Inequalities and Linear Programming: Applied Mathematics for Business

Inequalities in One Variable

Introduction to Inequalities

Inequalities are mathematical statements that compare two expressions using symbols such as <, >, ≤, or ≥. Unlike equations, which have a finite set of solutions, inequalities often have infinitely many solutions forming intervals on the real number line.

• Definition: An inequality is an expression of the form A < B, A > B, A ≤ B, or A ≥ B, where A and B are real numbers or algebraic expressions.

• Example: $4x + 7 < 19$

• Solution Set: The set of all values of the variable that make the inequality true.

• Interval Notation: Solutions are often expressed as intervals, e.g., $x > 3$ is $(3, +\infty)$.

Solving Inequalities

To solve inequalities, isolate the variable using algebraic operations, similar to solving equations. However, special care must be taken when multiplying or dividing by negative numbers.

• Equivalent Inequalities: Two inequalities are equivalent if they have the same solution set.

• Key Properties:

Property	Example
If $a < b$, then $a + c < b + c$	If $x - 3 < 0$, then $(x - 3) + 3 < 0 + 3$, so $x < 3$
If $a < b$ and $c$ is positive, then $ac < bc$	If $x < 4$, then $2x < 2(4)$, so $2x < 8$
If $a < b$ and $c$ is negative, then $ac > bc$	If $-x/5 < 6$, then $-5(-x/5) > -5(6)$, so $x > -30$
Transitive Property: If $a < b$ and $b < c$, then $a < c$	If $a < x$ and $x < 2$, then $a < 2$

Examples of Solving Inequalities

• Example 1: Solve $3x - 9x < 4$ $3x - 9x < 4$ $-6x < 4$ $x > -\frac{2}{3}$ (after dividing by -6 and reversing the inequality sign) Solution Set: $(-\frac{2}{3}, +\infty)$

• Example 2: Solve $1 < x < 5$ Solution Set: $(1, 5)$

• Example 3: Solve $5t + 8 \leq 7(1 + t)$ $5t + 8 \leq 7 + 7t$ $8 - 7 \leq 7t - 5t$ $-2t \leq -1$ $t \geq \frac{1}{2}$ Solution Set: $[\frac{1}{2}, +\infty)$

Absolute Value and Inequalities

Theorem: Absolute Value Inequalities

Absolute value inequalities can be rewritten as compound inequalities.

• If $a > 0$, then $|u| < a$ if and only if $-a < u < a$

• If $a > 0$, then $|u| > a$ if and only if $u < -a$ or $u > a$

Examples

• Example 1: Solve $|x| < 1$ $-1 < x < 1$ Solution Set: $(-1, 1)$

• Example 2: Solve $|x| \geq 1$ $x \leq -1$ or $x \geq 1$ Solution Set: $(-\infty, -1] \cup [1, +\infty)$

• Example 3: Solve $|2x - 1| \leq 3$ $-3 \leq 2x - 1 \leq 3$ $-3 + 1 \leq 2x \leq 3 + 1$ $-2 \leq 2x \leq 4$ $-1 \leq x \leq 2$ Solution Set: $[-1, 2]$

• Example 4: Solve $|x - 3| > 1$ $x - 3 < -1$ or $x - 3 > 1$ $x < 2$ or $x > 4$ Solution Set: $(-\infty, 2) \cup (4, +\infty)$

Graphing Linear Inequalities in Two Variables

Introduction

Linear inequalities in two variables have solution sets that are regions in the coordinate plane. The boundary of the region is a straight line, and the solution set is one side of the line (including or excluding the line depending on the inequality).

• General Form: $Ax + By > C$, $Ax + By < C$, $Ax + By \geq C$, $Ax + By \leq C$

• Boundary Line: The line $Ax + By = C$ separates the plane into two regions.

• Dashed vs. Solid Line: Use a dashed line for strict inequalities ($<$ or $>$), and a solid line for inclusive inequalities ($\leq$ or $\geq$).

• Test Point: Use a test point (often (0,0)) to determine which side of the boundary line is the solution region.

Example

• Graph $y > x + 4$: Draw the line $y = x + 4$ as a dashed line. Shade the region above the line, since $y$ is greater than $x + 4$.

Applications of Linear Inequalities in Two Variables

Modeling Real-World Problems

Linear inequalities can be used to model constraints in business and economics, such as budgets, production limits, and resource allocation.

• Example: Hilaria works two jobs, earning $10/hour$ in food service and $15/hour$ tutoring. She needs at least $240/week$. Let $x$ be hours in food service, $y$ be hours tutoring. Model: $10x + 15y \geq 240$

• Finding Solutions: Any $(x, y)$ pair that satisfies the inequality is a feasible work schedule.

Systems of Linear Inequalities

Solving by Graphing

A system of linear inequalities consists of two or more inequalities. The solution is the region where all inequalities are satisfied simultaneously.

• Steps:

  1. Graph each inequality on the same coordinate plane.

  2. Shade the solution region for each inequality.

  3. The overlapping (intersection) region is the solution to the system.

  4. Test points can be used to verify solutions.

• Example: Solve the system by graphing: $y \geq 2x - 1$ $y < x + 1$ Solution: The region above the solid line $y = 2x - 1$ and below the dashed line $y = x + 1$ is the solution.

Applications of Systems of Inequalities

Business Example

Systems of inequalities can model multiple constraints in business scenarios, such as inventory, cost, and production requirements.

• Example: Christy wants to display at least 25 photos at a booth, with small photos costing $4$and large photos $10$, and a budget of $200$.

• Let: $x$ = number of small photos, $y$ = number of large photos

• System:

  • $x + y \geq 25$ (at least 25 photos)

  • $4x + 10y \leq 200$ (budget constraint)

• Feasible Solutions: Any $(x, y)$ pair in the intersection region of the two inequalities.

Linear Programming Application in Business

Introduction to Linear Programming

Linear programming is a mathematical technique used to optimize an objective function subject to linear constraints. It is widely used in business for maximizing profit or minimizing cost.

• Objective Function: The function to be maximized or minimized, e.g., profit or cost.

• Decision Variables: Variables representing quantities to be determined, e.g., number of products to produce.

• Constraints: Linear inequalities representing limitations, such as resources or budgets.

Business Example: Toy Production

• Company produces Space Ray and Zapper dolls.

• Decision Variables: $X_1$ = Space Rays produced per week, $X_2$ = Zappers produced per week

• Objective Function: Maximize weekly profit: $8X_1 + 5X_2$

• Constraints:

  • Plastic: $a_1X_1 + a_2X_2 \leq 1000$ (where $a_1$, $a_2$ are pounds of plastic per unit)

  • Labor: $b_1X_1 + b_2X_2 \leq 2400$ (where $b_1$, $b_2$ are minutes of labor per unit)

  • Non-negativity: $X_1 \geq 0$, $X_2 \geq 0$

General Linear Programming Model

• Standard Form:

Maximize or Minimize: $Z = c_1x_1 + c_2x_2 + \cdots + c_nx_n$ Subject to: $a_{11}x_1 + a_{12}x_2 + \cdots + a_{1n}x_n \leq b_1$ $a_{21}x_1 + a_{22}x_2 + \cdots + a_{2n}x_n \leq b_2$ $\vdots$ $x_j \geq 0$ for all $j$

Optimal Solution

• If a linear programming problem has an optimal solution, it occurs at an extreme point (vertex) of the feasible region.

Example: Minimizing Cost in Mineral Extraction

Ore Type	Mineral A (lbs/ton)	Mineral B (lbs/ton)	Cost per ton
Ore 1	40	15	$140
Ore 2	25	30	$155

• Constraints:

  • At least 4000 lbs of A: $40x_1 + 25x_2 \geq 4000$

  • At least 2000 lbs of B: $15x_1 + 30x_2 \geq 2000$

• Objective Function: Minimize cost: $140x_1 + 155x_2$

• Decision Variables: $x_1$ = tons of Ore 1, $x_2$ = tons of Ore 2

Additional info: These topics are foundational for microeconomics, especially in areas such as resource allocation, production optimization, and cost minimization.