Applications of Inequalities

Inequalities are especially useful because many real-world situations do not have one exact answer. Instead, there may be a maximum, minimum, acceptable range, or other constraint.

For example, suppose you have $50 to spend on movie tickets that cost $8 each.

If n represents the number of tickets:

8n ≤ 50

The inequality is used instead of an equation because you do not have to spend exactly $50. You simply cannot spend more than $50.

Solving:

n ≤ 6.25

But you cannot buy 6.25 tickets, so the practical solution is:

n ≤ 6

This example demonstrates an important idea:

A mathematical solution must always be interpreted in the context of the problem.

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From Real Situations to Mathematics

Creating a mathematical model begins by identifying:

• the unknown quantity
• the variable
• important numerical information
• operations connecting the quantities
• restrictions or constraints
• whether the boundary is included

For example:

"A concert hall can hold no more than 800 people."

Let:

p = number of people

"No more than" means less than or equal to.

Therefore:

p ≤ 800

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Recognizing Inequality Language

Certain words and phrases often indicate inequalities.

The wording must be interpreted carefully because the presence or absence of equality changes the solution set.

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At Least and At Most

Two phrases appear especially often in applications.

At least means:

≥

For example:

"You need at least 60 points."

If p represents points:

p ≥ 60

At most means:

≤

For example:

"You can carry at most 15 kg."

If m represents mass:

m ≤ 15

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Constraints

A constraint is a condition that restricts possible values.

Examples include:

• maximum cost
• minimum age
• maximum mass
• minimum score
• available time
• storage capacity
• acceptable temperature
• required production
• available materials

Inequalities provide a mathematical way of expressing these restrictions.

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Defining Variables

Before writing an inequality, clearly define the variable.

Suppose:

"A bus can carry at most 60 passengers."

A good definition is:

p = number of passengers

Then:

p ≤ 60

Defining variables makes mathematical models easier to understand and communicate.

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Translating Words into Expressions

Suppose a gym charges:

• $20 registration fee
• $8 per visit

If v represents the number of visits, total cost is:

20 + 8v

If you can spend no more than $100:

20 + 8v ≤ 100

This inequality models the situation.

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Solving the Gym Problem

Start with:

20 + 8v ≤ 100

Subtract 20:

8v ≤ 80

Divide by 8:

v ≤ 10

Therefore, you can make at most:

10 visits

The solution must be interpreted in context because v represents a number of visits.

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A General Modelling Process

A reliable process for inequality applications is:

1. Read the situation carefully.
2. Identify the unknown quantity.
3. Define a variable.
4. Identify the constraint.
5. Translate the situation into an inequality.
6. Solve the inequality.
7. Check whether the mathematical solution makes sense in context.
8. State the answer using appropriate units and language.

The final step is important. A numerical inequality alone may not completely answer the real-world question.

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Budget Constraints

Budgets naturally produce inequalities because spending normally cannot exceed the available amount.

Suppose you have:

$120

A museum trip has:

• $30 transportation cost
• $15 admission per person

Let n represent the number of people.

Total cost:

30 + 15n

Budget constraint:

30 + 15n ≤ 120

Solve:

15n ≤ 90

n ≤ 6

Therefore:

At most 6 people can attend within the $120 budget.

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Why the Boundary Is Included

In the previous example:

30 + 15n ≤ 120

The symbol ≤ is used because spending exactly $120 is allowed.

If the problem instead stated:

"The total cost must be less than $120"

we would write:

30 + 15n < 120

Small changes in wording can change the inequality.

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Minimum Requirements

Inequalities also model minimum requirements.

Suppose a student needs at least:

70 points

to qualify for an award.

The student already has:

52 points

Let x represent additional points.

Then:

52 + x ≥ 70

Subtract 52:

x ≥ 18

The student needs at least 18 additional points.

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Maximum Capacity

Suppose an elevator has a maximum mass capacity of:

900 kg

There are already people inside with a combined mass of:

620 kg

Let m represent additional mass.

Then:

620 + m ≤ 900

Subtract 620:

m ≤ 280

No more than 280 kg of additional mass can be added.

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Transportation Example

A truck can carry no more than:

5000 kg

The truck already contains:

1400 kg

Each crate has mass:

120 kg

Let c represent the number of additional crates.

Then:

1400 + 120c ≤ 5000

Subtract 1400:

120c ≤ 3600

Divide by 120:

c ≤ 30

Therefore, the truck can carry at most:

30 additional crates

under this simplified model.

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Whole-Number Constraints

Suppose solving a problem gives:

n ≤ 7.8

If n represents:

• people
• cars
• books
• tickets

then n must normally be a whole number.

Therefore, the largest possible value is:

7

But if n represents:

• distance
• time
• mass
• temperature

then a decimal such as 7.8 may be perfectly reasonable.

Context determines how the mathematical solution should be interpreted.

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Rounding in Inequality Problems

Rounding requires special care.

Suppose:

n ≤ 12.6

and n represents boxes.

The maximum number of complete boxes is:

12

Rounding normally to 13 would violate the inequality.

Now suppose:

n ≥ 12.6

and n represents the number of workers required.

If workers must be whole people, at least:

13 workers

are required.

Do not automatically use ordinary rounding rules. Instead, choose the nearest whole value that still satisfies the constraint.

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Time Constraints

Suppose you have at most:

3 hours

to complete a task.

You have already worked for:

45 minutes

Let t represent additional working time in minutes.

Three hours is:

180 minutes

Therefore:

45 + t ≤ 180

Subtract 45:

t ≤ 135

You can work for at most another:

135 minutes

or:

2 hours 15 minutes

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Distance Constraints

Suppose a runner wants to travel at least:

10 km

and has already completed:

3.5 km

Let d represent additional distance.

Then:

3.5 + d ≥ 10

Subtract 3.5:

d ≥ 6.5

The runner must travel at least another:

6.5 km

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Speed Limits

Suppose a road has a maximum permitted speed of:

80 km/h

If v represents speed:

v ≤ 80

In context, speed also cannot be negative:

v ≥ 0

Together:

0 ≤ v ≤ 80

This is a compound inequality representing the realistic range.

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Temperature Ranges

A refrigerator should operate between:

1°C and 5°C

including both limits.

Let T represent temperature.

Then:

1 ≤ T ≤ 5

The acceptable solution region is the interval:

[1, 5]

A temperature of 3°C satisfies the condition.

A temperature of 7°C does not.

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Manufacturing Tolerances

Engineering and manufacturing frequently use inequalities to describe acceptable measurements.

Suppose a component must have a diameter between:

19.95 mm and 20.05 mm

inclusive.

Let d represent diameter.

Then:

19.95 ≤ d ≤ 20.05

Any diameter within this range is acceptable.

For example:

19.98 mm → acceptable

20.00 mm → acceptable

20.05 mm → acceptable

20.08 mm → not acceptable

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Evaluating Whether a Value Satisfies a Constraint

Suppose a package must satisfy:

m ≤ 25 kg

Does a 23.4 kg package satisfy the condition?

Substitute:

23.4 ≤ 25

True.

Therefore, it satisfies the constraint.

Does a 25.6 kg package satisfy it?

25.6 ≤ 25

False.

Therefore, it does not satisfy the constraint.

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Multiple Constraints

Real situations often involve more than one restriction.

Suppose a ride requires passengers to be:

• at least 120 cm tall
• no taller than 200 cm

Let h represent height.

Then:

120 ≤ h ≤ 200

A person must satisfy both conditions.

This is an "and" compound inequality.

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Separate Acceptable Regions

Sometimes either of two conditions can trigger an action.

Suppose a machine gives a warning if temperature is:

• below 5°C
• above 80°C

Then:

T < 5 or T > 80

The warning region consists of two separate intervals:

(−∞, 5) ∪ (80, ∞)

This is an "or" compound inequality.

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Solution Regions

A solution region represents all values satisfying the inequality or inequalities.

For:

x ≥ 4

the solution region contains every value from 4 upward.

For:

2 ≤ x ≤ 7

the solution region is between 2 and 7, including both endpoints.

For:

x < −3 or x > 5

there are two separate solution regions.

Number lines help make these regions visible.

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Interpreting a Solution Region

Suppose:

5 ≤ t ≤ 12

where t represents operating time in hours.

Mathematically, the solution includes every real value between 5 and 12.

In context, this means:

The system can operate for at least 5 hours and at most 12 hours.

A complete answer should explain what the solution means rather than simply reporting the inequality.

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Ticket-Sales Problem

A school wants to raise at least:

$1200

Tickets cost:

$15 each

Let n represent tickets sold.

Then:

15n ≥ 1200

Divide by 15:

n ≥ 80

Therefore:

At least 80 tickets must be sold.

Check:

15(80) = 1200

The boundary value satisfies the requirement.

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Fundraising with a Starting Amount

A club already has:

$350

and wants at least:

$1000

Each fundraiser ticket generates:

$13

Let n represent tickets sold.

Model:

350 + 13n ≥ 1000

Subtract 350:

13n ≥ 650

Divide by 13:

n ≥ 50

Therefore:

At least 50 tickets must be sold.

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Comparing Plans

Inequalities can determine when one option becomes better than another.

Suppose:

Plan A:

$30 + $4 per use

Plan B:

$60 + $2 per use

Let x represent the number of uses.

When is Plan A cheaper?

Write:

30 + 4x < 60 + 2x

Subtract 2x:

30 + 2x < 60

Subtract 30:

2x < 30

Divide by 2:

x < 15

Plan A is cheaper for fewer than 15 uses.

At exactly 15 uses, the plans cost the same.

For more than 15 uses, Plan B is cheaper.

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Checking the Comparison

At:

x = 10

Plan A:

30 + 4(10) = 70

Plan B:

60 + 2(10) = 80

Plan A is cheaper.

At:

x = 20

Plan A:

30 + 4(20) = 110

Plan B:

60 + 2(20) = 100

Plan B is cheaper.

This agrees with the inequality solution.

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Profit Constraints

Suppose a business has fixed costs of:

$500

and earns:

$20

for each item sold.

To earn more than $1000 after covering the fixed cost:

20n − 500 > 1000

Add 500:

20n > 1500

Divide by 20:

n > 75

Because n represents whole items:

At least 76 items must be sold.

Notice that the original condition says more than $1000, so exactly 75 items is not enough.

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Break-Even Problems

Suppose:

Cost:

C = 400 + 8x

Revenue:

R = 18x

For the business to avoid a loss:

R ≥ C

Therefore:

18x ≥ 400 + 8x

Subtract 8x:

10x ≥ 400

x ≥ 40

The business must sell at least:

40 units

to break even or make a profit.

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Resource Constraints

Suppose a factory has:

600 kg

of material.

Each product requires:

7.5 kg

Let n represent the number of products.

Then:

7.5n ≤ 600

Divide by 7.5:

n ≤ 80

The factory can produce at most:

80 products

from the available material.

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Constraints with More Than One Resource

Suppose a product requires:

• 3 kg of material A
• 2 kg of material B

The factory has:

• 120 kg of material A
• 100 kg of material B

If x products are made:

Material A:

3x ≤ 120

so:

x ≤ 40

Material B:

2x ≤ 100

so:

x ≤ 50

Both constraints must be satisfied.

Therefore:

x ≤ 40

Material A is the limiting resource.

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Identifying the Limiting Constraint

In the previous example:

Material A permits at most 40 products.

Material B permits at most 50 products.

Since both conditions must be satisfied, the more restrictive condition controls production.

Therefore:

maximum production = 40 products

This idea is important in:

• manufacturing
• budgeting
• scheduling
• transportation
• engineering
• business planning

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Geometry Applications

Suppose the perimeter of a rectangular garden cannot exceed:

50 m

Its length is:

8 m

and width is w.

Perimeter:

P = 2L + 2W

Therefore:

2(8) + 2w ≤ 50

16 + 2w ≤ 50

2w ≤ 34

w ≤ 17

The garden's width can be at most:

17 m

assuming positive dimensions.

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Minimum Area or Size Requirements

Suppose a rectangular display has width:

4 m

and must have an area of at least:

24 m²

Let L represent length.

Area:

A = L × W

Therefore:

4L ≥ 24

Divide by 4:

L ≥ 6

The display must be at least:

6 m long

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Science Application

Suppose an experiment requires water temperature to remain:

above 20°C but no higher than 30°C

Let T represent temperature.

"Above 20" means:

T > 20

"No higher than 30" means:

T ≤ 30

Together:

20 < T ≤ 30

Interval notation:

(20, 30]

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Engineering Safety Application

A machine component is designed to operate safely when rotational speed is:

• at least 1000 rpm
• no greater than 2500 rpm

Let ω represent rotational speed in rpm.

Then:

1000 ≤ ω ≤ 2500

Values outside this interval do not satisfy the stated operating constraint.

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Evaluating Several Possible Values

Suppose:

1000 ≤ ω ≤ 2500

Check:

ω = 900

Fails the lower limit.

ω = 1000

Acceptable.

ω = 1800

Acceptable.

ω = 2500

Acceptable.

ω = 2600

Fails the upper limit.

This is a practical way of evaluating whether measurements meet specifications.

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Sports Application

Suppose an athlete must complete a race in no more than:

12.5 seconds

Let t represent finishing time.

Then:

t ≤ 12.5

A time of 12.3 seconds qualifies.

A time of exactly 12.5 seconds qualifies.

A time of 12.6 seconds does not.

Notice that a smaller time is better in this situation.

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Data and Technology Application

Suppose a digital file must be smaller than:

25 MB

Let s represent file size.

Then:

s < 25

A 24.9 MB file satisfies the condition.

A 25 MB file does not because the requirement says smaller than, not "25 MB or smaller."

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Interpreting Negative Solutions

Negative values may be perfectly reasonable depending on the variable.

Suppose:

T ≤ −5

where T is temperature in °C.

Negative solutions make sense.

But if:

n ≤ −5

where n is the number of students in a classroom, the mathematical result would not make sense physically.

Always ask:

What does the variable represent?

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Domain Restrictions

The domain describes the values that make sense for a variable.

If x represents distance:

x ≥ 0

If n represents the number of people:

n = 0, 1, 2, 3, ...

If T represents temperature in Celsius:

negative values may be allowed.

If t represents elapsed time:

t ≥ 0

A mathematical inequality should be interpreted together with its realistic domain.

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When the Algebraic Answer Is Not the Final Answer

Suppose:

18n ≥ 100

Solve:

n ≥ 5.56

If n represents hours, a value such as 5.56 hours may be meaningful.

If n represents buses, however, you cannot use 5.56 buses.

You need:

at least 6 buses

The interpretation depends on the variable.

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Checking Whether a Solution Is Reasonable

After solving a real-world inequality, ask:

• Does the sign make sense?
• Can the variable be negative?
• Can the variable contain decimals?
• Must the answer be a whole number?
• Does the boundary satisfy the original condition?
• Are there additional practical restrictions?
• Does the answer have the correct units?

These questions help distinguish a correct algebraic answer from a complete practical solution.

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Worked Example: Event Budget

An event has:

$1500

available.

The venue costs:

$600

and food costs:

$18 per person

Let n represent the number of people.

The cost must not exceed the budget:

600 + 18n ≤ 1500

Subtract 600:

18n ≤ 900

Divide by 18:

n ≤ 50

Because n represents people:

A maximum of 50 people can attend within the budget.

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Worked Example: Minimum Production

A company has already produced:

125 units

and needs at least:

500 units

Each production cycle makes:

25 units

Let c represent additional cycles.

Model:

125 + 25c ≥ 500

Subtract 125:

25c ≥ 375

Divide by 25:

c ≥ 15

At least:

15 additional production cycles

are required.

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Worked Example: Delivery Weight

A delivery van can carry:

1800 kg

The driver and equipment have a combined mass of:

240 kg

Each package has mass:

32 kg

Let p represent packages.

Model:

240 + 32p ≤ 1800

Subtract 240:

32p ≤ 1560

Divide:

p ≤ 48.75

Since p must be a whole number:

p ≤ 48

The van can carry at most:

48 packages

Check:

240 + 32(48) = 1776 kg

This is acceptable.

For 49 packages:

240 + 32(49) = 1808 kg

This exceeds the limit.

Therefore, 48 is the correct practical maximum.

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Worked Example: Acceptable Measurement Range

A machine produces rods that should have lengths between:

99.7 cm and 100.3 cm

inclusive.

Let L represent length.

Model:

99.7 ≤ L ≤ 100.3

A rod measuring:

100.1 cm

satisfies both conditions.

A rod measuring:

100.5 cm

does not satisfy the upper condition.

The solution region represents every acceptable rod length.

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Worked Example: Two Pricing Plans

Company A charges:

$40 + $6 per hour

Company B charges:

$70 + $3 per hour

When is Company A cheaper?

Let h represent hours.

40 + 6h < 70 + 3h

Subtract 3h:

40 + 3h < 70

Subtract 40:

3h < 30

h < 10

Therefore, Company A is cheaper for fewer than 10 hours.

At exactly 10 hours, both companies charge the same amount.

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Representing Practical Solutions on Number Lines

Suppose a person's luggage must satisfy:

m ≤ 23

A number-line model would have:

• closed circle at 23
• shading left

But context adds:

m ≥ 0

Therefore, the realistic range is:

0 ≤ m ≤ 23

This demonstrates why mathematical graphs and real-world interpretations are not always identical unless all relevant constraints are included.

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Solution Regions with Two Variables

Some inequality applications involve two variables.

For example:

x + y ≤ 10

might represent two types of items sharing a capacity of 10 units.

Unlike a one-variable inequality, the solutions are ordered pairs such as:

(2, 5)

(4, 6)

(0, 10)

The solutions form a region on a coordinate plane rather than a section of a number line.

This idea leads to linear inequalities in two variables.

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Evaluating an Ordered Pair

Suppose:

2x + y ≤ 12

Does:

(x, y) = (4, 3)

satisfy the constraint?

Substitute:

2(4) + 3 ≤ 12

11 ≤ 12

True.

Therefore:

(4, 3) satisfies the inequality.

Now test:

(5, 4)

2(5) + 4 ≤ 12

14 ≤ 12

False.

Therefore, it does not satisfy the constraint.

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Inequalities in Decision-Making

Real-world decision-making often involves several constraints simultaneously.

A business might need to consider:

• budget
• employee hours
• materials
• storage
• production capacity
• minimum demand

An engineer might consider:

• mass
• temperature
• pressure
• speed
• stress
• safety factors

Inequalities allow these restrictions to be represented mathematically and evaluated systematically.

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A Reliable Strategy for Inequality Applications

When solving a real-world inequality problem:

1. Identify the unknown.
2. Define the variable clearly.
3. Identify important numerical information.
4. Identify words indicating minimums or maximums.
5. Write an expression representing the situation.
6. Choose the correct inequality symbol.
7. Solve the inequality.
8. Check whether the boundary should be included.
9. Apply realistic domain restrictions.
10. Check whether rounding is necessary.
11. Verify important values using the original situation.
12. Write a final sentence explaining the solution in context.

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Common Mistakes

Using an equation when the situation describes a limit

A maximum or minimum usually requires an inequality.

Confusing "at least" and "at most"

At least means ≥.

At most means ≤.

Ignoring fixed costs

For a cost such as "$20 plus $5 per item," the model is:

20 + 5x

not simply 5x.

Rounding in the wrong direction

If n ≤ 7.8 and n represents people, the maximum is 7, not 8.

Ignoring the variable's domain

Negative numbers or decimals may not make sense for some quantities.

Giving only the algebraic result

The final answer should explain what the solution means.

Ignoring equality

A phrase such as "no more than 50" includes exactly 50.

Using the wrong units

Always interpret the final value using the units from the problem.

Failing to check the boundary

Testing the limiting value is a useful way to confirm the model.

Ignoring additional constraints

For example, m ≤ 25 might mathematically include negative values, but mass also requires m ≥ 0.

Assuming every decimal answer can be used directly

Discrete quantities such as people, vehicles, and packages require whole-number interpretations.

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Key Terms

Inequality — mathematical statement comparing quantities using <, >, ≤, or ≥.

Constraint — condition limiting the possible values of a variable.

Mathematical model — mathematical representation of a real-world situation.

Variable — symbol representing an unknown or changing quantity.

Solution set — all values satisfying an inequality.

Solution region — graphical region containing values that satisfy one or more inequalities.

Boundary value — limiting value separating acceptable and unacceptable solutions.

Maximum — greatest permitted value, often represented using ≤.

Minimum — smallest permitted value, often represented using ≥.

Domain — set of values that are meaningful for a variable.

Discrete quantity — quantity that takes separate values, often whole numbers, such as people or objects.

Continuous quantity — quantity that can take values throughout an interval, such as time, mass, or distance.

Compound inequality — combination of multiple inequality conditions.

Tolerance — acceptable range of variation from a desired measurement.

Limiting constraint — restriction that places the strongest limit on possible solutions.

Feasible solution — solution satisfying all relevant constraints.

Verification — checking that a proposed solution satisfies the original conditions.

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Key Takeaways

• Inequalities are used when real-world quantities have limits, ranges, or restrictions.
• A variable should be clearly defined before constructing a mathematical model.
• "At least" usually corresponds to ≥.
• "At most" usually corresponds to ≤.
• "More than" corresponds to >.
• "Less than" corresponds to <.
• Constraints determine which values are acceptable.
• Budget problems often use inequalities because spending cannot exceed available money.
• Minimum-requirement problems commonly use ≥.
• Capacity problems commonly use ≤.
• Compound inequalities can represent acceptable ranges.
• Multiple constraints must all be considered when determining feasible solutions.
• The most restrictive condition may become the limiting constraint.
• Mathematical solutions must be interpreted according to the variable's domain.
• People, objects, vehicles, and similar quantities usually require whole-number solutions.
• Continuous quantities such as mass, time, and distance can often use decimal solutions.
• Rounding must preserve the inequality rather than simply following ordinary rounding rules.
• Substitution can determine whether a proposed value satisfies a constraint.
• Boundary values should be checked carefully.
• Realistic restrictions such as non-negative values may need to be added to the mathematical model.
• A complete answer should state what the solution means in the original context.

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Check Your Understanding

1. A concert venue can hold no more than 750 people. Define a variable and write an inequality representing the capacity.
2. A student needs at least 65 points to pass and currently has 48 points. Write and solve an inequality for the additional points needed.
3. You have $90. A delivery fee is $10, and each item costs $8. Write and solve an inequality for the maximum number of items you can purchase.
4. A truck has a maximum capacity of 4000 kg. It already carries 1150 kg, and each crate has a mass of 95 kg. Determine the maximum number of additional crates it can carry.
5. A machine operates safely between 15°C and 65°C, inclusive. Write a compound inequality and interval notation representing the safe operating range.
6. A warning activates when pressure is below 20 kPa or above 120 kPa. Write a compound inequality representing the warning conditions.
7. A company charges $25 plus $7 per hour. You have no more than $95. Determine the maximum number of hours you can purchase.
8. A club has already raised $425 and wants at least $1000. Each ticket sold adds $15. Determine the minimum number of additional tickets that must be sold.
9. A van can carry 1600 kg. The driver and equipment have a combined mass of 280 kg. Each package weighs 27 kg. Write and solve an inequality, then determine the greatest whole number of packages that can be carried.
10. A runner must finish a race in no more than 14.0 seconds. Determine whether times of 13.8 s, 14.0 s, and 14.2 s satisfy the condition. Explain why.
11. A manufactured component is acceptable when its diameter is between 29.95 mm and 30.05 mm, inclusive. Determine whether measurements of 29.93 mm, 29.95 mm, 30.02 mm, 30.05 mm, and 30.07 mm are acceptable.
12. Plan A costs $35 + $5x and Plan B costs $65 + $2x. Determine when Plan A is cheaper than Plan B and explain what happens at the boundary value.
13. A rectangular garden has a length of 12 m and a perimeter that cannot exceed 50 m. Write and solve an inequality to determine the maximum possible width.
14. A factory has 500 kg of material A and 360 kg of material B. Each product requires 8 kg of A and 6 kg of B. Write both constraints, determine the maximum whole number of products that can be manufactured, and identify the limiting resource.
15. An event has a $2400 budget. The venue costs $750, security costs $250, and food costs $22 per guest. Write and solve an inequality for the number of guests. Interpret the mathematical solution, apply the appropriate whole-number restriction, verify the maximum number of guests using the original cost model, and explain why one additional guest would violate the constraint.