Study Guide · Inequalities & Optimization

Inequalities & Optimization

From a single inequality to finding the best outcome within constraints

6 sections 19 worked examples 7 checkpoints Intuition-first

Solving Linear Inequalities

Almost like equations — with one critical difference

Solving inequalities follows almost the same steps as equations. You can add, subtract, and multiply/divide by positive numbers freely. The one special rule:

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Flip the inequality sign when multiplying or dividing by a negative number.
−2x > 6 → divide by −2 → x < −3 (sign reversed!)
This is the single most common error. Always ask: did I divide by a negative?

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Why does the sign flip? On the number line: 3 > 1, but −3 < −1. Multiplying by −1 reverses the order. The inequality sign must follow that reversal.

Types of Solutions

Simple: x < 6 — a single ray on the number line
Compound (AND): 9 ≤ x ≤ 13 — a segment; both conditions must hold simultaneously
Compound (OR): x < −3 OR x > 5 — two rays; either condition suffices
Notation: x ∈ (−∞, 6) means x < 6; x ∈ [−3, 5] means −3 ≤ x ≤ 5

Basic

Solving with sign flip

Solve: 4 − 3x > 13

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Step 1 — Subtract 4: −3x > 9

Step 2 — Divide by −3 (flip sign!): x < −3
Dividing by negative −3 reverses the inequality

x < −3

Basic

Multi-step inequality

Solve: 2(x + 3) ≤ 5x − 6

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Step 1 — Expand: 2x + 6 ≤ 5x − 6

Step 2 — Collect x terms: 6 + 6 ≤ 5x − 2x → 12 ≤ 3x

Step 3 — Divide by +3 (no flip): 4 ≤ x, i.e. x ≥ 4

x ≥ 4

Medium

Double (compound) inequality

Solve: −1 ≤ 3x − 4 < 11

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Step 1 — Add 4 to all three parts: −1 + 4 ≤ 3x − 4 + 4 < 11 + 4 → 3 ≤ 3x < 15

Step 2 — Divide all by 3 (positive, no flip): 1 ≤ x < 5

1 ≤ x < 5 (includes 1, excludes 5)

Medium

Inequality with fractions

Solve: (2x − 1)/3 > (x + 2)/2

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Step 1 — Multiply both sides by 6 (LCM, positive): 2(2x−1) > 3(x+2)

Step 2 — Expand: 4x − 2 > 3x + 6

Step 3 — Solve: x > 8 → x > 8

x > 8

Checkpoint 1 — Solving Inequalities

Solve: −4x + 7 ≥ 19
Solve: 3(2x − 1) < 2(x + 5)
Solve: −3 < 2x + 1 ≤ 9

−4x ≥ 12 → x ≤ −3 (flip!) → x ≤ −3
6x − 3 < 2x + 10 → 4x < 13 → x < 13/4 = 3.25
Subtract 1: −4 < 2x ≤ 8; divide by 2: −2 < x ≤ 4

Graphing 1-Variable Inequalities

On the number line

A one-variable inequality describes a set of numbers on the number line. There are two things to mark: the endpoint and the direction.

Number Line Rules

Closed dot ● at the endpoint — use for ≤ or ≥ (endpoint included)
Open dot ○ at the endpoint — use for < or > (endpoint excluded)
Shade the ray in the direction that satisfies the inequality
For AND compound (e.g. 1 ≤ x < 5): shade the segment between endpoints
For OR compound (e.g. x < −3 OR x > 2): shade both rays outward

Basic

Simple number line graph

Describe how to graph x ≤ 3 on a number line

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Endpoint: x = 3. The inequality is ≤ (includes 3) → closed dot ● at 3

Direction: All x ≤ 3 lie to the left of 3 → shade left

Closed dot at 3, ray shaded to the left

Basic

Compound AND inequality graph

Describe how to graph −2 < x ≤ 4

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Left endpoint x = −2: Strict < → open dot ○ at −2

Right endpoint x = 4: ≤ includes 4 → closed dot ● at 4

Shade: The segment between −2 and 4

Open dot at −2, closed dot at 4, segment shaded between them

Medium

Solve and graph

Solve then describe the graph: 2|x − 1| ≥ 6

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Step 1 — Divide by 2: |x − 1| ≥ 3

Step 2 — Absolute value rule (≥): x − 1 ≥ 3 OR x − 1 ≤ −3

Step 3 — Solve each: x ≥ 4 OR x ≤ −2

Graph: Closed dot at −2, ray left; closed dot at 4, ray right. The middle segment is NOT shaded.

x ≤ −2 OR x ≥ 4 (two outward rays, closed dots)

Checkpoint 2 — Number Line Graphs

Describe the graph of x > −1
Describe the graph of −5 ≤ x < 0
Solve and describe graph: |2x + 3| < 7

Open dot at −1, ray shaded right
Closed dot at −5, open dot at 0, segment between them shaded
|2x+3| < 7 → −7 < 2x+3 < 7 → −10 < 2x < 4 → −5 < x < 2. Open dots at −5 and 2, segment shaded.

Graphing 2-Variable Inequalities

Half-planes on the Cartesian plane

A 2-variable inequality like y ≥ 2x + 1 describes a half-plane — one side of the boundary line (plus the line itself if the inequality is ≤ or ≥).

Solid line for ≤/≥, dashed for </>. Test (0,0) to decide which side to shade.

5-Step Graphing Method

Step 1: Rearrange to isolate y if helpful
Step 2: Graph the boundary line (y = mx + b)
Step 3: Solid line for ≤ or ≥; dashed line for < or >
Step 4: Test the point (0, 0) in the original inequality
Step 5: True → shade origin side; False → shade opposite side

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(0, 0) shortcut: Almost always use (0, 0) as the test point — it's the easiest to substitute. Only use a different point if the boundary line passes through (0, 0).

Basic

Graph a half-plane

Describe how to graph: 2x − y < 4

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Step 1 — Rearrange: −y < 4 − 2x → y > 2x − 4 (flip sign, dividing by −1)

Step 2 — Boundary line: y = 2x − 4. Slope = 2, y-intercept = −4

Step 3 — Line type: Strict < → dashed line

Step 4 — Test (0, 0): 2(0) − 0 < 4 → 0 < 4 ✓ True

Step 5 — Shade: Origin (0,0) satisfies it → shade the side containing the origin (above the line)

Dashed line y = 2x − 4, shade above (origin side)

Basic

Graph y ≥ 2x + 1

Describe how to graph: y ≥ 2x + 1

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Boundary line: y = 2x + 1. Slope = 2, y-intercept = 1. The ≥ means solid line.

Test (0, 0): 0 ≥ 2(0) + 1 → 0 ≥ 1. FALSE.

Shade: Origin fails → shade the side NOT containing the origin (above and to the left of the line).

Solid line y = 2x + 1, shade the half-plane NOT containing the origin

Medium

Find the boundary line passes through origin — alternate test point

Describe how to graph: y ≤ 3x

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Boundary line: y = 3x. Passes through the origin — cannot use (0, 0) as test point!

Alternative test point: Use (1, 0). Test: 0 ≤ 3(1) → 0 ≤ 3 ✓ True.

Shade: (1, 0) satisfies it → shade the side containing (1, 0), which is below the line y = 3x.

Line type: ≤ → solid line.

Solid line y = 3x, shade below (the side containing (1, 0))

Checkpoint 3 — Graphing Half-Planes

Describe the graph of 3x − 2y > 6
Which side of y = x + 2 is shaded for y < x + 2?
For x + y ≤ 5, test the point (2, 2). Is it in the shaded region?

Rearrange: y < (3x−6)/2. Dashed line y = 1.5x−3. Test (0,0): 0 < −3? No → shade below the line (away from origin).
Test (0,0): 0 < 0+2 ✓ → shade the side containing the origin (below the line)
2+2=4 ≤ 5 ✓ → yes, (2,2) is in the shaded region

Systems of Linear Inequalities

The feasible region

A system of inequalities defines a feasible region — the set of all points satisfying ALL constraints simultaneously. This is usually a polygon (or sometimes unbounded).

Finding the Feasible Region

Graph each inequality separately on the same axes
The feasible region is the area where all shadings overlap
Find all vertices (corner points) by solving pairs of boundary lines as simultaneous equations
Include the non-negativity constraints x ≥ 0, y ≥ 0 when specified (restricts to the first quadrant)
If a boundary is dashed, its vertices are open points — not included in the region

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Open vertices: If a vertex lies on a dashed boundary line, it is NOT part of the feasible region. Mark it with an open circle.

Medium

Find vertices of a feasible region

Find all vertices of the region: x ≥ 0, y ≥ 0, x + y ≤ 6, 2x + y ≤ 8

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Identify boundary lines: x = 0, y = 0, x + y = 6, 2x + y = 8

Vertex A — intersection of x = 0 and y = 0: (0, 0)

Vertex B — x-axis (y=0) with 2x+y=8: 2x = 8 → (4, 0). Check: x+y = 4 ≤ 6 ✓

Vertex C — intersection of x+y=6 and 2x+y=8:
Subtract: (2x+y) − (x+y) = 8−6 → x = 2; then y = 6−2 = 4 → (2, 4)

Vertex D — y-axis (x=0) with x+y=6: y = 6 → (0, 6). Check: 2(0)+6 = 6 ≤ 8 ✓

Vertices: (0,0), (4,0), (2,4), (0,6)

Medium

Check if a point is in the feasible region

Constraints: x ≥ 0, y ≥ 0, x + 2y ≤ 10, 3x + y ≤ 12. Is (3, 3) feasible?

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Check each constraint at (3, 3):

x ≥ 0: 3 ≥ 0 ✓

y ≥ 0: 3 ≥ 0 ✓

x + 2y ≤ 10: 3 + 6 = 9 ≤ 10 ✓

3x + y ≤ 12: 9 + 3 = 12 ≤ 12 ✓

Yes, (3, 3) satisfies all constraints — it is in the feasible region

Checkpoint 4 — Feasible Regions

Find the vertices of: x ≥ 0, y ≥ 0, x + y ≤ 5, x + 2y ≤ 8
Is (4, 1) feasible for: x ≥ 0, y ≥ 0, 2x + y ≤ 9, x + 3y ≤ 7?

Intersections: (0,0); y=0,x+y=5 → (5,0); x=0,x+2y=8 → (0,4); x+y=5 and x+2y=8: subtract→y=3,x=2 → (2,3). Vertices: (0,0), (5,0), (2,3), (0,4)
Check (4,1): 2(4)+1=9≤9✓; 4+3(1)=7≤7✓; x,y≥0✓ → Yes, (4,1) is feasible

Linear Programming

Finding the best outcome within constraints

Linear programming is the method of finding the maximum or minimum of a linear objective function subject to a set of linear constraint inequalities. This is the mathematical basis for most real-world resource optimization.

The optimal value always occurs at a vertex — never in the interior of the feasible region.

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The Fundamental Theorem of Linear Programming: If the feasible region is bounded (a closed polygon), the objective function takes its maximum and minimum values at vertices of the polygon. You never need to check interior points.

The 7-Step Method

Step 1: Define variables — what do x and y represent?
Step 2: Write the objective function (maximise or minimise what?)
Step 3: Write all constraints (including x ≥ 0, y ≥ 0 if needed)
Step 4: Graph constraints and shade the feasible region
Step 5: Find all vertices (corner points)
Step 6: Evaluate the objective function at every vertex
Step 7: State the maximum (or minimum) and interpret in context

Medium

Evaluate objective function at vertices

Feasible region has vertices (0,0), (5,0), (3,4), (0,6). Maximise P = 3x + 5y.

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Evaluate P at each vertex:
P(0, 0) = 3(0) + 5(0) = 0
P(5, 0) = 3(5) + 5(0) = 15
P(3, 4) = 3(3) + 5(4) = 9 + 20 = 29
P(0, 6) = 3(0) + 5(6) = 30

Maximum P = 30 at vertex (0, 6)

Medium

Both maximise and minimise

Vertices: (1,2), (4,1), (5,4), (2,5). Objective: C = 2x + 3y. Find max and min.

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Evaluate C at each vertex:
C(1, 2) = 2 + 6 = 8
C(4, 1) = 8 + 3 = 11
C(5, 4) = 10 + 12 = 22
C(2, 5) = 4 + 15 = 19

Maximum C = 22 at (5, 4); Minimum C = 8 at (1, 2)

Checkpoint 5 — Linear Programming Setup

Vertices: (0,0), (6,0), (4,3), (0,5). Maximise Z = 4x + 2y. Which vertex gives the maximum?
Write the objective function and constraints for: A factory makes chairs ($30 profit) and tables ($50 profit). At most 20 items total; chairs need 2 hours, tables 5 hours, 70 hours available.

Z(0,0)=0; Z(6,0)=24; Z(4,3)=16+6=22; Z(0,5)=10 → Maximum Z=24 at (6,0)
Let x=chairs, y=tables. Objective: P = 30x + 50y. Constraints: x+y ≤ 20; 2x+5y ≤ 70; x ≥ 0; y ≥ 0

Finding the Optimal Value

Full optimization problems from start to finish

Now we apply all 7 steps to complete problems. The key challenge is translating a word problem into variables, an objective function, and constraints — then working through the geometry.

Hard

Bakery optimization

A bakery makes cakes ($15 profit) and pies ($20 profit). At most 10 items; cakes 2 h, pies 3 h, 24 h available. Maximize profit.

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Step 1 — Variables: x = cakes, y = pies

Step 2 — Objective: Maximise P = 15x + 20y

Step 3 — Constraints:
x + y ≤ 10 (total items)
2x + 3y ≤ 24 (hours)
x ≥ 0, y ≥ 0

Step 5 — Vertices:
A = (0, 0)
B = (10, 0) [y=0 in x+y=10]
C = (0, 8) [x=0 in 2x+3y=24]
D: solve x+y=10 and 2x+3y=24 → substitute y=10−x: 2x+3(10−x)=24 → −x=−6 → x=6, y=4 → D=(6,4)

Step 6 — Evaluate P:
P(0,0) = 0 P(10,0) = 150 P(0,8) = 160 P(6,4) = 90+80 = 170 ← Maximum

Maximum profit $170 — make 6 cakes and 4 pies

Hard

Minimize cost

A diet requires at least 8 units of protein and 6 units of iron. Food A costs $2 and provides 4 protein, 1 iron. Food B costs $3 and provides 2 protein, 3 iron. Minimize cost.

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Variables: x = servings of A, y = servings of B

Objective: Minimise C = 2x + 3y

Constraints:
4x + 2y ≥ 8 (protein, simplifies to 2x + y ≥ 4)
x + 3y ≥ 6 (iron)
x ≥ 0, y ≥ 0

Vertices (of the feasible region above both lines):
A: x = 0 in 2x+y=4 → (0, 4)
B: y = 0 in x+3y=6 → (6, 0)
C: intersection of 2x+y=4 and x+3y=6:
From first: y=4−2x; substitute: x+3(4−2x)=6 → x+12−6x=6 → −5x=−6 → x=6/5, y=4−12/5=8/5 → C=(1.2, 1.6)

Evaluate C:
C(0,4) = 0+12 = 12 C(6,0) = 12+0 = 12 C(1.2,1.6) = 2.4+4.8 = 7.2 ← Minimum

Minimum cost $7.20 — use 1.2 servings of A and 1.6 servings of B

Hard

Production planning

A company makes product X ($40 profit) and product Y ($60 profit). Each X needs 3 hours machine time and 2 hours labour. Each Y needs 2 hours machine time and 4 hours labour. Machine: 120 h, Labour: 160 h. Maximise profit.

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Variables & Objective: x = units X, y = units Y; Maximise P = 40x + 60y

Constraints:
3x + 2y ≤ 120 (machine hours)
2x + 4y ≤ 160 → x + 2y ≤ 80 (labour)
x ≥ 0, y ≥ 0

Vertices:
A = (0, 0)
B: y=0 in 3x+2y=120 → (40, 0)
C: x=0 in x+2y=80 → (0, 40)
D: solve 3x+2y=120 and x+2y=80: subtract → 2x=40 → x=20; y=(80−20)/2=30 → (20, 30)

Evaluate P:
P(0,0)=0 P(40,0)=1600 P(0,40)=2400 P(20,30)=800+1800=2600 ← Maximum

Maximum profit $2600 — produce 20 units of X and 30 units of Y

Checkpoint 6 — Full Optimization

For the bakery problem: if the constraint changes to at most 8 items total (hours unchanged), find the new optimal solution.
A farmer plants wheat (profit $100/ha) and corn (profit $150/ha). Max 20 ha total. Wheat needs 2 h/ha work, corn needs 4 h/ha work, 60 h labour available. Wheat needs 10 L water, corn 30 L, 400 L available. Write the objective function and all constraints.

New constraints: x+y≤8, 2x+3y≤24. Vertices: (0,0),(8,0),(0,8),(solve: x+y=8,2x+3y=24→x=0,y=8 same point). Evaluate P=15x+20y: P(0,0)=0; P(8,0)=120; P(0,8)=160 ← Maximum. Make all 8 pies.
x=wheat ha, y=corn ha. Objective: P=100x+150y. Constraints: x+y≤20; 2x+4y≤60 (→x+2y≤30); 10x+30y≤400 (→x+3y≤40); x≥0; y≥0

Checkpoint 7 — Mixed Review

Solve: −3(x − 2) > 2(x + 1) + 7
Describe the feasible region for: x ≥ 0, y ≥ 0, y ≤ 4, x + y ≤ 6. List all vertices.
For objective function Z = 5x + 8y and vertices (0,0), (4,0), (3,3), (0,4): find the maximum and where it occurs.

−3x+6 > 2x+2+7 → −3x+6 > 2x+9 → −5x > 3 → x < −3/5 (flip!) → x < −0.6
Constraints form a quadrilateral in Q1. Vertices: (0,0), (6,0), (2,4), (0,4)
— (6,0): x+y=6 meets y=0
— (2,4): x+y=6 meets y=4 → x=2
Z(0,0)=0; Z(4,0)=20; Z(3,3)=15+24=39; Z(0,4)=32 → Maximum Z=39 at (3,3)

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Exam tip: In any linear programming problem, always verify your vertices satisfy ALL constraints. A common mistake is to include an intersection point that lies outside the feasible region. Substitute each vertex into every inequality before evaluating the objective function.