Linear Inequalities & Programming

Linear programming uses inequalities to describe limits, then finds the best possible outcome inside those limits. The shaded region is not decoration; it represents all values that satisfy every condition at the same time.

For CSEC, these problems usually involve resources such as time, money, materials, or production limits. Define the variables, write the inequalities from the wording, shade the feasible region, and test corner points when maximising or minimising. Each step explains the model.

Inequalities in One Variable

An inequality shows a relationship where something is greater than, less than, greater than or equal to, or less than or equal to something else.

Symbols:

$>$ : greater than
$<$ : less than
$\geq$ : greater than or equal to
$\leq$ : less than or equal to

Solving One-Variable Linear Inequalities

Solve like equations, BUT flip the inequality sign when multiplying or dividing by a negative number.

Example

Solve $3x - 2 > 7$ :

Step 1: Add 2 to both sides $3x > 9$

Step 2: Divide by 3 (positive, so don't flip) $x > 3$

Answer: All numbers greater than 3. Set notation: $\{x : x > 3\}$

Number line representation:

x > 3

Example

Solve $-2x + 5 \leq 1$ :

Step 1: Subtract 5 from both sides $-2x \leq -4$

Step 2: Divide by -2 (negative, so FLIP the sign!) $x \geq 2$

Answer: All numbers greater than or equal to 2. Set notation: $\{x : x \geq 2\}$

Number line representation:

x ≥ 2

Inequalities in Two Variables

An inequality like $y > 2x + 1$ represents a region (not just a line).

Graphing Two-Variable Inequalities

Step 1: Graph the boundary line $y = 2x + 1$

Use dashed line if inequality is $<$ or $>$ (not included)
Use solid line if inequality is $\leq$ or $\geq$ (included)

Step 2: Shade the appropriate region

For $y > 2x + 1$ : shade ABOVE the line
For $y < 2x + 1$ : shade BELOW the line
Test a point if unsure: pick $(0, 0)$ and check if it satisfies the inequality

Example

Graph $y \leq -x + 4$ :

Step 1: Graph boundary line $y = -x + 4$

Y-intercept: $(0, 4)$
Slope: $-1$
Use SOLID line (inequality includes $=$ )

Step 2: Shade region

Test point $(0, 0)$ : $0 \leq -(0) + 4 = 4$ ✓ TRUE
Since $(0, 0)$ makes it true and it's below the line, shade BELOW

Region y ≤ -x + 4 (shaded below the line)

What Is Linear Programming?

Linear programming is a method to find the best solution (maximum or minimum) when you have:

An objective function (what you want to maximize or minimize)
Constraints (limitations or restrictions, usually linear inequalities)

Real-world applications:

Maximize profit subject to resource limits
Minimize cost subject to production requirements
Optimize resource allocation

The Feasible Region

The feasible region is the area that satisfies all constraints simultaneously.

To find it:

Graph each constraint as a linear inequality
Shade the region that satisfies ALL constraints
The overlapping region is the feasible region

Example

Graph the constraints: $x + y \leq 8$ $2x + y \leq 12$ $x \geq 0, y \geq 0$

Step 1: Graph each line

Line 1: $x + y = 8$ (intercepts at $(0,8)$ and $(8,0)$ )
Line 2: $2x + y = 12$ (intercepts at $(0,12)$ and $(6,0)$ )
Lines 3 & 4: axes (since $x \geq 0$ and $y \geq 0$ )

Step 2: Shade regions below/right of lines for $\leq$ and $\geq$

Step 3: Overlapping region = feasible region (usually a polygon)

Feasible region (intersection of x+y≤6, x≥0, y≥0, y≤4)

Optimization

The optimal solution occurs at a vertex (corner point) of the feasible region.

Method:

Find all corner points of the feasible region
Evaluate the objective function at each corner point
Choose the point that gives the maximum or minimum value

Example

Maximize profit: $P = 3x + 2y$ subject to: $x + y \leq 8$ $2x + y \leq 12$ $x \geq 0, y \geq 0$

Step 1: Find corner points of feasible region

$(0, 0)$ (origin)
$(6, 0)$ (where line 2 meets x-axis)
$(4, 4)$ (where lines 1 and 2 intersect)
$(0, 8)$ (where line 1 meets y-axis)

Step 2: Evaluate $P = 3x + 2y$ at each corner

At $(0, 0)$ : $P = 3(0) + 2(0) = 0$
At $(6, 0)$ : $P = 3(6) + 2(0) = 18$
At $(4, 4)$ : $P = 3(4) + 2(4) = 12 + 8 = 20$ ← MAXIMUM
At $(0, 8)$ : $P = 3(0) + 2(8) = 16$

Optimal solution: $(4, 4)$ gives maximum profit of 20

This means: produce 4 of product $x$ and 4 of product $y$ for maximum profit.

Inequalities in One Variable

An inequality shows a relationship where something is greater than, less than, greater than or equal to, or less than or equal to something else.

Symbols:

$>$ : greater than
$<$ : less than
$\geq$ : greater than or equal to
$\leq$ : less than or equal to

Solving One-Variable Linear Inequalities

Solve like equations, BUT flip the inequality sign when multiplying or dividing by a negative number.

Example

Solve $3x - 2 > 7$ :

Step 1: Add 2 to both sides $3x > 9$

Step 2: Divide by 3 (positive, so don't flip) $x > 3$

Answer: All numbers greater than 3. Set notation: $\{x : x > 3\}$

Number line representation:

x > 3

Example

Solve $-2x + 5 \leq 1$ :

Step 1: Subtract 5 from both sides $-2x \leq -4$

Step 2: Divide by -2 (negative, so FLIP the sign!) $x \geq 2$

Answer: All numbers greater than or equal to 2. Set notation: $\{x : x \geq 2\}$

Number line representation:

x ≥ 2

Inequalities in Two Variables

An inequality like $y > 2x + 1$ represents a region (not just a line).

Graphing Two-Variable Inequalities

Step 1: Graph the boundary line $y = 2x + 1$

Use dashed line if inequality is $<$ or $>$ (not included)
Use solid line if inequality is $\leq$ or $\geq$ (included)

Step 2: Shade the appropriate region

For $y > 2x + 1$ : shade ABOVE the line
For $y < 2x + 1$ : shade BELOW the line
Test a point if unsure: pick $(0, 0)$ and check if it satisfies the inequality

Example

Graph $y \leq -x + 4$ :

Step 1: Graph boundary line $y = -x + 4$

Y-intercept: $(0, 4)$
Slope: $-1$
Use SOLID line (inequality includes $=$ )

Step 2: Shade region

Test point $(0, 0)$ : $0 \leq -(0) + 4 = 4$ ✓ TRUE
Since $(0, 0)$ makes it true and it's below the line, shade BELOW

Region y ≤ -x + 4 (shaded below the line)

What Is Linear Programming?

Linear programming is a method to find the best solution (maximum or minimum) when you have:

An objective function (what you want to maximize or minimize)
Constraints (limitations or restrictions, usually linear inequalities)

Real-world applications:

Maximize profit subject to resource limits
Minimize cost subject to production requirements
Optimize resource allocation

The Feasible Region

The feasible region is the area that satisfies all constraints simultaneously.

To find it:

Graph each constraint as a linear inequality
Shade the region that satisfies ALL constraints
The overlapping region is the feasible region

Example

Graph the constraints: $x + y \leq 8$ $2x + y \leq 12$ $x \geq 0, y \geq 0$

Step 1: Graph each line

Line 1: $x + y = 8$ (intercepts at $(0,8)$ and $(8,0)$ )
Line 2: $2x + y = 12$ (intercepts at $(0,12)$ and $(6,0)$ )
Lines 3 & 4: axes (since $x \geq 0$ and $y \geq 0$ )

Step 2: Shade regions below/right of lines for $\leq$ and $\geq$

Step 3: Overlapping region = feasible region (usually a polygon)

Feasible region (intersection of x+y≤6, x≥0, y≥0, y≤4)

Optimization

The optimal solution occurs at a vertex (corner point) of the feasible region.

Method:

Find all corner points of the feasible region
Evaluate the objective function at each corner point
Choose the point that gives the maximum or minimum value

Example

Maximize profit: $P = 3x + 2y$ subject to: $x + y \leq 8$ $2x + y \leq 12$ $x \geq 0, y \geq 0$

Step 1: Find corner points of feasible region

$(0, 0)$ (origin)
$(6, 0)$ (where line 2 meets x-axis)
$(4, 4)$ (where lines 1 and 2 intersect)
$(0, 8)$ (where line 1 meets y-axis)

Step 2: Evaluate $P = 3x + 2y$ at each corner

At $(0, 0)$ : $P = 3(0) + 2(0) = 0$
At $(6, 0)$ : $P = 3(6) + 2(0) = 18$
At $(4, 4)$ : $P = 3(4) + 2(4) = 12 + 8 = 20$ ← MAXIMUM
At $(0, 8)$ : $P = 3(0) + 2(8) = 16$

Optimal solution: $(4, 4)$ gives maximum profit of 20

This means: produce 4 of product $x$ and 4 of product $y$ for maximum profit.