation

Linear Inequality Properties & Transformations

Fundamental Properties:

• Addition Property: If $a < b$ , then $a + c < b + c$ for any real number c

• Subtraction Property: If $a < b$ , then $a - c < b - c$ for any real number c

• Multiplication Property: If $a < b$ and $c > 0$ , then $ac < bc$

• Division Property: If $a < b$ and $c > 0$ , then $\frac{a}{c} < \frac{b}{c}$

• Sign Reversal: If $a < b$ and $c < 0$ , then $ac > bc$ (inequality reverses)

Feasible Region & Vertex Principle

Key Theorems:

• Feasible Region: The intersection of all constraint inequalities forms a convex polygon (or unbounded region)

• Vertex Principle: For linear objective functions, optimal values occur at vertices of the feasible region

• Corner Point Method: Evaluate objective function at all vertices to find maximum/minimum

• Bounded vs. Unbounded: Feasible regions can be bounded (finite area) or unbounded (infinite area)

Test-Point Method for Graphing

Systematic Approach:

1. Graph boundary line: Replace inequality with equality and plot the line

2. Determine line type: Solid line for ≤ or ≥, dashed line for < or >

3. Choose test point: Select a point not on the line (usually (0,0) if possible)

4. Test inequality: Substitute test point into original inequality

5. Shade region: Shade the side containing the test point if inequality is true

Detailed Worked Examples

Example 1: Complex Linear Inequality

Solve the inequality: $\frac{2(x-1)}{3} - \frac{x+2}{4} \geq 1$ and represent the solution on a number line.

Step 1: Clear denominators (multiply by LCD = 12)

$12 \cdot \left[\frac{2(x-1)}{3} - \frac{x+2}{4}\right] \geq 12 \cdot 1$

$12 \cdot \frac{2(x-1)}{3} - 12 \cdot \frac{x+2}{4} \geq 12$

$4 \cdot 2(x-1) - 3 \cdot (x+2) \geq 12$

$8(x-1) - 3(x+2) \geq 12$

Step 2: Distribute and simplify

$8x - 8 - 3x - 6 \geq 12$

$5x - 14 \geq 12$

Step 3: Solve for x

$5x \geq 12 + 14$

$5x \geq 26$

$x \geq \frac{26}{5} = 5.2$

Step 4: Number line representation

Closed circle at 5.2, arrow pointing right (since x ≥ 5.2)

Solution: $x \geq 5.2$ or $[5.2, \infty)$ in interval notation.

Example 2: Inequality System & Feasible Region

Graph the system of inequalities and find the vertices of the feasible region:

$\begin{cases} x + 2y \leq 8 \\ x - y \geq 1 \\ y \geq 0 \\ x \geq 0 \end{cases}$

Step 1: Graph each inequality

Inequality 1: $x + 2y \leq 8$ → Line: $y = -\frac{1}{2}x + 4$ , shade below

Inequality 2: $x - y \geq 1$ → Line: $y = x - 1$ , shade below

Inequality 3: $y \geq 0$ → Line: $y = 0$ (x-axis), shade above

Inequality 4: $x \geq 0$ → Line: $x = 0$ (y-axis), shade right

Step 2: Find intersection points (vertices)

Vertex A: Intersection of $x = 0$ and $y = 0$ → (0, 0)

Vertex B: Intersection of $x = 0$ and $x + 2y = 8$ → (0, 4)

Vertex C: Intersection of $x + 2y = 8$ and $x - y = 1$

• Solving: $x = 1 + y$ , substitute: $(1 + y) + 2y = 8$ → $3y = 7$ → $y = \frac{7}{3}$

• $x = 1 + \frac{7}{3} = \frac{10}{3}$ → $\left(\frac{10}{3}, \frac{7}{3}\right)$

Vertex D: Intersection of $x - y = 1$ and $y = 0$ → (1, 0)

Step 3: Verify feasibility

Check each vertex in all inequalities to ensure they satisfy all constraints.

Vertices: A(0,0), B(0,4), C(10/3, 7/3), D(1,0). The feasible region is a quadrilateral.

Example 3: Linear Programming Optimization

A factory produces two products: Widget A and Widget B. Each Widget A requires 2 hours of labor and 1 hour of machine time, yielding a profit of $5. Each Widget B requires 1 hour of labor and 3 hours of machine time, yielding a profit of $7. Available: 12 hours labor, 15 hours machine time. Find the optimal production mix.

Step 1: Define variables and objective

Let $x$ = number of Widget A, $y$ = number of Widget B

Objective: Maximize $P = 5x + 7y$ (profit)

Step 2: Set up constraints

Labor constraint: $2x + y \leq 12$

Machine constraint: $x + 3y \leq 15$

Non-negativity: $x \geq 0, y \geq 0$

Step 3: Find vertices of feasible region

Vertex 1: (0, 0) - intersection of $x = 0$ and $y = 0$

Vertex 2: (6, 0) - intersection of $2x + y = 12$ and $y = 0$

Vertex 3: (0, 5) - intersection of $x = 0$ and $x + 3y = 15$

Vertex 4: Intersection of $2x + y = 12$ and $x + 3y = 15$

• From first equation: $y = 12 - 2x$

• Substitute: $x + 3(12 - 2x) = 15$ → $x + 36 - 6x = 15$ → $-5x = -21$ → $x = \frac{21}{5} = 4.2$

• $y = 12 - 2(4.2) = 3.6$ → (4.2, 3.6)

Step 4: Evaluate objective function at vertices

P(0,0) = 5(0) + 7(0) = $0

P(6,0) = 5(6) + 7(0) = $30

P(0,5) = 5(0) + 7(5) = $35

P(4.2,3.6) = 5(4.2) + 7(3.6) = 21 + 25.2 = $46.2

Optimal Solution: Produce 4.2 Widget A and 3.6 Widget B for maximum profit of $46.20. In practice, round to whole numbers: 4 Widget A and 4 Widget B for profit of $48.

Advanced Techniques & Optimization Strategies

Sensitivity Analysis in Linear Programming

Shadow Prices: The rate of change in the objective function value when a constraint is relaxed by one unit.

• Labor shadow price: How much profit increases per additional hour of labor

• Machine shadow price: How much profit increases per additional hour of machine time

• Binding constraints: Constraints that are satisfied as equalities at the optimal solution

• Non-binding constraints: Constraints with slack (not fully utilized)

Integer Programming Considerations

When solutions must be whole numbers (e.g., number of products):

• Rounding method: Round fractional solutions to nearest integers

• Feasibility check: Verify rounded solutions satisfy all constraints

• Alternative solutions: Check nearby integer points for better objective values

• Branch and bound: Advanced method for guaranteed optimal integer solutions

Multiple Objective Optimization

When optimizing multiple criteria simultaneously:

• Weighted sum method: Combine objectives with weights: $w_1 f_1 + w_2 f_2$

• Goal programming: Minimize deviations from target values

• Pareto optimality: Solutions where no objective can be improved without worsening another

• Trade-off analysis: Examine how changing one objective affects others

Common Pitfalls & Error Prevention

Pitfall 1: Sign Reversal Errors

Error: Forgetting to reverse inequality when multiplying/dividing by negative numbers.

Solution: Always check the sign of the coefficient when isolating variables.

Pitfall 2: Incorrect Test Point Selection

Error: Choosing test points that lie on boundary lines or outside the feasible region.

Solution: Always choose points clearly on one side of the boundary line.

Pitfall 3: Missing Vertices in Optimization

Error: Not finding all intersection points of constraint boundaries.

Solution: Systematically find intersections of all pairs of constraint lines.

Pitfall 4: Unbounded Feasible Regions

Error: Assuming all linear programming problems have bounded feasible regions.

Solution: Check if the feasible region extends infinitely in any direction.

Pitfall 5: Infeasible Systems

Error: Not recognizing when constraint systems have no feasible solutions.

Solution: Check if constraint boundaries create an empty feasible region.

Comprehensive Practice Problems

Problem 1: Complex Inequality

Solve: $\frac{3x-2}{4} - \frac{2x+1}{3} < \frac{x-1}{2}$

Show Solution

Step 1: Find LCD = 12 and multiply both sides

Step 2: $3(3x-2) - 4(2x+1) < 6(x-1)$

Step 3: $9x - 6 - 8x - 4 < 6x - 6$

Step 4: $x - 10 < 6x - 6$ → $-5x < 4$ → $x > -\frac{4}{5}$

Problem 2: Inequality System

Graph and find vertices: $\begin{cases} 2x + y \leq 10 \\ x + 2y \leq 8 \\ x \geq 0, y \geq 0 \end{cases}$

Show Solution

Vertices:

• (0,0): intersection of x=0 and y=0

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

• (0,4): intersection of x=0 and x+2y=8

• (4,2): intersection of 2x+y=10 and x+2y=8

Problem 3: Linear Programming

Maximize $z = 3x + 4y$ subject to: $\begin{cases} x + y \leq 6 \\ 2x + y \leq 8 \\ x \geq 0, y \geq 0 \end{cases}$

Show Solution

Vertices and z-values:

• (0,0): z = 0

• (4,0): z = 12

• (0,6): z = 24

• (2,4): z = 22

Maximum: z = 24 at (0,6)

Problem 4: Real-World Application

A farmer has 100 acres and can plant corn (yields $200/acre, needs 2 hours labor) or soybeans (yields $150/acre, needs 1 hour labor). Available: 120 hours labor. How should the farmer allocate land for maximum profit?

Show Solution

Variables: x = acres of corn, y = acres of soybeans

Objective: Maximize P = 200x + 150y

Constraints:

• $x + y \leq 100$ (land constraint)

• $2x + y \leq 120$ (labor constraint)

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

Optimal solution: (20, 80) with profit $34,000

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