Unit #2 Study Guide - Linear Inequalities

Formulating and Solving Single Linear Inequalities

• Formulating Budget Constraints from Scenario Descriptions:

  • Octavia's Dessert Budget:
  • Octavia plans to buy desserts for a party and has a maximum total budget of $\$55$.
  • Let $x$ represent the number of pies purchased, with each pie costing $\$4$.
  • Let $y$ represent the number of cakes purchased, with each cake costing $\$8$.
  • The total cost of purchasing $x$ pies and $y$ cakes is given by $4x + 8y$.
  • Because the total spend must be no more than $\$55$, the scenario is modeled by the linear inequality:       $4x + 8y \le 55$

• Algebraic Isolation of Variables and Inequality Direction Rules:

  • When solving a linear inequality for a given variable, dividing or multiplying both sides by a negative number reverses the direction of the inequality sign.
  • Luca's Algebraic Task:
  • Luca is given the linear inequality $-15P + 10D < 50$ in math class and must solve for $P$.
  • Step 1: Subtract $10D$ from both sides:       $-15P < 50 - 10D$
  • Step 2: Divide both sides by $-15$ and reverse the inequality sign from < to >:       $P > \frac{50 - 10D}{-15}$
  • Step 3: Simplify the right-hand side expression into individual terms:       $P > \frac{50}{-15} - \frac{10D}{-15}$$P > -\frac{10}{3} + \frac{2}{3}D$
  • Graph Boundary Line Characteristics:
  • Inequalities containing strict inequalities (< or >) use a dashed boundary line on a coordinate plane, indicating points on the line are not included in the solution set.
  • Inequalities containing non-strict inequalities (\le or \ge) use a solid boundary line, indicating points on the line are included in the solution set.
  • Because Luca's solved inequality $P > -\frac{10}{3} + \frac{2}{3}D$ utilizes a strict > operator, its graph requires a dashed boundary line.

Real-World Time and Production Constraints

• Modeling Crafting Constraints:
  • Marcy's Production Time Limit:
  • Marcy creates bracelets ($x$) and keychains ($y$) and has a maximum total of $12\,\text{hours}$ available to work on both items.
  • The situation is represented by the inequality:       $2x + 3y \le 12$
  • Because the relation contains \le, the graph of this inequality features a solid boundary line (no dotted lines).
  • Testing Candidate Production Combinations against $2x + 3y \le 12$:
  • Candidate A: $2\,\text{bracelets}$ ($x = 2$) and $4\,\text{keychains}$ ($y = 4$):       $2(2) + 3(4) = 4 + 12 = 16 \nleq 12$       This combination is invalid.
  • Candidate B: $3\,\text{bracelets}$ ($x = 3$) and $2\,\text{keychains}$ ($y = 2$):       $2(3) + 3(2) = 6 + 6 = 12 \le 12$       This combination is valid and true.
  • Candidate C: $4\,\text{bracelets}$ ($x = 4$) and $3\,\text{keychains}$ ($y = 3$):       $2(4) + 3(3) = 8 + 9 = 17 \nleq 12$       This combination is invalid.
  • Candidate D: $5\,\text{bracelets}$ ($x = 5$) and $2\,\text{keychains}$ ($y = 2$):       $2(5) + 3(2) = 10 + 6 = 16 \nleq 12$       This combination is invalid.

Multi-Constraint Systems of Linear Inequalities

• Owen's Orchard Scenario (Cost and Tree Count Constraints):

  • Owen is planting lemon trees ($x$) and orange trees ($y$) in his orchard.
  • Cost parameters: Each lemon tree costs $\$5$ and each orange tree costs $\$3$. Owen has a maximum budget of $\$30$.
  • Quantity parameters: Owen needs to ensure the total number of trees planted does not exceed $10\,\text{trees}$.
  • System of linear inequalities:     $5x + 3y \le 30$$x + y \le 10$
  • Evaluation of Coordinate Pairs $(x, y)$:
  • Option A $(1, 12)$: $1\,\text{lemon tree}$ and $12\,\text{orange trees}$.       $1 + 12 = 13 > 10$       Fails total tree constraint.
  • Option B $(2, 5)$: $2\,\text{lemon trees}$ and $5\,\text{orange trees}$.       $5(2) + 3(5) = 10 + 15 = 25 \le 30$$2 + 5 = 7 \le 10$       Valid and true.
  • Option C $(3, 4)$: $3\,\text{lemon trees}$ and $4\,\text{orange trees}$.       $5(3) + 3(4) = 15 + 12 = 27 \le 30$$3 + 4 = 7 \le 10$       Valid and true.
  • Option D $(5, 4)$: $5\,\text{lemon trees}$ and $4\,\text{orange trees}$.       $5(5) + 3(4) = 25 + 12 = 37 > 30$       Fails budget constraint.

• Adaya's Nutritional and Financial Constraints (Bananas and Strawberries):

  • Adaya has a budget of $\$40$ per month to spend on bananas ($b$ pounds) and strawberries ($s$ pounds).
  • Cost rates: Bananas cost $\$0.80\,\text{per pound}$ and strawberries cost $\$2.50\,\text{per pound}$.
  • Vitamin C content: One pound of bananas provides $100\,\text{mg}$ of Vitamin C, and one pound of strawberries provides $150\,\text{mg}$ of Vitamin C.
  • Nutritional requirement: Adaya needs to consume at least $1,500\,\text{mg}$ of Vitamin C from these fruits each month.
  • System of linear inequalities:     $0.80b + 2.50s \le 40$$100b + 150s \ge 1,500$

• School Event Snack Planning (Apples and Watermelons):

  • A school is purchasing apples ($a$) and watermelons ($w$) for an event.
  • Cost rates: Each apple costs $\$0.50$ and each watermelon costs $\$0.75$. The school has a total budget of $\$60$.
  • Quantity limit: The school wants to buy no more than $40\,\text{watermelons}$.
  • System of linear inequalities:     $0.50a + 0.75w \le 60$$w \le 40$

• Agricultural Resource Limits (Corn and Wheat):

  • A farmer plans crops of corn ($x$) and wheat ($y$) under a maximum water usage restriction of $13\,\text{water units}$.
  • Water usage inequality: $2x + 3y \le 13$
  • Verification of coordinate solutions $(x, y)$:
  • Point $(4, 1)$: $2(4) + 3(1) = 8 + 3 = 11 \le 13$ (Possible solution).
  • Point $(3, 2)$: $2(3) + 3(2) = 6 + 6 = 12 \le 13$ (Possible solution).
  • Point $(2, 3)$: $2(2) + 3(3) = 4 + 9 = 13 \le 13$ (Possible solution).

• Geometric Fencing Limits (Jeff's Garden):

  • Jeff is constructing a rectangular garden with fencing represented by $2x + 2y \le 30$, where $x$ is the length and $y$ is the width in feet.
  • Given a specified length $x = 10\,\text{feet}$:     $2(10) + 2y \le 30$$20 + 2y \le 30$$2y \le 10$$y \le 5\,\text{feet}$
  • The maximum allowable width satisfying the fencing constraint is $5\,\text{feet}$.

• Catering Order System (Sandwiches and Salads):

  • A catering company orders sandwiches ($x$) and salads ($y$).
  • Cost parameters: Sandwiches cost $\$4$ each, and salads cost $\$15$ each, with a total spending cap of $\$500$.
  • Weight parameters: Sandwiches weigh $1\,\text{pound}$ each, and salads weigh $0.35\,\text{pounds}$ each, with a maximum weight capacity of $30\,\text{pounds}$.
  • System A inequalities:     $4x + 15y \le 500$$x + 0.35y \le 30$

Optimization Under Budget Constraints

• Package Shipping Option Comparison:
  • Amal has a budget limit of $\$200$ and needs to find which shipping service allows shipping the GREATEST number of packages ($x$) for less than or equal to $\$200$
  • Quick-Ship Company:
  • Rate: $\$4\,\text{per package}$ plus a $\$50\,\text{service fee}$.
  • Inequality: $4x + 50 \le 200$
  • Solving for $x$:       $4x \le 150$$x \le 37.5$
  • Maximum complete packages: $37\,\text{packages}$.
  • Fast-Track Company:
  • Rate: $\$6\,\text{per package}$ plus a $\$20\,\text{service fee}$.
  • Inequality: $6x + 20 \le 200$
  • Solving for $x$:       $6x \le 180$$x \le 30$
  • Maximum complete packages: $30\,\text{packages}$.
  • Reliable Shop Company:
  • Rate: $\$7\,\text{per package}$ with no service fee.
  • Inequality: $7x \le 200$
  • Solving for $x$:       $x \le 28.57$$Maximum complete packages: 28\,\text{packages}$.
  • Optimal Decision: Quick-Ship Company allows the greatest package quantity of $37\,\text{packages}$ within the $\$200$ budget.

Graphical Systems and Coordinate Point Testing

• Point Verification Method for Graphical Systems:

  • To determine if an ordered pair $(x, y)$ is a valid solution to a system of inequalities, substitute the coordinates into all system equations.
  • System under test:     $y \le 2x + 1$$y > -x + 3$
  • Testing candidate point $(4, 3)$:
  • First inequality: $3 \le 2(4) + 1 \implies 3 \le 9$ (True).
  • Second inequality: $3 > -(4) + 3 \implies 3 > -1$ (True).
  • Because $(4, 3)$ satisfies both conditions, it is contained in the solution set region.

• Graph Shading Analysis ($y = -x - 1$):

  • An inequality shaded above the line $y = -x - 1$ represents points satisfying $y > -x - 1$ or $y \ge -x - 1$.
  • Ordered pairs situated in the shaded region above the boundary line possess $y$-values greater than the evaluated line expression $-x - 1$.