Inequalities and Linear Programming: Your Ultimate Study Guide

Hey everyone! Welcome to the study notes for Inequalities and Linear Programming. Don't let the long name scare you! This chapter is all about making smart decisions using maths. We'll start with simple comparisons (like 'is x bigger than 5?'), move on to graphing regions, and finally, learn how to find the BEST possible solution to real-world problems, like how a company can make the most profit. It's a super useful skill, so let's get started!


Part 1: Inequalities in One Unknown (Working on a Number Line)

Prerequisite Quick-Check: Simple Linear Inequalities

Remember these symbols? They are the heart of this topic!

  • > : Greater than
  • < : Less than
  • ≥ : Greater than or equal to
  • ≤ : Less than or equal to

The most important rule to remember is the Golden Rule of Inequalities:
When you multiply or divide both sides of an inequality by a NEGATIVE number, you MUST flip the inequality sign.

Example: Solve $$-2x > 6$$
To solve for x, we divide by -2. Because we're dividing by a negative number, we flip the sign!
$$x < -3$$

8.1: Compound Linear Inequalities

This is when we have more than one inequality to deal with at the same time. There are two types: "AND" and "OR".

The "AND" type (Intersection)

This means the solution must satisfy ALL the conditions at the same time. Think of it as looking for the overlap on the number line.

Example: Solve $$x > 3$$ and $$x \le 7$$
We need numbers that are bigger than 3 AND less than or equal to 7. Looking at a number line, the overlap is between 3 and 7.
The solution is $$3 < x \le 7$$.

The "OR" type (Union)

This means the solution can satisfy EITHER ONE of the conditions (or both). You just combine all the possible solution areas.

Example: Solve $$x < -1$$ or $$x \ge 4$$
We need numbers that are either smaller than -1 OR are greater than or equal to 4. There's no overlap, so we just state both conditions as the final answer.
The solution is $$x < -1 \text{ or } x \ge 4$$.

A Real-World Application: Triangle Inequality

Did you know that for any triangle with side lengths a, b, and c, the sum of any two sides must be greater than the third side? This gives us three compound inequalities: $$a+b > c$$, $$a+c > b$$, and $$b+c > a$$. You can use this to find the possible range for a missing side length!

Common Mistake Alert: Don't mix up "and" and "or"! A solution like "$$x > 5 \text{ and } x < 2$$" is impossible because there's no number that is both bigger than 5 and smaller than 2. The overlap is empty!

8.2 & 8.3: Tackling Quadratic Inequalities

Now let's look at inequalities with an $$x^2$$ term, like $$x^2 - 5x + 4 > 0$$. Don't worry, there's a clear process for this. The key is to first find where the expression equals zero.

Quick Recap: Quadratic Graphs (Parabolas)

The graph of $$y = ax^2 + bx + c$$ is a parabola.

  • Solving $$ax^2 + bx + c > 0$$ is the same as asking: "Where is the parabola ABOVE the x-axis?"
  • Solving $$ax^2 + bx + c < 0$$ is the same as asking: "Where is the parabola BELOW the x-axis?"
Method 1: The Graphical Method (LO 8.2)

This method is great for visual learners.

Example: Solve $$x^2 - x - 6 > 0$$

  1. Find the roots: First, solve the equation $$x^2 - x - 6 = 0$$.
    $$(x-3)(x+2) = 0$$
    So, the roots are $$x = 3$$ and $$x = -2$$. These are the x-intercepts.
  2. Sketch the graph: Since the coefficient of $$x^2$$ is positive (it's 1), the parabola opens upwards. Sketch a U-shaped curve that crosses the x-axis at -2 and 3.
  3. Find the solution: The question asks for $$x^2 - x - 6 > 0$$, which means "where is the graph above the x-axis?". Looking at your sketch, the graph is above the x-axis when x is to the left of -2, or to the right of 3.
  4. Write the answer: $$x < -2 \text{ or } x > 3$$
Method 2: The Algebraic Method (LO 8.3)

This method uses a number line and is very reliable.

Example: Solve $$x^2 - x - 6 > 0$$ again.

  1. Find the critical values: These are just the roots we found before. Set $$x^2 - x - 6 = 0$$ to get $$x = 3$$ and $$x = -2$$.
  2. Divide the number line: Use these critical values to split the number line into three regions: $$x < -2$$, $$-2 < x < 3$$, and $$x > 3$$.
  3. Test each region: Pick a simple test number from each region and plug it into the original inequality ($$x^2 - x - 6 > 0$$) to see if it's true or false.
    • Region 1: $$x < -2$$. Let's test $$x=-3$$.
      $$(-3)^2 - (-3) - 6 = 9 + 3 - 6 = 6$$. Is $$6 > 0$$? Yes (True).
    • Region 2: $$-2 < x < 3$$. Let's test $$x=0$$.
      $$(0)^2 - (0) - 6 = -6$$. Is $$-6 > 0$$? No (False).
    • Region 3: $$x > 3$$. Let's test $$x=4$$.
      $$(4)^2 - (4) - 6 = 16 - 4 - 6 = 6$$. Is $$6 > 0$$? Yes (True).
  4. Write the answer: The inequality is true for the first and third regions. So the solution is $$x < -2 \text{ or } x > 3$$.
Key Takeaways for Part 1
  • Always flip the sign when multiplying/dividing by a negative number.
  • "AND" means overlap/intersection. "OR" means combine/union.
  • For quadratic inequalities, find the roots first. Then either sketch the graph or test regions on a number line.

Part 2: Into the 2D Plane & Linear Programming

8.4: Graphing Linear Inequalities in Two Unknowns

We're now working on the x-y coordinate plane. An equation like $$y = 2x + 1$$ is a straight line. An inequality like $$y > 2x + 1$$ represents an entire region on one side of that line.

Step-by-Step Guide to Graphing

Example: Represent the inequality $$2x + y \le 4$$ graphically.

  1. Draw the Boundary Line: First, pretend it's an equation and draw the line $$2x + y = 4$$. A good way is to find the intercepts:
    • When $$x=0$$, $$y=4$$. Point is (0, 4).
    • When $$y=0$$, $$2x=4$$, so $$x=2$$. Point is (2, 0).
    Plot these points and draw the line connecting them.
  2. Solid or Dotted Line? This is important!
    • Use a solid line for ≤ and ≥ (because the line itself is included in the solution).
    • Use a dotted line for < and > (because the line is a boundary but not part of the solution).
    For our example $$2x + y \le 4$$, we use a solid line.
  3. Shade the Correct Region: Pick a simple test point that is NOT on the line. The point (0, 0) is usually the best choice.
    Substitute (0, 0) into the original inequality: $$2(0) + (0) \le 4$$ which gives $$0 \le 4$$.
    Is this statement true? Yes!
    Memory Aid: If the test point is TRUE, you shade TOWARDS that point. If it were false, you would shade the other side.
    Since (0, 0) is true, we shade the region that includes the origin.

8.5: Systems of Linear Inequalities

This sounds complex, but it just means graphing several inequalities on the same set of axes. The solution is the area where ALL the shaded regions overlap. This overlapping area has a special name: the feasible region.

Example: Find the feasible region for the system:

  • $$x \ge 0$$
  • $$y \ge 0$$
  • $$x + y \le 5$$

You would graph all three inequalities.

  1. $$x \ge 0$$ is a solid vertical line at the y-axis, shaded to the right.
  2. $$y \ge 0$$ is a solid horizontal line at the x-axis, shaded upwards.
  3. $$x + y \le 5$$ is a solid line passing through (5,0) and (0,5), shaded towards the origin.

The feasible region is the triangle in the first quadrant formed by the points (0,0), (5,0), and (0,5).

Note: The syllabus only requires you to solve these systems graphically. No need for algebraic methods here!

8.6: Solving Linear Programming Problems

This is the final boss of the chapter, and it's where everything comes together. Linear programming is a method to find the maximum or minimum value of something, given a set of limits or rules.

Real-World Analogy: The Baker

Imagine you're a baker making cakes (x) and cookies (y).

  • Constraints: You have limited flour and sugar. These are your inequalities (e.g., $$2x + 1y \le 10$$ kg of flour). You also can't make negative cakes, so $$x \ge 0, y \ge 0$$.
  • Objective Function: You want to maximize your profit. If you make $30 profit per cake and $10 per cookie, your objective is to maximize Profit $$P = 30x + 10y$$.
The Magic Key: The Vertex Theorem

Here's the most important concept: The maximum or minimum value of the objective function will ALWAYS occur at one of the vertices (corners) of the feasible region.

Step-by-Step Guide to Solving
  1. Define Variables: State clearly what x and y represent. (e.g., Let x be the number of cakes...)
  2. Write Constraints: Translate the problem's limits into a system of linear inequalities.
  3. Write the Objective Function: Write the equation for the quantity you want to maximize or minimize (like Profit P or Cost C).
  4. Graph and Find the Feasible Region: Graph your constraints and shade the overlapping area.
  5. Find the Vertices: Identify the coordinates of all the corners of your feasible region. You might need to solve simultaneous equations for where two lines intersect.
  6. Test the Vertices: Substitute the coordinates of each vertex into your objective function.
  7. State the Conclusion: Identify which vertex gives the maximum or minimum value and answer the question in a full sentence. (e.g., "The maximum profit is $150 when 5 cakes and 0 cookies are made.")
Did you know?

Linear programming was developed during World War II to solve logistical problems, like how to deploy troops and supplies most efficiently. Today, it's used everywhere, from airline scheduling to financial planning and manufacturing!

Key Takeaways for Part 2
  • For two-variable inequalities, draw the boundary line (solid/dotted) and test a point to find the shading region.
  • The feasible region is the overlap of all inequality solutions.
  • In Linear Programming, the best (max/min) solution is always at a vertex of the feasible region.
  • Follow the 7-step process to solve linear programming problems systematically.

And that's a wrap! The best way to master these concepts is through practice. Work through problems, draw your graphs carefully, and soon you'll be solving these like a pro. Good luck!