Applications and Function Analysis

 

Interpretation of parameters:

• a = −5 determines the downward curvature. In the ideal constant-acceleration model, a = −g/2.

• b = 20 represents the initial vertical velocity in m/s.

• c = 2 represents the initial height in metres.

The model is physically meaningful from launch until the ball reaches the ground, approximately 4.10 seconds after launch.

Worked Example 3: Building a Rational Model

A manufacturer has a fixed production cost and a constant cost per item.

The following average costs are recorded:

We expect the average cost to have the form:

A(x) = a/x + b

where:

• a represents fixed production cost.

• b represents variable cost per item.

Step 1: Use two data points.

For x = 20:

50 = a/20 + b

For x = 40:

35 = a/40 + b

Step 2: Subtract the equations.

50 − 35 = a/20 − a/40

15 = a/40

a = 600

Step 3: Find b.

50 = 600/20 + b

50 = 30 + b

b = 20

Final model:

Step 4: Verify the other data points.

A(60) = 600/60 + 20 = 30

A(100) = 600/100 + 20 = 26

Both match.

Interpretation:

The fixed cost is $600.

The variable cost is $20 per item.

As production increases, average cost approaches $20.

The horizontal asymptote is:

y = 20

For this application, x is a positive whole number.

Worked Example 4: Building a Cubic Model from Geometry

A company manufactures open-top boxes by cutting squares of side length x from each corner of a rectangular sheet measuring 30 cm by 20 cm.

After folding the sides upward:

• Height = x

• Length = 30 − 2x

• Width = 20 − 2x

The volume is:

V(x) = x(30 − 2x)(20 − 2x)

Expand:

(30 − 2x)(20 − 2x) = 600 − 100x + 4x²

Therefore:

This is a cubic polynomial model.

Determine the practical domain.

All dimensions must be positive:

x > 0

30 − 2x > 0 gives x < 15.

20 − 2x > 0 gives x < 10.

Therefore:

0 < x < 10

Although the polynomial is mathematically defined for all real x, only this interval represents a physical box.

Calculate the volume when x = 4.

V(4) = 4(30 − 8)(20 − 8)

= 4(22)(12)

= 1056 cm³

This example shows how geometry naturally produces polynomial models and how practical restrictions differ from a function's mathematical domain.

Interpreting Model Parameters

Parameters are constants within an equation that control the model's behaviour.

Understanding them allows us to explain the relationship rather than simply perform calculations.

Worked Example 5: Interpreting Changes in Parameters

Consider two average-cost models:

Company A:

A(x) = 400/x + 15

Company B:

B(x) = 700/x + 12

Company A has lower fixed costs but higher variable costs.

Company B has higher fixed costs but lower variable costs.

Find the production quantity where their average costs are equal.

400/x + 15 = 700/x + 12

15 − 12 = 700/x − 400/x

3 = 300/x

3x = 300

x = 100

At 100 items, both companies have the same average cost:

A(100) = 4 + 15 = $19

B(100) = 7 + 12 = $19

For fewer than 100 items, Company A has the lower average cost.

For more than 100 items, Company B has the lower average cost.

This illustrates how parameters can influence real business decisions.

Evaluating the Accuracy of Mathematical Models

A model is not necessarily reliable simply because it produces an equation.

We must compare its predictions with observed data.

One of the most important ideas is the residual.

A residual is the difference between an observed value and the value predicted by a model.

Residual = Observed value − Predicted value

Or:

e = y − ŷ

where:

• y is the observed value.

• ŷ is the predicted value.

• e is the residual.

A positive residual means the model predicted too low.

A negative residual means the model predicted too high.

Worked Example 6: Calculating Residuals

Suppose a model predicts the number of visitors to a website using:

V(t) = 100t + 500

where t is the number of weeks after a campaign begins.

The actual data are:

For week 2:

Residual = 730 − 700 = +30

The model underestimated visitors by 30.

For week 3:

Residual = 780 − 800 = −20

The model overestimated visitors by 20.

Visualizing Residuals

Interpreting Residual Patterns

A useful model should generally produce residuals that are small relative to the scale of the data.

We also examine whether residuals follow a pattern.

• Residuals scattered around zero suggest the model captures the main trend.

• A curved residual pattern suggests the chosen function may be missing curvature.

• Residuals that increase in magnitude suggest changing prediction accuracy.

• A single unusually large residual may indicate an outlier or measurement problem.

Residuals alone cannot prove that a model is correct, but they are valuable for identifying weaknesses.

Measuring Model Accuracy

Several numerical measures can be used to compare models.

Mean Absolute Error (MAE)

The mean absolute error measures the average size of the prediction errors.

where n is the number of observations.

For the visitor data:

Absolute residuals:

20, 10, 30, 20, 10

MAE = (20 + 10 + 30 + 20 + 10)/5

MAE = 90/5

MAE = 18 visitors.

This means the model's predictions differ from the observations by an average of 18 visitors.

Mean Squared Error (MSE)

Mean squared error gives more weight to large errors.

For the same data:

MSE = (400 + 100 + 900 + 400 + 100)/5

MSE = 1900/5

MSE = 380 visitors².

Root Mean Squared Error (RMSE)

RMSE is the square root of MSE:

RMSE = √380

RMSE ≈ 19.49 visitors.

RMSE has the same units as the original output variable, making it easier to interpret than MSE.

Coefficient of Determination (R²)

R² measures how much of the variation in the observed outputs is accounted for by a model relative to predicting their mean.

For a model evaluated on a dataset:

where ȳ is the mean observed output.

Values close to 1 often indicate a strong fit on the evaluated dataset.

However, R² does not establish that a model is scientifically correct, nor does it guarantee accurate future predictions. R² can also be negative when predictions are worse than using the observed mean.

Worked Example 7: Comparing Two Models

Suppose the following data describe a company's revenue over five years.

We consider two models.

Model A: Linear

L(t) = 8t + 8

Model B: Quadratic

Q(t) = t² + 4t + 10

Calculate predictions.

Evaluate Model A.

Residuals:

2, −1, −2, −1, 2

MAE:

(2 + 1 + 2 + 1 + 2)/5 = 1.6

Evaluate Model B.

All residuals are zero.

MAE = 0

Therefore, Model B fits the observed data exactly.

Does this guarantee Model B is the better long-term model?

No.

The quadratic model perfectly fits these five observations, but future revenue may be affected by competition, economic conditions, market saturation, or other factors.

An exact fit to past data does not guarantee accurate extrapolation.

When evaluating models, consider mathematical accuracy, simplicity, assumptions, and whether the relationship makes sense in context.

Interpolation and Extrapolation

Mathematical models are often used to estimate values that were not directly measured.

There are two important types of prediction.

Interpolation means predicting within the range of observed input values.

Extrapolation means predicting outside the observed range.

Suppose measurements were collected for x-values between 0 and 10.

Predicting at x = 6.5 is interpolation.

Predicting at x = 25 is extrapolation.

Extrapolation is generally more uncertain because the relationship may change beyond the observed range.

For example, a quadratic model of company revenue might fit five years of data but predict unrealistic revenue after 50 years.

A good modelling report should clearly distinguish between predictions supported by observed data and predictions that extend beyond it.

Solving Real-World Problems Using Models

After creating and evaluating a model, we can use it to answer practical questions.

These often involve solving equations, identifying maximum or minimum values, or comparing different functions.

Worked Example 8: Break-Even Analysis

A company sells each product for $40.

Its total cost is:

C(x) = 500 + 15x

Revenue is:

R(x) = 40x

Profit is revenue minus cost:

P(x) = R(x) − C(x)

P(x) = 40x − (500 + 15x)

P(x) = 25x − 500

Find the break-even quantity.

At break-even:

P(x) = 0

25x − 500 = 0

25x = 500

x = 20

The company breaks even when it sells 20 items.

Find the profit from 80 items.

P(80) = 25(80) − 500

= 2000 − 500

= $1,500

Interpretation: Selling more than 20 items produces a positive profit under this model.

Worked Example 9: Solving a Rational Cost Model

A manufacturer's average cost per item is:

A(x) = 900/x + 18

The company wants average cost to be no more than $30.

Solve:

900/x + 18 ≤ 30

For the practical domain x > 0:

900/x ≤ 12

Multiply both sides by positive x:

900 ≤ 12x

x ≥ 75

Therefore, the company must produce at least 75 items to achieve an average cost of $30 or less.

The positivity of x matters when multiplying an inequality by x.

Worked Example 10: Optimizing a Polynomial Model

A farmer has 80 metres of fencing to create a rectangular enclosure along a straight wall. The wall forms one side, so fencing is required for only three sides.

Let x be the width perpendicular to the wall.

The two widths use 2x metres of fencing.

The remaining fenced length is:

80 − 2x

Area:

A(x) = x(80 − 2x)

Expand:

A(x) = −2x² + 80x

This is a downward-opening quadratic.

Find the maximum area.

Vertex x-coordinate:

x = −80/[2(−2)]

x = 20

Length:

80 − 2(20) = 40

Maximum area:

A(20) = 20 × 40

= 800 m²

Conclusion: The farmer should construct an enclosure 20 metres wide and 40 metres long to maximize area.

The practical domain is 0 < x < 40.

Evaluating the Limitations of Models

Every mathematical model relies on assumptions.

For example:

A projectile model may ignore air resistance.

A production-cost model may assume fixed costs and constant material prices.

A population model may assume stable environmental conditions.

A travel-time model may assume constant speed.

These assumptions simplify calculations, but they may reduce accuracy.

Questions to Ask When Evaluating a Model

• Does the model match the observed data reasonably well?

• Are its parameters meaningful?

• Does it predict realistic values?

• Are the domain and range appropriate for the situation?

• Are there outliers or systematic residual patterns?

• Is the model unnecessarily complicated?

• Are predictions being made outside the observed data?

• Could important external factors change the relationship?

A useful model is not necessarily the most complicated model. Often, a simpler model is preferable when it explains the data adequately and is easier to interpret.

Common Mistakes

Mistake 1: Choosing a Model Based Only on Its Graph

A graph may resemble a quadratic curve, but this does not prove the relationship is quadratic.

Examine the variables, data patterns, and underlying mathematical relationship.

Mistake 2: Assuming All Curved Relationships Are Quadratic

A curved relationship could be cubic, rational, exponential, logarithmic, or another function type.

For example, travel time for a fixed distance follows t = d/v, not a quadratic model.

Mistake 3: Ignoring Units

A parameter has meaning only when interpreted with appropriate units.

In C(x) = 5x + 100, the coefficient 5 might represent dollars per item, while 100 represents dollars.

Mistake 4: Using an Incorrect Domain

A production quantity cannot be negative.

A time after launch cannot be negative.

A physical length must be positive.

The mathematical domain and practical domain may be different.

Mistake 5: Confusing a Perfect Fit with a Reliable Model

A polynomial can fit a small number of data points exactly while giving unrealistic predictions outside the observed range.

Mistake 6: Confusing Residual Signs

Residual = Observed − Predicted.

A positive residual means the model underestimated the observed value.

A negative residual means the model overestimated it.

Mistake 7: Ignoring Asymptotes in Rational Models

An asymptote may represent an important limiting value.

For A(x) = 500/x + 12, the horizontal asymptote y = 12 represents the variable cost per item.

Mistake 8: Treating Predictions as Certainties

Models are simplified representations of reality. Their predictions depend on assumptions, data quality, and the stability of the relationship.

Mistake 9: Forgetting to Interpret the Answer

Solving x = 75 is not enough.

A complete conclusion should explain that 75 items must be produced to meet the target average cost.

Key Terms

• Mathematical Model: A mathematical representation of a real-world situation.

• Polynomial Model: A model using a polynomial function.

• Rational Model: A model using a quotient of polynomial functions.

• Independent Variable: The input quantity.

• Dependent Variable: The output quantity.

• Parameter: A constant in a model that controls its behaviour.

• Linear Model: A model with a constant rate of change.

• Quadratic Model: A second-degree polynomial model.

• Cubic Model: A third-degree polynomial model.

• Inverse Variation: A relationship where one quantity is proportional to the reciprocal of another.

• Regression: A method of estimating a model's parameters from data.

• Residual: The difference between an observed value and a predicted value.

• Mean Absolute Error (MAE): The average absolute prediction error.

• Mean Squared Error (MSE): The average squared prediction error.

• Root Mean Squared Error (RMSE): The square root of MSE.

• Coefficient of Determination (R²): A measure comparing prediction error with variation around the observed mean.

• Interpolation: Estimating a value within the observed input range.

• Extrapolation: Estimating a value outside the observed input range.

• Asymptote: A line approached by a graph in a limiting situation.

• Practical Domain: The input values that make sense in the real-world context.

• Break-Even Point: The quantity at which revenue equals cost.

• Optimization: Finding the greatest or least possible value under given conditions.

• Assumption: A simplifying condition accepted when constructing a model.

• Model Validation: Evaluating how well a model performs against relevant observations, preferably including data not used to fit it.

Key Takeaways

1. Mathematical modelling connects real-world situations with mathematical functions.

2. Polynomial models are useful for relationships involving constant rates, area, volume, motion, and curved trends.

3. Rational models are useful for ratios, rates, average costs, and inverse relationships.

4. Data tables and graphs help identify possible models.

5. Constant first differences suggest a linear relationship for equally spaced inputs.

6. Constant second differences suggest a quadratic relationship under suitable conditions.

7. Model parameters should be interpreted using real-world units.

8. Residuals measure differences between observed and predicted values.

9. MAE, MSE, RMSE, and R² can help evaluate model accuracy.

10. A model that fits past data perfectly may still make poor future predictions.

11. Interpolation is generally more reliable than extrapolation.

12. Practical domains may be more restricted than mathematical domains.

13. Models can be used to find break-even points, optimize quantities, and make predictions.

14. Good mathematical modelling requires interpretation, evaluation, and awareness of limitations.

Check Your Understanding

Question 1: Choosing a Model

Choose a suitable polynomial or rational model for each situation and explain why.

a) The height of a ball thrown vertically upward.

b) The time needed to travel 300 km at different constant speeds.

c) The area of a square as its side length changes.

d) The average cost per item when there is a fixed production cost.

Question 2: Identifying a Linear Model

The following data show the cost of a service.

Find a linear model and interpret its parameters.

Question 3: Creating a Quadratic Model

A quadratic function passes through:

(0, 3), (1, 8), and (2, 11).

Find an equation of the form:

f(x) = ax² + bx + c

Question 4: Creating a Rational Model

The average cost of producing x items follows:

A(x) = a/x + b

Given:

A(20) = 40

A(40) = 30

Find a and b, then calculate A(100).

Question 5: Evaluating Residuals

A model predicts the following values:

a) Calculate the residuals.

b) Calculate the MAE.

c) Explain what the MAE means.

Question 6: Interpolation and Extrapolation

A scientist collects data for temperatures between 10°C and 40°C.

Classify the following predictions as interpolation or extrapolation.

a) Predicting at 25°C.

b) Predicting at 60°C.

c) Predicting at 5°C.

Which predictions require greater caution?

Question 7: Interpreting Parameters

A company uses:

C(x) = 750 + 18x

where x is the number of items produced.

a) What does 750 represent?

b) What does 18 represent?

c) Calculate the cost of producing 150 items.

d) Write an average-cost function.

Question 8: Solving a Rational Model

A factory's average cost is:

A(x) = 1200/x + 25

How many items must be produced for average cost to be no more than $40?

Question 9: Optimization

A rectangular garden has a perimeter of 60 m.

a) Write an area function in terms of one side length x.

b) Find the maximum area.

c) Determine the garden dimensions that produce this area.

Question 10: Challenge — Comparing Two Models

Two companies have average-cost functions:

A(x) = 300/x + 18

B(x) = 600/x + 15

a) Find the production quantity where average costs are equal.

b) Calculate the average cost at that quantity.

c) Determine which company is cheaper for 50 items.

d) Determine which company is cheaper for 200 items.

e) Explain why the preferred company changes as production increases.

Fully Worked Answers

Answer 1: Choosing Models

a) Quadratic polynomial.

Under constant gravitational acceleration, height changes quadratically with time.

b) Rational.

Time = Distance / Speed, so t(v) = 300/v.

c) Quadratic polynomial.

A(s) = s².

d) Rational.

Average cost is total cost divided by production quantity.

Answer 2: Linear Model

The cost increases by $25 for each additional hour.

Therefore:

m = 25

Use:

C = 25h + b

Substitute (1, 45):

45 = 25 + b

b = 20

Final model:

C = 25h + 20

The hourly charge is $25.

The fixed fee is $20.

Answer 3: Quadratic Model

Use:

f(x) = ax² + bx + c

At (0, 3):

c = 3

At (1, 8):

a + b + 3 = 8

a + b = 5

At (2, 11):

4a + 2b + 3 = 11

4a + 2b = 8

2a + b = 4

Subtract:

(2a + b) − (a + b) = 4 − 5

a = −1

Then:

−1 + b = 5

b = 6

Final model:

f(x) = −x² + 6x + 3

Answer 4: Rational Model

Given:

A(x) = a/x + b

Using A(20) = 40:

40 = a/20 + b

Using A(40) = 30:

30 = a/40 + b

Subtract:

10 = a/40

a = 400

Substitute:

40 = 400/20 + b

40 = 20 + b

b = 20

Final model:

A(x) = 400/x + 20

At x = 100:

A(100) = 400/100 + 20

= 4 + 20

= $24 per item.

Answer 5: Residuals

Residual = Observed − Predicted.

Residuals:

12 − 10 = +2

18 − 20 = −2

25 − 24 = +1

31 − 34 = −3

Absolute residuals:

2, 2, 1, 3

MAE = (2 + 2 + 1 + 3)/4

MAE = 8/4

MAE = 2.

The model's predictions differ from the observations by an average of 2 units.

Answer 6: Interpolation and Extrapolation

Observed range: 10°C to 40°C.

a) 25°C is interpolation.

b) 60°C is extrapolation.

c) 5°C is extrapolation.

Predictions at 60°C and 5°C require greater caution because they are outside the observed range.

Answer 7: Interpreting Parameters

Given:

C(x) = 750 + 18x

a) 750 represents the fixed cost of $750.

b) 18 represents the variable cost of $18 per item.

c) At 150 items:

C(150) = 750 + 18(150)

= 750 + 2700

= $3,450

d) Average cost:

A(x) = C(x)/x

A(x) = (750 + 18x)/x

Final model:

A(x) = 750/x + 18

for x > 0.

Answer 8: Rational Model

Given:

A(x) = 1200/x + 25

We require:

1200/x + 25 ≤ 40

Subtract 25:

1200/x ≤ 15

Since x > 0:

1200 ≤ 15x

x ≥ 80

Answer: At least 80 items must be produced.

Answer 9: Optimization

Perimeter:

2L + 2W = 60

Divide by 2:

L + W = 30

Let W = x.

Then:

L = 30 − x

Area:

A(x) = x(30 − x)

Expand:

A(x) = −x² + 30x

The parabola opens downward.

Vertex:

x = −30/[2(−1)]

x = 15

Length:

30 − 15 = 15

Maximum area:

15 × 15 = 225 m²

Answer: A 15 m by 15 m square gives the maximum area of 225 m².

Answer 10: Comparing Companies

Given:

A(x) = 300/x + 18

B(x) = 600/x + 15

a) Equal average costs

300/x + 18 = 600/x + 15

18 − 15 = 600/x − 300/x

3 = 300/x

3x = 300

x = 100

b) Average cost at 100 items

A(100) = 300/100 + 18

= $21

B(100) = 600/100 + 15

= $21

c) At 50 items

A(50) = 300/50 + 18 = $24

B(50) = 600/50 + 15 = $27

Company A is cheaper.

d) At 200 items

A(200) = 300/200 + 18 = $19.50

B(200) = 600/200 + 15 = $18

Company B is cheaper.

e) Interpretation

Company A has lower fixed costs but higher variable costs.

Company B has higher fixed costs but lower variable costs.

At low production quantities, Company A's lower fixed costs make it cheaper.

At high production quantities, Company B's lower variable cost becomes more important.

The crossover occurs at 100 items.

This completes Real-World Modelling, covering all five learning targets: selecting polynomial and rational models, constructing models from data, evaluating accuracy, interpreting parameters, and solving practical problems using mathematical functions.