Graphing Linear Inequalities and Applications

Graphing Linear Inequalities and Systems of Inequalities

Key idea: An inequality in the plane defines a region of points (x, y) that satisfy a relation between y and x. We often compare y to a linear function of x, i.e., a line of the form
$y \,\ge\, m x + b\quad\text{or}\quad y \,\le\, m x + b.$
The boundary is the line $y = m x + b.$ The region is the set of points on or above that line (for (\ge)) or on or below that line (for (\le)).
Example check (from the transcript): Consider the inequality
$y \ge -\frac{2}{3}x + 4.$
- The line is (y = -\frac{2}{3}x + 4).
- The point ((3,4)) lies above the line because plugging in gives $4 \ge -\frac{2}{3}(3) + 4 = 2,\$ so the inequality is satisfied.</li><li>The point ((0,0)) does not satisfy the inequality since (0 \ge -\frac{2}{3}(0) + 4 = 4) is false.</li><li>Intuition: The region that satisfies (y \ge -\frac{2}{3}x + 4) is all points on or above that blue line.</li></ul></li><li>How to graph a linear inequality (general method): 1) Graph the boundary line by treating the inequality as an equation: (y = m x + b). 2) Decide which side to shade: for (y \ge m x + b), shade above the line; for (y \le m x + b), shade below the line. 3) Note the book’s convention discussed in class: graph the line first, then “throw away” (shade out) the region you do not want, leaving the region that satisfies the inequality. This means:<ul><li>For "above the line" shading, you discard the points below the line.</li><li>For "below the line" shading, you discard the points above the line. 4) This throwing-away approach can be convenient when dealing with systems of inequalities: you can apply the shading one inequality at a time to find the feasible region. 5) The exam may not require graphing every inequality, but understanding the idea helps with applications and solving systems.</li></ul></li><li>Quick recap: graph the equality boundary, then shade the appropriate side. If you’re working with multiple inequalities (a system), the feasible region is the intersection of the individual half-planes (the points that survive all “throws” across inequalities).</li><li>Worked example: graph the inequality $ 4x - 2y \ge 12. $<ul><li>Put into standard slope-intercept form by isolating y:</li></ul>$ 4x - 2y \ge 12\quad\Rightarrow\quad -2y \ge -4x + 12\quad\Rightarrow\quad y \le 2x - 6. $<ul><li>Boundary line: (y = 2x - 6).</li><li>Since the inequality is "less than or equal to", we shade below the line.</li><li>Pick a point on the line to verify: when (x = 1), (y = 2(1) - 6 = -4). The point ((1,-4)) is on the boundary; shading below includes the region containing that point.</li><li>The book’s shading convention would require throwing away the points above the line; the remaining region is below the line where the inequality holds.</li></ul></li><li>Systems of inequalities and the feasible region:<ul><li>A system is a list of inequalities that a solution point must satisfy simultaneously.</li><li>A solution to the system is a point that satisfies every inequality in the system.</li><li>Process: apply inequalities one by one, discarding points that fail any inequality; the leftovers form the feasible set (the intersection of all half-planes).</li><li>Vertex points (corner points) of the feasible region are often important for applications (e.g., optimization problems).</li></ul></li><li>Application: optimization with a diet problem (rice and soybeans)<ul><li>Decision variables: let (x) be cups of rice per day and (y) be cups of soybeans per day.</li><li>Nutritional constraints (minimums):</li><li>Protein: (15 x + 22.5 y \ge 90) grams/day.</li><li>Calories: (810 x + 270 y \ge 1620) calories/day.</li><li>Vitamin B2: (per transcript) an inequality of the form (a x + b y \ge 1) milligram/day, since we need at least 1 mg/day of B2; (a, b) are the mg amounts per cup of rice and soybeans, respectively.</li><li>Non-negativity: (x \ge 0) and (y \ge 0).</li><li>Objective (cost minimization): cost per day is $ \text{Cost} = 0.21\,x + 0.14\,y $ dollars per day (21¢ per cup of rice and 14¢ per cup of soybeans).</li><li>Approach (conceptual):</li><li>First, find the feasible region for the system (where all inequalities hold).</li><li>The cheapest diet tends to lie on the boundary (at or near a vertex) of the feasible region because increasing beyond the minimums costs more.</li><li>If you can hit a vertex that satisfies multiple minimums simultaneously, you save money on both constraints.</li><li>The transcript notes that the cheapest feasible diet is found at one of the four vertices of the feasible region, with the minimum cost computed by evaluating the cost function at each vertex; the example yields a minimum around 66¢ per day (exact coordinates of the optimal vertex are not given in the transcript and are stated as to be determined in the next section).</li><li>Practical takeaway: in real-world optimization, you typically identify candidate vertices by solving intersection points of boundary lines (the lines corresponding to the equalities at the constraints), then evaluate the objective function at these points to find the minimum cost.</li></ul></li><li>Intersections of lines (finding a point that lies on both boundary lines)<ul><li>Concept: For two lines to intersect, their boundary equations must be satisfied simultaneously.</li><li>Example 1 (simple intersection): find the intersection of</li></ul>$ y = 2x - 3, $ $ y = x + 1.
 - Solve by setting the right-hand sides equal: (2x - 3 = x + 1).
 - Solve: subtract (x) from both sides to get (x - 3 = 1) -> (x = 4).
 - Substitute into either equation to find (y): in (y = x + 1), (y = 4 + 1 = 5).
 - Intersection point: ((4, 5)).
 - Example 2 (convert to standard form, then solve):
 - Given lines not in standard form, convert first to standard form. Suppose the lines are
 y = -\frac{1}{2}x + 3, $ $ y = -\frac{5}{2}x + 9. $</li><li>Set them equal: (-\tfrac{1}{2}x + 3 = -\tfrac{5}{2}x + 9).</li><li>Solve for (x). Move terms to collect x: add (\tfrac{5}{2}x) to both sides to get (2x + 3 = 9) (after clearing denominators, the transcript shows a sequence that leads to (x = 3)).</li><li>Then substitute back to find (y): e.g., in (y = -\tfrac{1}{2}x + 3), with (x = 3) we get (y = -\tfrac{1}{2}(3) + 3 = -\tfrac{3}{2} + 3 = \tfrac{3}{2}.</li><li>Intersection point: ((3, \tfrac{3}{2})).</li><li>Takeaway: Intersections of boundary lines give potential candidate points for vertices of the feasible region in a system of inequalities; evaluating the objective function at these vertices finds the optimum in linear programming-style problems.</li></ul></li><li>Connections to broader concepts<ul><li>Vertex/minimum principle: For linear objectives over a convex polygonal feasible region, an optimal solution lies at a vertex (or along an edge if the objective is parallel to that edge).</li><li>Real-world relevance: Systems of inequalities model constraints in production, diet planning, resource allocation, and cost minimization; the feasible region represents all admissible plans, and optimization finds the best among them.</li><li>Ethical/practical notes: In optimization, constraints reflect non-negativity and minimum requirements (e.g., minimum calories, protein, vitamins); overshooting minimums increases cost or resource use, so solutions near the boundary are often preferred.</li></ul></li><li>Important clarifications about shading conventions (classroom discussion)<ul><li>There are at least two valid, consistent ways to handle shading:</li><li>Shade the region that satisfies the inequality directly (above for (\ge), below for (\le)).</li><li>Follow the book’s convention of graphing the boundary line and then crossing out the region that does not satisfy the inequality. The points that remain are the solutions.</li><li>Either approach is acceptable if you are consistent and clear about which region you are identifying as the solution set. The instructor notes that the exam will focus on applying the techniques rather than mastering the shading convention alone.</li></ul></li><li>Quick recap of notation and forms<ul><li>Boundary line form:$ y = m x + b $where m is the slope and b is the y-intercept.</li><li>Standard form:$ a x + b y = c $with possible rearrangements to solve for y or x.</li><li>Inequality forms:</li><li>$ y \ge m x + b $(region above or on the line)</li><li>$ y \le m x + b$$ (region below or on the line)
 - Feasible region: the set of all (x, y) that satisfy every inequality in a system.
 - Vertex/corner point: intersection point of two boundary lines that forms a corner of the feasible region.
- Practical study tips inspired by the transcript
 - When dealing with systems of inequalities, practice graphing each boundary line and then determining the feasible region by removing the non-satisfying region one inequality at a time.
 - For optimization problems, identify the feasible region, locate its vertices (intersections of boundary lines), and evaluate the objective function at those vertices to find the minimum (or maximum).
 - Remember to include non-negativity constraints when modeling real-world quantities (e.g., quantities of food, materials, or products cannot be negative).
 - In assessments, you may be asked to convert inequalities to a standard form, graph the boundary, and discuss which side is included in the solution.
- Concluding notes from the lecture
 - The instructor emphasizes understanding the process and its applications, rather than only the mechanics of shading for a single inequality.
 - Friday’s planned discussion will connect these ideas to supply and demand contexts and show additional examples.
 - Recitations are encouraged to reinforce the concepts.