Modeling with Inequalities

Many real-world situations involve limits, requirements, ranges, and constraints rather than one exact value. In these situations, an equation may not be enough.

For example, suppose a school bus can carry at most 48 students.

If n represents the number of students, the situation can be modeled as:

n ≤ 48

This inequality does not describe one possible value. It describes an entire solution set:

0, 1, 2, 3, ..., 48

In a realistic context, n must also be a whole number because we cannot have part of a student.

Inequality models help us answer questions such as:

• How much can we spend without exceeding a budget?
• How many items must be sold to reach a target?
• What measurements are within an acceptable range?
• How much weight can a vehicle safely carry?
• How many hours can someone work?
• Which possible solutions satisfy all of the restrictions?

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Equations and Inequalities in Models

An equation represents two quantities that are exactly equal.

For example:

C = 5n + 20

might describe a cost consisting of a $20 starting fee plus $5 per item.

An inequality describes a range of possible values.

If the maximum budget is $100:

5n + 20 ≤ 100

The inequality now asks:

Which values of n keep the total cost at or below $100?

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What Is a Constraint?

A constraint is a condition that limits the possible solutions to a problem.

Common constraints include:

• maximum cost
• minimum score
• available time
• limited materials
• weight capacity
• minimum age
• maximum occupancy
• acceptable temperature range
• required production level

Constraints are often represented using inequalities.

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Translating Words into Inequality Symbols

Understanding the language of constraints is essential.

The phrases at least and at most are especially important.

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At Least

At least 12

means:

12 is allowed, and values greater than 12 are allowed.

Therefore:

x ≥ 12

For example:

"A student must score at least 70%."

Let s represent the score.

s ≥ 70

A score of exactly 70% satisfies the requirement.

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At Most

At most 12

means:

12 is allowed, and values below 12 are allowed.

Therefore:

x ≤ 12

For example:

"A bag may weigh at most 20 kg."

Let w represent its mass.

w ≤ 20

Exactly 20 kg is acceptable.

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Strict Inequalities

Sometimes the boundary itself is not allowed.

"Temperature must be below 30°C."

T < 30

"Applicants must be older than 16."

a > 16

The symbols < and > indicate that the boundary value is excluded.

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Defining Variables

Before creating an inequality, define the unknown quantity.

Suppose:

"A concert ticket costs $15, and you have $90."

Define:

n = number of tickets purchased

The total cost is:

15n

The available money creates the constraint:

15n ≤ 90

Clearly defining the variable makes the model easier to interpret.

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Creating an Inequality Model

A useful modeling process is:

1. identify the unknown quantity
2. define a variable
3. identify the constraint
4. decide which inequality symbol represents the constraint
5. write the mathematical expression
6. solve the inequality
7. interpret the solution in context
8. check whether additional restrictions apply

The final interpretation is just as important as the algebra.

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Budget Constraints

Suppose you have:

$80

A movie ticket costs:

$12

Let:

n = number of tickets

The constraint is:

12n ≤ 80

Solve:

n ≤ 80/12

n ≤ 6.67 approximately

Mathematically:

n ≤ 6.67

But tickets are discrete objects.

Therefore, the realistic conclusion is:

You can buy at most 6 tickets.

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Fixed and Variable Costs

Many cost problems include both:

• a fixed cost
• a variable cost

Suppose a club charges:

• $25 registration fee
• $8 per visit

You have:

$100

Let:

v = number of visits

Total cost:

25 + 8v

Budget constraint:

25 + 8v ≤ 100

Solve:

8v ≤ 75

v ≤ 9.375

Because visits are whole numbers:

v ≤ 9

Therefore, you can afford at most 9 visits.

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Minimum Requirements

Inequalities also model situations where a quantity must reach or exceed a minimum.

Suppose a student needs at least:

240 points

to earn an award.

The student already has:

165 points

and earns:

15 points per activity

Let:

a = number of additional activities

Model:

165 + 15a ≥ 240

Subtract 165:

15a ≥ 75

Divide by 15:

a ≥ 5

The student must complete:

at least 5 additional activities.

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Why Rounding Depends on the Situation

Suppose a calculation gives:

n ≥ 4.2

and n represents the number of workers required.

Using 4 workers would not satisfy the requirement.

Therefore:

at least 5 workers are required

We round upward because the constraint is a minimum.

Now suppose:

n ≤ 4.8

and n represents the number of tickets that can be purchased.

The greatest possible whole number is:

4

We cannot round to 5 because 5 would violate the constraint.

When interpreting inequalities, do not simply use ordinary rounding rules. Choose a value that still satisfies the inequality.

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Capacity Constraints

Suppose an elevator can safely carry:

900 kg

Six passengers and their luggage already have a combined mass of:

510 kg

Additional boxes each have a mass of:

35 kg

Let:

b = number of boxes

The model is:

510 + 35b ≤ 900

Subtract 510:

35b ≤ 390

Divide by 35:

b ≤ 11.14 approximately

Because b must be a whole number:

b ≤ 11

The elevator can carry at most 11 additional boxes under this model.

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Checking the Conclusion

Check 11 boxes:

510 + 35(11)

= 510 + 385

= 895 kg

This is acceptable.

Check 12 boxes:

510 + 35(12)

= 510 + 420

= 930 kg

This exceeds the limit.

Therefore, 11 is the greatest feasible number.

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Feasible Solutions

A feasible solution is a value that satisfies all of the constraints in a problem.

For:

510 + 35b ≤ 900

b = 8 is feasible.

Check:

510 + 35(8) = 790

790 ≤ 900

True.

But:

b = 15

is not feasible.

510 + 35(15) = 1035

1035 ≤ 900

False.

A solution can be mathematically possible in general but infeasible under the stated constraints.

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The Feasible Solution Set

The feasible solution set contains all values that satisfy the constraints.

Suppose:

x ≥ 4

and x can be any real number.

The feasible set includes:

4, 4.1, 5, 7.8, 100, ...

In interval notation:

[4, ∞)

If x instead represents a number of people, then only appropriate whole-number values are meaningful:

4, 5, 6, 7, ...

Context determines which values in the mathematical solution set are actually feasible.

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Graphing Inequality Solutions

Solutions to one-variable inequalities can be represented on a number line.

For:

x > 3

use:

• an open circle at 3
• shading to the right

For:

x ≥ 3

use:

• a closed circle at 3
• shading to the right

For:

x < 3

use:

• an open circle at 3
• shading to the left

For:

x ≤ 3

use:

• a closed circle at 3
• shading to the left

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Open and Closed Circles

Use an open circle when the boundary is not included.

Examples:

x < 5

x > −2

Use a closed circle when the boundary is included.

Examples:

x ≤ 5

x ≥ −2

A useful question is:

"Is the boundary value allowed?"

If yes, use a closed circle.

If no, use an open circle.

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Direction on a Number Line

For:

x > 4

the solutions are greater than 4, so the graph extends right.

For:

x < 4

the solutions are less than 4, so the graph extends left.

Do not rely only on memorized arrow rules. Read the inequality:

x > 4

means:

x is greater than 4.

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Compound Constraints

Many real-world situations have both a minimum and maximum.

Suppose a machine operates safely between:

40°C and 70°C, inclusive

The model is:

40 ≤ T ≤ 70

The feasible solution set contains all temperatures from 40°C through 70°C.

Both endpoints are included.

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Graphing a Range

For:

40 ≤ T ≤ 70

the number line has:

• closed circle at 40
• closed circle at 70
• shading between them

In interval notation:

[40, 70]

This represents a bounded feasible region.

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When Boundaries Are Excluded

Suppose a chemical process requires:

pH greater than 5 but less than 9

Then:

5 < pH < 9

The endpoints are excluded.

Interval notation:

(5, 9)

The number line uses open circles at both 5 and 9.

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Multiple Constraints

Suppose a competition requires participants to be:

• at least 12 years old
• no older than 17

Let:

a = age

The constraints are:

a ≥ 12

and:

a ≤ 17

Combined:

12 ≤ a ≤ 17

The feasible ages are those satisfying both conditions.

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"And" Constraints

When two conditions must both be satisfied, we use the idea of and.

For example:

x ≥ 2

and:

x ≤ 8

Combined:

2 ≤ x ≤ 8

The feasible region is the overlap of the two solution sets.

A useful reminder is:

AND = OVERLAP

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"Or" Constraints

Sometimes either condition may be satisfied.

Suppose a machine automatically shuts down when:

T < 5°C or T > 90°C

The model is:

T < 5 or T > 90

The solution contains two separate regions.

A useful reminder is:

OR = COMBINE

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Solving an Inequality Model

Suppose a delivery company charges:

$18 fixed fee

plus:

$4 per kilometre

You can spend no more than:

$70

Let:

d = distance in kilometres

Model:

18 + 4d ≤ 70

Subtract 18:

4d ≤ 52

Divide by 4:

d ≤ 13

The maximum distance you can afford is:

13 km

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Verifying the Boundary

Check d = 13:

18 + 4(13)

= 18 + 52

= 70

This exactly matches the budget.

Because the inequality uses ≤, the boundary is included.

Therefore:

13 km is feasible.

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A Minimum Earnings Problem

A worker needs to earn at least:

$500

The worker receives a fixed payment of:

$80

plus:

$28/hour

Let:

h = hours worked

Model:

80 + 28h ≥ 500

Subtract 80:

28h ≥ 420

Divide by 28:

h ≥ 15

The worker must work at least:

15 hours

to meet the target.

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A Time Constraint

A student has:

120 minutes

to complete homework.

Math requires approximately:

35 minutes

and each science problem takes:

5 minutes

Let:

p = number of science problems

Model:

35 + 5p ≤ 120

Subtract 35:

5p ≤ 85

Divide by 5:

p ≤ 17

The student can complete at most 17 science problems within the available time under this model.

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Inequalities with Negative Rates

Suppose a freezer begins at:

12°C

and cools by:

3°C per hour.

Its temperature is:

T = 12 − 3h

Suppose the freezer must reach:

T ≤ −6°C

Model:

12 − 3h ≤ −6

Subtract 12:

−3h ≤ −18

Divide by −3.

Because we divide by a negative number, reverse the inequality:

h ≥ 6

The freezer needs at least:

6 hours

to reach −6°C or colder.

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Why the Inequality Reverses

Whenever both sides of an inequality are multiplied or divided by a negative number, the inequality sign must reverse.

For example:

3 < 7

Multiply both sides by −1:

−3 > −7

The order reverses on the number line.

This rule applies to inequality models exactly as it does to purely algebraic inequalities.

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Realistic Restrictions

Suppose algebra gives:

n ≤ 8.5

If n represents the number of boxes:

n cannot be:

• negative
• fractional

Therefore, the realistic feasible set might be:

n = 0, 1, 2, ..., 8

The algebraic solution and the contextual solution are related but not always identical.

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Non-Negative Constraints

Many quantities cannot realistically be negative.

Examples include:

• number of people
• number of tickets
• distance travelled
• mass
• number of products
• time after an event begins

Therefore, a model may need an additional constraint such as:

x ≥ 0

Suppose:

x ≤ 12

and x represents items produced.

A more complete model is:

0 ≤ x ≤ 12

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Whole-Number Constraints

Some variables must be whole numbers.

Suppose:

n = number of buses

Then:

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

A solution such as:

n ≥ 3.4

must be interpreted as:

n ≥ 4

for the practical situation.

This is different from a continuous quantity such as time, where 3.4 hours may be perfectly meaningful.

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Continuous Constraints

Suppose:

t = time in hours

and:

t ≤ 3.4

A value such as:

3.25 hours

is possible.

Therefore, the entire interval may be feasible.

This distinction between discrete and continuous variables is important when interpreting inequality solutions.

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Modeling Acceptable Measurement Ranges

Suppose a manufactured rod should be:

50 cm

long, with an acceptable variation of:

±0.2 cm

The minimum acceptable length is:

50 − 0.2 = 49.8 cm

The maximum acceptable length is:

50 + 0.2 = 50.2 cm

Therefore:

49.8 ≤ L ≤ 50.2

A rod measuring:

50.1 cm

is feasible.

A rod measuring:

50.4 cm

is not.

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Testing a Feasible Value

For:

49.8 ≤ L ≤ 50.2

test:

L = 50.15

Check:

49.8 ≤ 50.15 ≤ 50.2

True.

Therefore:

50.15 cm is acceptable.

Testing candidate values is an effective way to justify a conclusion.

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Testing an Infeasible Value

Test:

L = 49.6

Check:

49.8 ≤ 49.6 ≤ 50.2

False.

Therefore:

49.6 cm is outside the acceptable range.

A strong justification includes both the inequality and the comparison.

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Using Inequalities for Decision-Making

Suppose two venues have capacities:

Venue A: 250 people

Venue B: 400 people

An event expects:

320 people

Venue A requires:

n ≤ 250

But:

320 ≤ 250

is false.

Venue B requires:

n ≤ 400

and:

320 ≤ 400

is true.

Therefore, Venue B satisfies the capacity constraint while Venue A does not.

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Multiple Real-World Constraints

Suppose a delivery truck can carry:

• no more than 2,000 kg
• no more than 100 boxes

Let:

n = number of boxes

If each box weighs 25 kg, the mass constraint is:

25n ≤ 2000

Solve:

n ≤ 80

The box-count constraint is:

n ≤ 100

Both must be satisfied.

Therefore:

n ≤ 80

The mass restriction is the limiting constraint.

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Limiting Constraints

When several constraints apply, the limiting constraint is the one that restricts the feasible solution most strongly.

In the previous example:

mass allows at most 80 boxes

space/count allows at most 100 boxes

Therefore:

80 boxes

is the practical maximum.

The truck may have space for more boxes, but it reaches its mass limit first.

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Comparing Constraints

Suppose a workshop has enough:

Material A for:

n ≤ 120

Material B for:

n ≤ 85

Labor time for:

n ≤ 100

All constraints must be satisfied.

Therefore:

n ≤ 85

Material B is the limiting resource.

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Feasible Regions with Two Variables

Some models involve two unknown quantities.

Suppose a student buys notebooks and pens.

Let:

n = number of notebooks

p = number of pens

A notebook costs $4.

A pen costs $2.

The student has no more than $20.

The budget constraint is:

4n + 2p ≤ 20

There are many possible solutions.

Examples include:

(0, 10)

(2, 6)

(3, 4)

(4, 2)

(5, 0)

as well as other combinations that cost less than $20.

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Boundary Lines

For a two-variable inequality such as:

4n + 2p ≤ 20

the related boundary equation is:

4n + 2p = 20

Points on this line use the entire $20 budget.

Points on the appropriate side of the line represent combinations that cost less than $20.

Together, these form a solution region.

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Testing a Point

Consider:

4n + 2p ≤ 20

Test:

(n, p) = (2, 4)

Substitute:

4(2) + 2(4) ≤ 20

8 + 8 ≤ 20

16 ≤ 20

True.

Therefore:

(2, 4) is feasible.

Now test:

(4, 4)

4(4) + 2(4) ≤ 20

16 + 8 ≤ 20

24 ≤ 20

False.

Therefore:

(4, 4) is not feasible.

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Graphical Solution Regions

For a two-variable inequality, the graph contains an entire region rather than just one line.

A typical process is:

1. graph the boundary line
2. determine whether the boundary is included
3. test a point
4. shade the side satisfying the inequality
5. consider additional contextual restrictions

For quantities such as numbers of notebooks and pens, we also require:

n ≥ 0

p ≥ 0

and normally whole-number values.

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Solid and Dashed Boundary Lines

For:

y ≤ 2x + 3

the boundary is included.

Use a solid line.

For:

y < 2x + 3

the boundary is excluded.

Use a dashed line.

Similarly:

≥ or ≤ → solid boundary

or < → dashed boundary

This is the two-dimensional equivalent of closed and open circles on a number line.

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Overlapping Constraints

Suppose a problem has:

y ≤ 2x + 6

and:

y ≥ x + 1

Each inequality creates its own solution region.

The feasible region is where the two shaded regions overlap.

A point must satisfy every constraint to be feasible.

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Justifying a Conclusion

A strong mathematical conclusion should do more than state an answer.

Instead of:

"They can buy 7."

write:

"The model gives n ≤ 7.5. Because n represents whole items, the greatest feasible value is 7. Buying 8 would exceed the budget, so 7 is the maximum."

This connects:

• the model
• the algebra
• the context
• the conclusion

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Justifying with Substitution

Suppose a van has a maximum load of:

750 kg

The van already contains:

430 kg

Each crate weighs:

45 kg

Model:

430 + 45c ≤ 750

Solving gives:

45c ≤ 320

c ≤ 7.11 approximately

Therefore:

c ≤ 7

Check 7:

430 + 45(7) = 745

Feasible.

Check 8:

430 + 45(8) = 790

Not feasible.

Therefore, 7 crates is the maximum, and substitution verifies the conclusion.

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Evaluating the Model Itself

Even a correctly solved inequality may rely on assumptions.

In the van example, the model assumes:

• every crate weighs exactly 45 kg
• the existing load is exactly 430 kg
• the stated maximum load is correct
• no other load is added

If actual crate masses vary, a safety margin may be necessary.

Mathematical models should always be interpreted within their assumptions.

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Safety Constraints

In safety-related situations, exceeding a limit may have serious consequences.

For example:

m ≤ 750 kg

means 750 kg is the maximum safe mass.

A calculated value of:

749.9 kg

may mathematically satisfy the model.

However, measurement uncertainty could make operating so close to the limit unwise.

Real engineering decisions may include additional safety factors beyond the simplified inequality.

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A Complete Modeling Example

A school is planning a trip.

A bus costs:

$250 fixed fee

plus:

$12 per student

The school has a maximum transportation budget of:

$850

Let:

n = number of students

Model:

250 + 12n ≤ 850

Subtract 250:

12n ≤ 600

Divide by 12:

n ≤ 50

Therefore:

at most 50 students can attend under this transportation budget.

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Checking the Boundary

For 50 students:

250 + 12(50)

= 250 + 600

= 850

This exactly matches the budget.

For 51 students:

250 + 12(51)

= 250 + 612

= 862

This exceeds the budget.

Therefore, 50 is the maximum feasible number.

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Adding Another Constraint

Suppose the bus itself can carry only:

46 students

Now there are two constraints:

Budget:

n ≤ 50

Capacity:

n ≤ 46

Both must be satisfied.

Therefore:

n ≤ 46

The bus capacity becomes the limiting constraint.

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Adding a Minimum Requirement

Suppose the trip will only run if at least:

30 students

participate.

Now:

n ≥ 30

and:

n ≤ 46

Combined:

30 ≤ n ≤ 46

Because n represents students, the feasible solution set consists of whole numbers:

30, 31, 32, ..., 46

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Interpreting the Feasible Set

The solution:

30 ≤ n ≤ 46

does not mean there is one correct number of students.

It means every whole-number value from 30 through 46 satisfies all of the constraints.

That entire collection is the feasible solution set.

This is one of the major differences between equation models and inequality models.

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A Reliable Modeling Strategy

When modeling a real-world situation with inequalities:

1. Read the situation carefully.
2. Identify the unknown quantity or quantities.
3. Define variables and units.
4. Identify every constraint.
5. Translate words such as "at least" and "at most" carefully.
6. Write an inequality for each constraint.
7. Solve the inequalities.
8. Reverse the inequality sign if multiplying or dividing by a negative.
9. Identify the feasible solution set.
10. Apply contextual restrictions such as non-negative or whole-number values.
11. Represent the solution graphically when useful.
12. Test important boundary values.
13. Identify the limiting constraint when several constraints apply.
14. Interpret the answer in words.
15. Verify that the conclusion satisfies every constraint.
16. Consider whether the model's assumptions are reasonable.

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Common Mistakes

Confusing "at least" and "at most"

At least 10:

x ≥ 10

At most 10:

x ≤ 10

Forgetting that the boundary may be included

"Maximum 50" includes 50:

x ≤ 50

Using ordinary rounding without considering the inequality

If:

n ≤ 6.8

and n represents whole items, the maximum is 6, not 7.

Forgetting to reverse the sign

When multiplying or dividing an inequality by a negative number, reverse the inequality symbol.

Ignoring realistic restrictions

A solution such as:

n = −4 people

may satisfy part of an algebraic expression but is not meaningful in context.

Ignoring whole-number restrictions

People, buses, boxes, and tickets are usually discrete.

Checking only one constraint

A feasible solution must satisfy all applicable constraints.

Confusing the boundary with the feasible region

For a two-variable inequality, the boundary line is only part of the solution when equality is allowed. The solution usually includes an entire region.

Using a solid boundary for a strict inequality

< and > use dashed boundaries.

≤ and ≥ use solid boundaries.

Giving an answer without justification

A conclusion should be supported using the inequality, solution, context, and preferably a boundary check.

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Key Terms

Inequality model — mathematical representation of a real-world constraint using inequality symbols.

Constraint — condition that limits possible solutions.

Feasible solution — value or combination of values satisfying all constraints.

Feasible solution set — collection of all values satisfying the constraints.

Feasible region — graphical region containing solutions satisfying all constraints.

Boundary — value separating solutions from non-solutions.

Boundary line — line separating solution and non-solution regions in a two-variable inequality.

Strict inequality — inequality using < or >, where the boundary is excluded.

Inclusive inequality — inequality using ≤ or ≥, where the boundary is included.

Compound inequality — combination of two inequality conditions.

Limiting constraint — constraint that places the strongest restriction on the feasible solutions.

Discrete variable — variable that takes separate countable values.

Continuous variable — variable that can take any value within an interval.

Domain restriction — limitation on values that are meaningful or allowed.

Solution region — part of a graph representing values satisfying an inequality.

Open circle — number-line symbol showing that a boundary value is excluded.

Closed circle — number-line symbol showing that a boundary value is included.

Solid boundary line — boundary included in a two-variable inequality.

Dashed boundary line — boundary excluded from a two-variable inequality.

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Key Takeaways

• Inequalities are useful for modeling limits, requirements, ranges, and constraints.
• Equations usually describe exact relationships, while inequalities describe sets of possible values.
• A constraint limits which solutions are acceptable.
• "At least" normally translates to ≥.
• "At most" normally translates to ≤.
• "Greater than" and "less than" exclude the boundary.
• Variables should be clearly defined before constructing a model.
• A feasible solution satisfies every applicable constraint.
• The feasible solution set contains all acceptable solutions.
• Context may restrict algebraic solutions to non-negative values, integers, or whole numbers.
• Discrete and continuous variables must be interpreted differently.
• Rounding must preserve the original inequality rather than simply follow ordinary rounding rules.
• Inequality solutions can be represented on number lines.
• Open circles represent excluded boundaries.
• Closed circles represent included boundaries.
• Compound inequalities can represent minimum and maximum limits simultaneously.
• "And" constraints require overlap between solution sets.
• "Or" constraints combine solution regions.
• Two-variable inequalities produce solution regions rather than a single line.
• Solid boundary lines are used when equality is included.
• Dashed boundary lines are used when equality is excluded.
• Multiple constraints create a feasible region where all conditions overlap.
• A limiting constraint determines the strongest practical restriction.
• Substitution can verify whether a proposed solution is feasible.
• A strong conclusion connects the mathematical solution directly to the real-world situation.
• Mathematical feasibility does not automatically guarantee practical safety or usefulness; assumptions and context still matter.

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Check Your Understanding

1. Write an inequality representing: A package may weigh no more than 25 kg.
2. Write an inequality representing: A student must score at least 60 points to pass.
3. Explain the difference between x < 8 and x ≤ 8, including how each would be represented on a number line.
4. You have $75 to spend on notebooks costing $8 each. Define a variable, create an inequality, solve it, and determine the maximum number of notebooks you can purchase.
5. A gym charges a $30 membership fee plus $6 per visit. You have no more than $120 to spend. Create and solve an inequality to determine the maximum number of visits.
6. A worker has already earned $180 and earns another $25 per hour. The worker wants to earn at least $500. Create and solve an inequality to determine the minimum number of additional whole hours required.
7. An elevator can carry a maximum of 1,000 kg. Its current load is 580 kg. Boxes weighing 32 kg each must be added. Determine the maximum number of whole boxes that can safely be added.
8. A machine operates safely when its temperature is at least 35°C and no more than 80°C. Write a compound inequality, describe its number-line graph, and give the corresponding interval notation.
9. A freezer begins at 15°C and cools by 4°C each hour. It must reach a temperature of −9°C or colder. Create and solve an inequality to determine how long the freezer must cool.
10. A calculation gives n ≤ 12.7, where n represents the number of passengers allowed on a small vehicle. Explain why 13 passengers would not be a valid interpretation.
11. A calculation gives n ≥ 6.2, where n represents the number of buses required to transport all students. Determine the minimum feasible number of buses and justify your conclusion.
12. A truck can carry at most 2,400 kg and at most 90 containers. Each container has a mass of 30 kg. Write both constraints, determine the maximum feasible number of containers, and identify the limiting constraint.
13. A student has $30 to purchase notebooks costing $5 each and pens costing $2 each. Let n represent notebooks and p represent pens. Write a two-variable inequality representing the budget. Determine whether (4, 5) is feasible and justify your answer.
14. A manufacturing process requires a component length to be 25.0 ± 0.15 mm. Write a compound inequality representing acceptable lengths. Determine whether components measuring 24.82 mm, 24.90 mm, and 25.14 mm are acceptable.
15. A school trip requires at least 24 students to run. A bus holds no more than 42 students. Transportation costs a fixed $180 plus $14 per student, and the school has a maximum budget of $750. Define a variable, create all relevant inequalities, solve each constraint, determine the complete feasible solution set, identify the limiting upper constraint, and justify the maximum number of students who can attend.