Advanced Mathematical Modeling

Lesson: Formulation of Mathematical Models

In many real-life situations, we want to understand a problem, make predictions, or choose the best decision. A mathematical model is a mathematical representation of a real-world situation. It turns words, quantities, and relationships into equations, inequalities, tables, graphs, or formulas.

The process of formulating a mathematical model means translating a context into mathematics. This requires us to decide what the important quantities are, name them with variables, identify fixed values called parameters, and make reasonable assumptions so the problem becomes manageable.

This skill is important because real-world problems are often messy. They include extra details, missing information, and quantities that change. A good model simplifies the situation enough to be useful while still matching reality closely enough to answer the question.

1. What is a mathematical model?

A mathematical model is a set of mathematical relationships that describes how quantities in a situation are connected. Depending on the problem, a model might be:

• an equation, such as \(C = 50 + 12m\),
• an inequality, such as \(3x + 2y \le 100\),
• a function, such as \(h(t) = -4.9t^2 + 20t + 1.5\),
• a geometric rule involving area, volume, or distance,
• or a statistical model based on data.

The goal is not just to do calculations. The goal is to represent the situation clearly so that the mathematics answers a real question.

2. Main parts of formulating a model

When building a mathematical model, there are four main parts to identify.

• Variables: quantities that can change. For example, time, distance, number of items sold, or temperature.
• Parameters: fixed values in the model. For example, a ticket price of \(12\) dollars or a constant speed of \(60\) km/h.
• Relationships: the mathematical connections between quantities. These may come from formulas, patterns, geometry, or data.
• Assumptions: simplifications that make the problem easier to model. For example, assuming a constant rate, ignoring air resistance, or treating demand as linear.

3. A step-by-step method for formulating models

You can use the following process whenever you are asked to create a model from a context.

1. Read the situation carefully. Identify what is happening and what question must be answered.
2. Choose the important quantities. Ignore details that do not affect the question.
3. Define variables clearly. State what each variable represents and include units when possible.
4. Identify parameters or known constants. These are values given in the problem or estimated from information.
5. Make simplifying assumptions. Write them clearly so the model is understandable.
6. Write the mathematical relationships. Use equations, formulas, or inequalities.
7. Check if the model makes sense. Think about units, signs, reasonable values, and whether the model matches the context.
8. Refine if needed. If the model is too simple or unrealistic, improve it.

4. Defining variables, parameters, and assumptions

One of the most important parts of mathematical modeling is being precise about symbols.

Suppose a problem says that a company charges a fixed monthly fee plus a cost for each gigabyte of data used. A student might quickly write \(C = 20 + 5x\). That equation may be correct, but the model is incomplete unless the symbols are defined.

A complete formulation would say:

• Let \(C\) be the monthly cost in dollars.
• Let \(x\) be the number of gigabytes used in the month.
• The fixed fee is \(20\) dollars.
• The cost per gigabyte is \(5\) dollars.
• Assume there are no taxes or extra fees.

Then the model is

$$C = 20 + 5x$$

Without definitions and assumptions, the equation alone may be unclear or incomplete.

5. Why assumptions matter

Real life is usually more complicated than school problems. A model cannot include every detail, so assumptions are necessary.

For example, if you model the distance traveled by a car using \(d = vt\), you are assuming the speed \(v\) stays constant. In reality, cars stop, turn, and speed up. Still, the model can be useful for short trips or average motion.

Good assumptions should be:

• reasonable for the situation,
• stated clearly,
• and connected to the purpose of the model.

If assumptions are unrealistic, the model may give misleading answers.

6. Types of relationships often used in Grade 11 modeling

In Grade 11, many models come from common mathematical patterns:

• Linear models: constant rate of change, such as \(y = mx + b\).
• Quadratic models: curved relationships, such as projectile motion or area problems.
• Geometric models: formulas for perimeter, area, volume, and the Pythagorean theorem.
• Statistical models: trend lines based on data, often used for prediction.
• Systems of equations or inequalities: situations with multiple conditions or constraints.

Formulating a model means choosing the type of relationship that best fits the context.

Worked Example 1: A linear cost model

Problem: A bike rental shop charges a basic fee of \(8\) dollars plus \(3\) dollars per hour. Formulate a mathematical model for the total cost.

Step 1: Define variables.

• Let \(h\) be the number of hours the bike is rented.
• Let \(C\) be the total cost in dollars.

Step 2: Identify parameters.

• Basic fee: \(8\)
• Hourly rate: \(3\)

Step 3: State an assumption.

• Assume the hourly rate stays the same for all rental hours.

Step 4: Write the model.

$$C = 8 + 3h$$

Interpretation: The total cost starts at \(8\) dollars and increases by \(3\) dollars for each additional hour.

Check: If the bike is rented for \(2\) hours, then

$$C = 8 + 3(2) = 14$$

This is reasonable because the cost should be more than the basic fee.

Worked Example 2: A geometric model with constraints

Problem: A farmer has \(200\) meters of fencing to build a rectangular pen. Formulate a model for the area of the pen in terms of its width.

Step 1: Define variables.

• Let \(w\) be the width of the rectangle in meters.
• Let \(l\) be the length of the rectangle in meters.
• Let \(A\) be the area in square meters.

Step 2: Use the perimeter condition.

A rectangle has perimeter

$$2l + 2w = 200$$

Solve for \(l\):

$$2l = 200 - 2w$$

$$l = 100 - w$$

Step 3: Write the area formula.

$$A = lw$$

Substitute \(l = 100 - w\):

$$A = w(100 - w)$$

$$A = 100w - w^2$$

Model:

$$A(w) = 100w - w^2$$

Assumptions:

• The pen is exactly rectangular.
• All \(200\) meters of fencing are used.
• Width and length are positive, so \(0 < w < 100\).

This example shows that a model can include both a formula and a domain restriction.

Worked Example 3: A model from a word problem involving revenue and profit

Problem: A school club sells T-shirts for \(15\) dollars each. The club pays \(120\) dollars as a fixed setup cost and \(6\) dollars to make each shirt. Formulate a model for profit when \(x\) shirts are sold.

Step 1: Define the variable.

• Let \(x\) be the number of T-shirts sold.
• Let \(P\) be the profit in dollars.

Step 2: Build revenue and cost models.

Revenue is the money earned from sales:

$$R = 15x$$

Cost is the setup cost plus production cost:

$$C = 120 + 6x$$

Step 3: Use the relationship for profit.

$$P = R - C$$

Substitute the expressions:

$$P = 15x - (120 + 6x)$$

$$P = 15x - 120 - 6x$$

$$P = 9x - 120$$

Model:

$$P(x) = 9x - 120$$

Assumptions:

• Every shirt produced is sold.
• The selling price stays at \(15\) dollars per shirt.
• The production cost stays at \(6\) dollars per shirt.

Interpretation: Profit increases by \(9\) dollars for each shirt sold after accounting for the variable cost, but the club must first recover the fixed setup cost.

Worked Example 4: A statistical model from data

Problem: A student records the number of hours studied and the corresponding test scores:

• \((1, 58)\)
• \((2, 64)\)
• \((3, 71)\)
• \((4, 77)\)

Formulate a simple mathematical model relating score to study time.

Step 1: Define variables.

• Let \(h\) be the number of hours studied.
• Let \(S\) be the test score.

Step 2: Look for a pattern.

The scores increase by about \(6\) or \(7\) points for each extra hour. A linear model is reasonable.

Step 3: Estimate the rate of change.

Using the first and last data points:

$$m = \frac{77 - 58}{4 - 1} = \frac{19}{3} \approx 6.33$$

Step 4: Write a linear model.

Use \(S = mh + b\). Substitute \((1,58)\):

$$58 = 6.33(1) + b$$

$$b \approx 51.67$$

So one possible model is

$$S \approx 6.33h + 51.67$$

Assumptions:

• The relationship between study time and score is approximately linear.
• Other factors affecting test scores are ignored.

Interpretation: The model predicts that each extra hour of study raises the score by about \(6.33\) points.

7. How to judge whether a model is good

After formulating a model, do not stop immediately. A model should be checked.

Ask these questions:

• Do the variables and units make sense?
• Are the assumptions clearly stated?
• Does the equation match the words in the problem?
• Are the outputs reasonable in real life?
• Is the model too simple, or is it appropriate for the purpose?

For example, if a cost model gives a negative cost, something is wrong. If an area model allows a negative width, the domain must be corrected. If a linear model is used for data that curves strongly, another model may be better.

8. Common mistakes when formulating models

• Not defining variables: Writing equations without saying what symbols mean.
• Ignoring units: Mixing meters and centimeters, or hours and minutes, without converting.
• Using the wrong relationship: Choosing a linear model when the situation is geometric or quadratic.
• Forgetting assumptions: Leaving out conditions such as constant rate or fixed price.
• Ignoring restrictions: Allowing impossible values like negative time or fractional people.

9. Helpful questions to ask yourself

When you build a model, ask:

• What am I trying to find or describe?
• Which quantities change?
• Which quantities stay fixed?
• What formula or pattern connects these quantities?
• What assumptions am I making?
• What values are realistic for the variables?

These questions help turn a confusing context into an organized mathematical structure.

10. Final summary

Formulating a mathematical model means translating a real-world problem into mathematics. To do this well, you must identify variables, parameters, and relationships, then make assumptions that simplify the situation without losing the main idea.

A strong model is clear, reasonable, and useful. It includes defined symbols, correct equations or inequalities, realistic restrictions, and stated assumptions. Once a model is written, it should always be checked and, if necessary, refined.

As you practice, remember that modeling is not just about finding an answer. It is about representing the real world with mathematics in a thoughtful and organized way.

Optimization under Constraints is about finding the largest or smallest possible value of a quantity when there are rules or limits you must follow.

In real life, we often want to make the best choice, but we cannot choose anything we want. A company may want to maximize profit with a limited budget. A farmer may want the largest possible area of a field using a fixed amount of fencing. A student may want to minimize travel time while staying within a budget. These are all optimization problems.

In math, the quantity we want to maximize or minimize is called the objective function. The rule or limit we must obey is called the constraint.

For example, if you want the largest area of a rectangle made with 40 meters of fencing, then:

• The objective is to maximize area.
• The constraint is that the perimeter must be 40 meters.

This lesson will show you how to turn words into equations, use the constraint to reduce the number of variables, and then find an absolute maximum or minimum on the allowed domain.

Why this matters in mathematical modeling: optimization is a major part of modeling real situations. You build a model using variables and equations, test what values are possible, and decide which value gives the best outcome.

Main idea: most optimization problems follow these steps.

1. Define the variables clearly.
2. Write the objective function.
3. Write the constraint equation or inequality.
4. Use the constraint to rewrite the objective in terms of one variable.
5. Identify the realistic domain.
6. Find the maximum or minimum value on that domain.
7. Interpret the result in the context of the problem.

Important vocabulary

• Objective function: the expression you want to maximize or minimize.
• Constraint: a condition that limits the possible values.
• Domain restriction: the set of values that make sense in the real situation.
• Absolute maximum: the greatest value of a function on a given domain.
• Absolute minimum: the smallest value of a function on a given domain.

A common strategy

Many optimization problems begin with two variables, such as length and width, or number of products A and B. But it is usually easier to optimize a function of only one variable.

So we use the constraint to solve for one variable in terms of the other. Then we substitute into the objective function.

For instance, if a rectangle has length \(l\) and width \(w\), and the perimeter is 40, then

$$2l + 2w = 40.$$

Solving for one variable gives

$$w = 20 - l.$$

If the area is \(A = lw\), then

$$A(l) = l(20 - l) = 20l - l^2.$$

Now the problem has been reduced to one variable.

How do we know where the maximum or minimum happens?

In 11th Grade math, you may find the best value by:

• looking at the graph,
• using the vertex of a quadratic,
• testing endpoints of a restricted domain,
• or comparing reasonable values in a table.

If the function is quadratic, the vertex often gives the highest or lowest value. If the domain is restricted, you must also check the endpoints.

Domain restrictions are essential. A value that works algebraically may not make sense in context. For example, a width cannot be negative, and the number of products made cannot be a fraction if only whole items are allowed.

So after writing the equations, always ask:

• Can the variable be negative?
• Must the value be a whole number?
• Is there a budget, time, or size limit?
• Are there endpoints that must be checked?

Worked Example 1: Maximizing the area of a rectangle with fixed perimeter

A rectangle uses 40 meters of fencing. What dimensions give the maximum area?

Step 1: Define variables.

Let the length be \(l\) meters and the width be \(w\) meters.

Step 2: Write the objective function.

We want to maximize area:

$$A = lw.$$

Step 3: Write the constraint.

The perimeter is 40 meters:

$$2l + 2w = 40.$$

Step 4: Rewrite using one variable.

Solve for \(w\):

$$w = 20 - l.$$

Substitute into the area formula:

$$A(l) = l(20 - l) = 20l - l^2.$$

Step 5: Identify the domain.

Both dimensions must be positive, so

$$0 < l < 20.$$

Step 6: Find the maximum.

The function

$$A(l) = -l^2 + 20l$$

is a downward-opening parabola, so its maximum occurs at the vertex.

For \(ax^2 + bx + c\), the vertex is at

$$x = -\frac{b}{2a}.$$

Here, \(a = -1\) and \(b = 20\), so

$$l = -\frac{20}{2(-1)} = 10.$$

Then

$$w = 20 - 10 = 10.$$

Step 7: Interpret.

The rectangle with maximum area is a 10 m by 10 m square.

The maximum area is

$$A = 10 \cdot 10 = 100 \text{ m}^2.$$

Key lesson: with a fixed perimeter, the rectangle of greatest area is a square.

Worked Example 2: Minimizing cost under a production constraint

A school is buying notebooks and folders for an event. Notebooks cost \(\$2\) each, and folders cost \(\$3\) each. The school needs a total of 100 items. How many notebooks and folders should it buy to minimize cost?

Step 1: Define variables.

Let \(n\) be the number of notebooks and \(f\) be the number of folders.

Step 2: Objective function.

The total cost is

$$C = 2n + 3f.$$

We want to minimize \(C\).

Step 3: Constraint.

The total number of items is 100:

$$n + f = 100.$$

Step 4: Rewrite with one variable.

Solve for \(f\):

$$f = 100 - n.$$

Substitute into the cost function:

$$C(n) = 2n + 3(100 - n).$$ $$C(n) = 2n + 300 - 3n = 300 - n.$$

Step 5: Domain.

Since the school cannot buy a negative number of items,

$$0 \leq n \leq 100.$$

Also, the values should be whole numbers.

Step 6: Find the minimum.

The function

$$C(n) = 300 - n$$

decreases as \(n\) increases. So the minimum cost happens when \(n\) is as large as possible.

The largest possible value is \(n = 100\). Then

$$f = 0.$$

Step 7: Interpret.

To minimize cost, the school should buy 100 notebooks and 0 folders.

The minimum cost is

$$C = 2(100) + 3(0) = 200.$$

Key lesson: when one item is always cheaper and there are no other restrictions, the minimum cost often happens at an endpoint of the domain.

Worked Example 3: Maximizing revenue with a price model

A student club sells tickets for a fundraiser. If the ticket price is \(\$10\), then 120 people will attend. For each \(\$1\) increase in price, attendance drops by 5 people. What ticket price gives the maximum revenue?

This is a realistic modeling problem because changing one variable, price, affects another variable, attendance.

Step 1: Define a variable.

Let \(x\) be the number of \(\$1\) price increases.

Then:

• New ticket price: \(10 + x\)
• New attendance: \(120 - 5x\)

Step 2: Objective function.

Revenue equals price times attendance:

$$R = (10 + x)(120 - 5x).$$

Step 3: Expand.

$$R(x) = 1200 - 50x + 120x - 5x^2$$ $$R(x) = -5x^2 + 70x + 1200.$$

Step 4: Domain.

Attendance cannot be negative:

$$120 - 5x \geq 0.$$

So

$$x \leq 24.$$

Also, in this situation, \(x \geq 0\) because we are only considering price increases. So the domain is

$$0 \leq x \leq 24.$$

Step 5: Find the maximum.

This is a downward-opening quadratic, so the maximum is at the vertex:

$$x = -\frac{b}{2a} = -\frac{70}{2(-5)} = 7.$$

Step 6: Interpret.

The best ticket price is

$$10 + 7 = 17.$$

The attendance will be

$$120 - 5(7) = 85.$$

The maximum revenue is

$$R = 17 \cdot 85 = 1445.$$

So the club should charge \(\$17\) per ticket, and the maximum revenue is \(\$1445\).

Key lesson: in many business models, revenue can be written as a quadratic, and the maximum occurs at the vertex.

Worked Example 4: Minimizing material for a rectangular pen beside a wall

A farmer wants to build a rectangular pen against a barn wall, so only three sides need fencing. The pen must have an area of 200 square meters. What dimensions use the least fencing?

Step 1: Define variables.

Let \(x\) be the width perpendicular to the wall, and let \(y\) be the length along the wall.

Step 2: Objective function.

Only three sides are fenced, so the total fencing is

$$F = 2x + y.$$

We want to minimize \(F\).

Step 3: Constraint.

The area must be 200 square meters:

$$xy = 200.$$

Step 4: Rewrite using one variable.

Solve for \(y\):

$$y = \frac{200}{x}.$$

Substitute into the fencing formula:

$$F(x) = 2x + \frac{200}{x}.$$

Step 5: Domain.

Since dimensions must be positive,

$$x > 0.$$

Step 6: Find the minimum.

This function is not quadratic, so a graph or table is helpful.

Try some values:

• If \(x = 5\), then \(y = 40\), so \(F = 2(5) + 40 = 50\)
• If \(x = 10\), then \(y = 20\), so \(F = 2(10) + 20 = 40\)
• If \(x = 8\), then \(y = 25\), so \(F = 16 + 25 = 41\)
• If \(x = 12\), then \(y \approx 16.67\), so \(F \approx 24 + 16.67 = 40.67\)

The smallest fencing amount from these values occurs near \(x = 10\).

Then

$$y = \frac{200}{10} = 20.$$

Step 7: Interpret.

The dimensions that use the least fencing are approximately 10 m by 20 m.

The minimum amount of fencing is

$$F = 40 \text{ m}.$$

Key lesson: not every optimization function is quadratic, but the same modeling process still works: define variables, use the constraint, reduce to one variable, check the realistic domain, and compare values carefully.

Tips for solving optimization problems

• Read carefully. Decide what quantity must be made largest or smallest.
• Name variables clearly. Include units if possible.
• Write the constraint exactly. Many mistakes happen here.
• Substitute to get one variable. This makes the problem easier to analyze.
• Check the domain. Real-world answers must make sense.
• Use the right tool. Vertex for quadratics, endpoints for restricted intervals, or a table/graph for other functions.
• Answer in words. State what the numbers mean in the context.

Common mistakes to avoid

• Optimizing the wrong quantity. For example, confusing area with perimeter.
• Forgetting the constraint.
• Not restricting the domain to realistic values.
• Finding the best value of the variable but forgetting to calculate the actual maximum or minimum quantity.
• Ignoring endpoints when the domain is limited.

How optimization connects to advanced mathematical modeling

In advanced modeling, you do more than solve an equation. You create a mathematical description of a real situation, examine what assumptions you made, and decide whether the model makes sense.

For example:

• In geometry, you may maximize area or minimize material.
• In algebra, you may model cost, revenue, or profit.
• In statistics, you may compare outcomes and decide which choice is most efficient.

A good model includes:

• reasonable variables,
• correct relationships,
• clear constraints,
• and answers that are realistic in context.

Brief Summary

Optimization under constraints means finding the greatest or least possible value of a quantity while following a rule or limit.

To solve these problems, define variables, write the objective function and constraint, substitute to get one variable, find the realistic domain, and then determine the absolute maximum or minimum.

Always check that your answer makes sense in the real situation. In mathematical modeling, the best answer is not just mathematically correct—it must also fit the context.

Modeling Periodic Phenomena means using mathematics to describe situations that repeat in a regular cycle.

Many real-world patterns are periodic: ocean tides rise and fall, daily temperatures warm and cool, and sound waves repeat rapidly. A very useful way to model these patterns is with sinusoidal functions, which are based on sine and cosine.

In this lesson, you will learn how to recognize periodic behavior, identify the important features of a cycle, and build a sinusoidal equation from data.

1. What is a periodic phenomenon?

A phenomenon is periodic if it repeats after a fixed amount of time or distance. That repeating amount is called the period.

• A tide may repeat every 12.4 hours.
• A temperature pattern may repeat every 24 hours.
• A seasonal pattern may repeat every 12 months.

If the graph rises and falls in a regular, wave-like way, a sinusoidal model may be appropriate.

2. The basic sinusoidal functions

The two most common periodic functions are:

$$y=A\sin(Bx)+D$$

and

$$y=A\cos(Bx)+D$$

These equations can also include a horizontal shift:

$$y=A\sin(B(x-C))+D$$

$$y=A\cos(B(x-C))+D$$

Each parameter has a meaning:

• Amplitude: \(|A|\) tells how far the graph moves above and below its middle line.
• Period: the length of one complete cycle.
• Midline: \(y=D\), the average value of the oscillation.
• Horizontal shift: \(C\), which moves the graph left or right.

3. Key measurements from data

When fitting a sinusoidal model to data, the first step is to find the main features of the cycle.

Suppose the data has a maximum value and a minimum value. Then:

$$\text{Amplitude}=\frac{\text{maximum}-\text{minimum}}{2}$$

$$\text{Midline}=\frac{\text{maximum}+\text{minimum}}{2}$$

If one full cycle takes \(P\) units, then the parameter \(B\) is:

$$B=\frac{2\pi}{P}$$

This is important because sine and cosine naturally repeat every \(2\pi\), so we scale the input to match the real-world period.

4. Choosing sine or cosine

You can often use either sine or cosine to model the same data, as long as you use the correct horizontal shift.

• Use cosine if the cycle starts at a maximum or minimum.
• Use sine if the cycle starts at the midline and moves upward or downward.

In practice, both are acceptable if they fit the data correctly.

5. A step-by-step strategy for building a model

1. Identify the maximum and minimum values.
2. Calculate the amplitude.
3. Find the midline.
4. Determine the period from the repeating pattern.
5. Compute \(B=\frac{2\pi}{P}\).
6. Choose sine or cosine based on where the cycle begins.
7. Add a horizontal shift if needed so the graph matches the data.
8. Check the model against key points in the data.

Worked Example 1: Modeling a Ferris Wheel

A rider on a Ferris wheel is 1 meter above the ground at the lowest point and 21 meters above the ground at the highest point. The wheel completes one rotation every 40 seconds. At time \(t=0\), the rider is at the highest point. Write a sinusoidal model for the rider's height.

Step 1: Find amplitude

$$A=\frac{21-1}{2}=10$$

Step 2: Find the midline

$$D=\frac{21+1}{2}=11$$

Step 3: Find \(B\)

The period is \(P=40\), so

$$B=\frac{2\pi}{40}=\frac{\pi}{20}$$

Step 4: Choose the function

Since the rider starts at the maximum height, cosine is a good choice.

$$h(t)=10\cos\left(\frac{\pi}{20}t\right)+11$$

Check: At \(t=0\), \(\cos(0)=1\), so

$$h(0)=10(1)+11=21$$

This matches the highest point.

Worked Example 2: Daily Temperature Model

Suppose the temperature in a city varies periodically through the day. The minimum temperature is \(12^\circ\text{C}\) and the maximum is \(28^\circ\text{C}\). The cycle repeats every 24 hours. Assume the temperature reaches its minimum at midnight, \(t=0\). Build a model.

Step 1: Amplitude

$$A=\frac{28-12}{2}=8$$

Step 2: Midline

$$D=\frac{28+12}{2}=20$$

Step 3: Period parameter

$$B=\frac{2\pi}{24}=\frac{\pi}{12}$$

Step 4: Choose a function

At \(t=0\), the temperature is at a minimum. A cosine graph with negative amplitude works well:

$$T(t)=-8\cos\left(\frac{\pi}{12}t\right)+20$$

Check:

At \(t=0\),

$$T(0)=-8(1)+20=12$$

which is the minimum temperature.

Worked Example 3: Fitting a Sinusoidal Model from Data

A tide is measured over time. The water level is highest at 7 meters and lowest at 3 meters. A full cycle takes 12 hours. The tide is at its average level and rising at \(t=0\). Find a model.

Step 1: Amplitude

$$A=\frac{7-3}{2}=2$$

Step 2: Midline

$$D=\frac{7+3}{2}=5$$

Step 3: Find \(B\)

$$B=\frac{2\pi}{12}=\frac{\pi}{6}$$

Step 4: Choose the function

The tide begins at the midline and is rising, which matches the basic sine pattern.

$$H(t)=2\sin\left(\frac{\pi}{6}t\right)+5$$

Check: At \(t=0\), \(\sin(0)=0\), so

$$H(0)=5$$

This is the average level, and the graph rises after that, as required.

6. Using a horizontal shift

Sometimes the cycle does not begin at a maximum, minimum, or midline. In that case, a horizontal shift helps line up the model with the data.

For example, suppose a wave has amplitude 4, midline 10, and period 8, and it reaches a maximum at \(t=3\).

First find \(B\):

$$B=\frac{2\pi}{8}=\frac{\pi}{4}$$

Since cosine reaches a maximum when its input is 0, shift the graph so that \(t=3\) gives input 0:

$$y=4\cos\left(\frac{\pi}{4}(t-3)\right)+10$$

This means the peak occurs at the correct time.

Worked Example 4: Acoustic Wave Model

A sound sensor records a wave with maximum value 6 and minimum value -2. One cycle lasts 0.5 seconds. The wave reaches a maximum at \(t=0.1\) seconds. Write a model.

Step 1: Amplitude

$$A=\frac{6-(-2)}{2}=4$$

Step 2: Midline

$$D=\frac{6+(-2)}{2}=2$$

Step 3: Period parameter

$$B=\frac{2\pi}{0.5}=4\pi$$

Step 4: Use cosine with a shift

Because the maximum occurs at \(t=0.1\), use:

$$y=4\cos(4\pi(t-0.1))+2$$

Check: When \(t=0.1\), the cosine input is 0, so

$$y=4(1)+2=6$$

That matches the maximum value.

7. Interpreting a sinusoidal model

Once you have a model, you should be able to explain what each part means in context.

For example, in

$$y=3\sin\left(\frac{\pi}{6}t\right)+12$$

• The amplitude is 3, so the quantity changes 3 units above and below the average.
• The midline is \(y=12\), so the average value is 12.
• The period is

$$P=\frac{2\pi}{\pi/6}=12$$

So the cycle repeats every 12 time units.

8. Evaluating whether a sinusoidal model makes sense

Not every repeating-looking pattern is perfectly sinusoidal. Real data may have noise, measurement error, or uneven cycles.

When deciding whether a sinusoidal model is reasonable, ask:

• Does the data repeat regularly?
• Are the highs and lows spaced about evenly?
• Is there a roughly constant midline?
• Is the change smooth rather than abrupt?

If the answers are mostly yes, a sinusoidal model is often useful, even if it is not perfect.

9. Common mistakes to avoid

• Confusing amplitude and maximum value: amplitude is half the distance from maximum to minimum, not the maximum itself.
• Forgetting the midline: many real situations do not oscillate around 0.
• Using the period directly as \(B\): remember that \(B=\frac{2\pi}{P}\).
• Ignoring the starting position: make sure the equation matches whether the data begins at a peak, trough, or midline.
• Forgetting units: period and horizontal shift must match the units of the input variable.

10. Quick guide for fitting a sinusoidal function from empirical data

• Find a high point and a low point from the data.
• Compute amplitude and midline.
• Estimate the cycle length to get the period.
• Convert period to \(B\).
• Pick sine or cosine based on the starting point.
• Shift the graph if the cycle starts later or earlier than the basic pattern.
• Test the model with known data points.

Summary

Periodic phenomena repeat in a regular cycle, and sinusoidal functions are powerful tools for modeling them.

To build a sinusoidal model, find the amplitude, midline, and period, then choose sine or cosine and apply any needed horizontal shift.

With practice, you can use these models to describe and predict real situations such as tides, temperatures, wheel motion, and waves.

Exponential and Logistic Growth Models

In many real-world situations, a quantity changes over time. Populations grow, diseases spread, rumors travel, and new technologies are adopted. Mathematics helps us describe these patterns using growth models.

Two important models are exponential growth and logistic growth. Exponential growth describes situations where the amount keeps increasing at a rate proportional to its current size, with no limit. Logistic growth describes situations where growth starts quickly but slows down because of limited resources, space, or other constraints.

In 11th Grade mathematical modeling, it is important not only to use these equations, but also to decide which model makes sense for a given situation and to interpret what the numbers mean in context.

1. Exponential Growth

Exponential growth happens when the rate of change depends on the current amount. The more there is now, the faster it grows.

A common exponential growth model is

$$P(t)=P_0(1+r)^t$$

where:

• \(P(t)\) is the amount after time \(t\),

• \(P_0\) is the initial amount,

• \(r\) is the growth rate written as a decimal,

• \(t\) is time.

For example, if a population starts at 500 and grows by 8% each year, then

$$P(t)=500(1.08)^t$$

This means each year, the population is multiplied by 1.08.

Key features of exponential growth:

• The graph rises faster and faster over time.

• There is no built-in maximum.

• The percent growth rate stays constant.

• It works well over short periods when resources are not yet limited.

When is exponential growth reasonable?

• Early stages of population growth in a new environment

• Early spread of an infection when few people are infected

• Money growing with compound interest

2. Logistic Growth

In real life, unlimited growth usually does not continue forever. Food, space, medical intervention, immunity, or social behavior can slow growth. This is where logistic growth is useful.

A logistic model includes a carrying capacity, which is the greatest amount the environment can support in the long run.

A common logistic function is

$$P(t)=\frac{K}{1+Ae^{-bt}}$$

where:

• \(K\) is the carrying capacity,

• \(A\) and \(b\) are positive constants,

• \(t\) is time.

The value of \(K\) is especially important because it represents the upper limit the model approaches.

Key features of logistic growth:

• Growth is slow at first.

• Then growth becomes rapid.

• Later, growth slows down again.

• The graph is S-shaped, also called a sigmoid curve.

• The output approaches \(K\), but does not exceed it in the model.

When is logistic growth reasonable?

• Population growth with limited food or habitat

• Spread of a disease in a population where many eventually become immune or recover

• Adoption of a product in a fixed market

3. Comparing the Models

• Exponential growth has no limit and keeps increasing faster.

• Logistic growth has a limit and eventually levels off.

• Exponential growth is often useful at the beginning of a process.

• Logistic growth is better when there are constraints or a maximum possible amount.

In modeling, choosing the right model is just as important as solving the equation.

4. Understanding Parameters in Context

When you see a model, each number should mean something in the real situation.

For exponential growth:

$$P(t)=P_0(1+r)^t$$

• \(P_0\): starting amount

• \(r\): percent increase per time period

• \(t\): number of time periods

For logistic growth:

$$P(t)=\frac{K}{1+Ae^{-bt}}$$

• \(K\): maximum sustainable population or total possible cases

• \(A\): related to the starting condition

• \(b\): how quickly the quantity grows

Even if you are not asked to derive these constants, you should be able to interpret them.

5. Worked Example 1: Basic Exponential Growth

A bacteria culture starts with 200 bacteria and grows by 25% each hour. Write a model and find the number of bacteria after 5 hours.

Step 1: Identify the initial amount and growth rate.

• Initial amount: \(P_0=200\)

• Growth rate: \(r=0.25\)

Step 2: Write the exponential model.

$$P(t)=200(1.25)^t$$

Step 3: Substitute \(t=5\).

$$P(5)=200(1.25)^5$$ $$P(5)=200(3.0517578125)\approx 610.35$$

Since population is usually counted in whole numbers, we round to about 610 bacteria.

Interpretation: Because the growth rate stays at 25% per hour, the culture grows faster as time goes on.

6. Worked Example 2: Finding Time in an Exponential Model

A town has 8,000 people. A flu outbreak begins with 120 infected people, and the number of infected people grows by 40% per day during the first few days. After how many days will the model predict about 1,000 infected people?

Step 1: Write the model.

$$I(t)=120(1.4)^t$$

We want to find \(t\) when \(I(t)=1000\).

$$1000=120(1.4)^t$$

Step 2: Divide by 120.

$$\frac{1000}{120}=(1.4)^t$$ $$8.333\ldots=(1.4)^t$$

Step 3: Solve using trial, calculator, or logarithms.

Using logarithms:

$$t=\frac{\log(8.333\ldots)}{\log(1.4)}$$ $$t\approx \frac{0.9208}{0.1461}\approx 6.30$$

So the model predicts about 1,000 infected people after 6.3 days.

Interpretation: This exponential model may work only at the start of the outbreak. It will eventually become unrealistic because the number of infected people cannot grow beyond the town's population.

7. Worked Example 3: Reading a Logistic Model

A deer population in a wildlife area is modeled by

$$P(t)=\frac{1200}{1+11e^{-0.4t}}$$

where \(t\) is measured in years.

Find:

1. the carrying capacity,

2. the initial population,

3. the population after 4 years.

Step 1: Identify the carrying capacity.

In a logistic model, the number on top is \(K\).

$$K=1200$$

So the carrying capacity is 1200 deer.

Step 2: Find the initial population.

The initial population is \(P(0)\).

$$P(0)=\frac{1200}{1+11e^{0}}$$ $$P(0)=\frac{1200}{1+11(1)}=\frac{1200}{12}=100$$

So the initial population is 100 deer.

Step 3: Find the population after 4 years.

$$P(4)=\frac{1200}{1+11e^{-0.4(4)}}$$ $$P(4)=\frac{1200}{1+11e^{-1.6}}$$

Using a calculator, \(e^{-1.6}\approx 0.2019\).

$$P(4)=\frac{1200}{1+11(0.2019)}$$ $$P(4)=\frac{1200}{1+2.2209}=\frac{1200}{3.2209}\approx 372.6$$

So after 4 years, the population is about 373 deer.

Interpretation: The population grows quickly at first, but it will gradually slow down and level off near 1200 deer.

8. Worked Example 4: Choosing Between Exponential and Logistic Growth

A new disease appears in a school district. At first, the number of reported cases rises rapidly. After several weeks, the increase slows because many students have recovered, stayed home, or become less likely to catch the disease.

Which model is more appropriate over the entire outbreak: exponential or logistic?

Answer: The better model is logistic growth.

Why?

• Exponential growth may describe the early stage, when cases rise quickly.

• But exponential growth assumes unlimited increase, which is not realistic in a school district with a limited number of students.

• Logistic growth accounts for a maximum number of possible cases and for the slowing of growth over time.

Modeling idea: In epidemiology, exponential growth can be useful for short-term predictions early on, but logistic growth is often better once limits and slowing effects matter.

9. How to Tell Which Model to Use

Ask yourself these questions:

• Is the quantity growing by a constant percent over equal time intervals?

• Is there no important limit in the short term?

If yes, an exponential model may fit.

• Does growth begin fast but then slow down?

• Is there a maximum possible value?

• Are resources or population size limited?

If yes, a logistic model may fit better.

10. Common Mistakes to Avoid

• Using percent as a whole number instead of a decimal: 12% should be written as \(0.12\), not 12.

• Forgetting the growth factor: If the rate is 12%, use \(1.12\), not just \(0.12\).

• Using exponential growth forever: Many real situations eventually hit limits.

• Misreading carrying capacity: In a logistic model, \(K\) is the long-term upper limit.

• Ignoring units: Time might be in days, months, or years, and the rate must match that unit.

11. Real-World Interpretation

Mathematical models are useful, but they are only approximations. A population may not grow exactly according to one equation forever. Weather, migration, medical treatment, policy changes, and human behavior can affect the results.

That is why advanced mathematical modeling includes not only writing equations, but also evaluating whether the model still makes sense and refining it when needed.

For example:

• An exponential model may fit early disease spread but fail later.

• A logistic model may fit a limited habitat better than an exponential one.

• Data collected over time can help decide whether the model should be changed.

12. Brief Summary

Exponential growth models describe situations where a quantity grows by a constant percent and has no built-in limit:

$$P(t)=P_0(1+r)^t$$

Logistic growth models describe situations where growth slows as the quantity approaches a maximum value called the carrying capacity:

$$P(t)=\frac{K}{1+Ae^{-bt}}$$

Use exponential growth for early, unrestricted growth and logistic growth when limits matter. In real-world modeling, always interpret the parameters and decide whether the model matches the situation.

Regression Analysis and Model Selection helps us answer an important question in mathematical modeling: Which equation best describes real-world data?

In many situations, data points do not lie exactly on a single line or curve. Instead, we use regression to find an equation that fits the data as closely as possible.

In 11th Grade Maths, you will often compare linear, quadratic, and exponential models. Then you will decide which model is best by looking at the graph, the residuals, and the value of R^2, called R-squared.

This lesson explains what regression is, how to use it, and how to choose the best model for a set of data.

1. What is regression analysis?

Regression analysis is the process of using data to find an equation that models the relationship between two variables.

If the input variable is called \(x\) and the output variable is called \(y\), regression finds an equation that predicts \(y\) from \(x\).

Common regression models in this course include:

• Linear regression: \(y = mx + b\)
• Quadratic regression: \(y = ax^2 + bx + c\)
• Exponential regression: \(y = ab^x\)

Each model has a different shape:

• A linear model forms a straight line and represents a constant rate of change.
• A quadratic model forms a parabola and is useful when data rises then falls, or falls then rises.
• An exponential model represents repeated percent growth or decay.

2. Why model selection matters

Different equations can fit the same data with different levels of accuracy. Choosing the wrong type of model can lead to poor predictions.

For example:

• If data grows by about the same amount each time, a linear model may work well.
• If data changes direction, a quadratic model may fit better.
• If data grows by the same percent each time, an exponential model is often best.

So, model selection means testing several reasonable models and comparing how well they fit the data.

3. Using technology for regression

In real data sets, calculating the best-fit equation by hand can be time-consuming. That is why students use technology such as:

• graphing calculators,
• spreadsheets,
• Desmos,
• statistical software.

These tools can quickly generate regression equations and values such as \(R^2\).

Even when technology does the calculations, you must still decide whether the model makes sense. A high \(R^2\) alone does not always guarantee the best choice.

4. Understanding residuals

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

The formula is:

$$ \text{residual} = \text{actual value} - \text{predicted value} $$

If a residual is positive, the actual point is above the model. If a residual is negative, the actual point is below the model.

Small residuals mean the model fits the data well. Large residuals mean the model misses the data by more.

When studying residuals, we often look at a residual plot. This graph shows residuals instead of original \(y\)-values.

A good model usually has residuals that:

• are close to 0,
• are scattered randomly,
• do not show a clear pattern.

If the residual plot shows a pattern, that suggests the model is not capturing the true relationship well.

5. Understanding \(R^2\)

The value \(R^2\), or R-squared, measures how well a regression model fits the data.

Its value is between 0 and 1, or sometimes written as a percent between 0% and 100%.

An \(R^2\) value closer to 1 means the model explains more of the variation in the data.

For example:

• \(R^2 = 0.95\) means the model fits very well.
• \(R^2 = 0.62\) means the fit is much weaker.

In general, when comparing models for the same data, a larger \(R^2\) suggests a better fit.

However, you should also check:

• whether the graph shape matches the data,
• whether the residuals are random,
• whether the model makes sense in context.

6. Recognizing which model might fit

Before using technology, it helps to look at the data and guess the possible model type.

• Linear: points lie around a straight line; first differences are roughly constant.
• Quadratic: points form a curved pattern with a turning point; second differences are roughly constant.
• Exponential: points increase or decrease faster over time; ratios are roughly constant.

These patterns help you choose which regressions to test first.

7. Worked Example 1: Choosing between linear and exponential

A town tracks the number of visitors to a nature center over 5 years.

Year \((x)\)	0	1	2	3	4
Visitors \((y)\)	120	150	180	210	240

Step 1: Look for a pattern.

The number of visitors increases by about 30 each year:

$$ 150-120=30, \quad 180-150=30, \quad 210-180=30, \quad 240-210=30 $$

A constant difference suggests a linear model.

Step 2: Write a likely model.

The starting value is 120 when \(x=0\), and the rate of change is 30, so a good model is:

$$ y = 30x + 120 $$

Step 3: Check fit.

This model predicts every value exactly, so all residuals are 0. That means the model is an excellent fit.

Conclusion: A linear regression is the best model.

8. Worked Example 2: Quadratic model

A ball is thrown upward, and its height is measured at different times.

Time \((x)\) seconds	0	1	2	3	4
Height \((y)\) meters	2	8	10	8	2

Step 1: Look at the shape.

The height rises, reaches a maximum, and then falls. That is not linear. It matches the shape of a quadratic model.

Step 2: Check differences.

First differences:

$$ 8-2=6, \quad 10-8=2, \quad 8-10=-2, \quad 2-8=-6 $$

Second differences:

$$ 2-6=-4, \quad -2-2=-4, \quad -6-(-2)=-4 $$

Since the second differences are constant, a quadratic model fits.

Step 3: Use technology.

A quadratic regression might give:

$$ y = -2x^2 + 8x + 2 $$

Step 4: Check one value.

When \(x=2\):

$$ y = -2(2)^2 + 8(2) + 2 = -8 + 16 + 2 = 10 $$

This matches the table exactly.

Conclusion: A quadratic regression is the best choice because the data has a turning point and the residuals would be 0 for this data set.

9. Worked Example 3: Exponential growth and model comparison

A bacteria culture is measured every hour.

Hour \((x)\)	0	1	2	3	4
Population \((y)\)	50	75	112	168	252

Step 1: Look for a pattern.

The differences are not constant:

$$ 75-50=25, \quad 112-75=37, \quad 168-112=56, \quad 252-168=84 $$

But the ratios are about constant:

$$ \frac{75}{50}=1.5, \quad \frac{112}{75}\approx 1.49, \quad \frac{168}{112}=1.5, \quad \frac{252}{168}=1.5 $$

A nearly constant ratio suggests an exponential model.

Step 2: Write the model form.

For exponential growth, use:

$$ y = ab^x $$

The initial amount is 50, so \(a=50\). The growth factor is about 1.5, so:

$$ y = 50(1.5)^x $$

Step 3: Check predictions.

At \(x=3\):

$$ y = 50(1.5)^3 = 50(3.375) = 168.75 $$

This is very close to the actual value 168.

Step 4: Compare models.

If technology gives a higher \(R^2\) for exponential regression than for linear regression, and the residual plot for the exponential model is more random, then the exponential model is the better choice.

Conclusion: The best-fit model is exponential.

10. Worked Example 4: Using residuals and \(R^2\) to choose the best model

Suppose technology tests the same data with three models and gives these results:

• Linear regression: \(R^2 = 0.89\)
• Quadratic regression: \(R^2 = 0.98\)
• Exponential regression: \(R^2 = 0.91\)

The residual plots show:

• Linear: clear curved pattern
• Quadratic: random scatter around 0
• Exponential: slight pattern

Which model should be selected?

The quadratic model should be selected because:

• it has the highest \(R^2\),
• its residuals are randomly scattered,
• the other models show patterns in their residuals, meaning those models miss some structure in the data.

Conclusion: The quadratic regression is the best-fit model.

11. A step-by-step process for model selection

When you are given a data set, use this process:

1. Graph the data. Look at the overall shape.
2. Guess possible models. Decide whether linear, quadratic, or exponential seems reasonable.
3. Use technology to run regressions. Find the equations and \(R^2\) values.
4. Compare residuals. Look for small, random scatter around 0.
5. Choose the best model. Select the one with a strong fit, reasonable shape, and meaningful context.
6. Use the model carefully. Make predictions only when they make sense for the situation.

12. Important notes about interpretation

A model is useful because it helps us make predictions, but all models have limits.

You should be careful when predicting values far beyond the data set. This is called extrapolation, and it can be unreliable because the pattern may change.

Predictions within the range of the data are usually more trustworthy. This is called interpolation.

Also, even if a model has a high \(R^2\), it should still make sense in the real-world context. For example, a model predicting a negative population would not be reasonable.

13. Common mistakes to avoid

• Choosing a model only because it has the highest \(R^2\), without checking residuals.
• Using a linear model when the data clearly curves.
• Confusing exponential growth with linear growth.
• Ignoring the meaning of the variables in context.
• Making predictions far outside the data range without caution.

14. Quick comparison of model types

• Linear: constant difference, straight-line trend, equation of form \(y=mx+b\)
• Quadratic: turning point, curved parabola, equation of form \(y=ax^2+bx+c\)
• Exponential: constant ratio or percent change, rapid growth or decay, equation of form \(y=ab^x\)

15. Brief summary

Regression analysis is used to find an equation that best fits a set of data. In this course, the most common choices are linear, quadratic, and exponential models.

To choose the best model, you should examine the graph, compare \(R^2\) values, and study residuals. A good model matches the shape of the data, has a high \(R^2\), and produces residuals with no clear pattern.

Model selection is not just about technology. It is about using mathematical reasoning to decide which equation best represents a real-world situation.

Lesson: Parameter Sensitivity and Validation

In mathematical modeling, we often create an equation or rule to represent a real-world situation. For example, we might model population growth, business profit, travel time, or the spread of a rumor. A model is useful because it helps us make predictions.

But real-world models are never perfect. They depend on parameters, which are fixed values in the model that represent important features of the situation. If those values are estimated incorrectly, the model’s prediction may change. This is why we study parameter sensitivity and validation.

Parameter sensitivity means checking how much a model’s output changes when one of its parameters changes a little.

Validation means testing whether the model gives results that are close to real data or reasonable for the situation.

These ideas help us answer important questions:

• Which parts of the model matter the most?
• Are small errors in the inputs causing large errors in the prediction?
• Can we trust the model enough to use it?

1. What is a parameter?

A parameter is a number in a model that stays constant for one version of the model, but could have a different value in another situation.

For example, in the linear model

$$y = mx + b$$

the values of \(m\) and \(b\) are parameters. They control the shape and position of the line.

In an exponential growth model

$$P(t) = P_0(1+r)^t$$

the values \(P_0\) and \(r\) are parameters. Here:

• \(P_0\) is the starting amount,
• \(r\) is the growth rate.

If we change a parameter, the model’s prediction changes. The key question is how much it changes.

2. Understanding parameter sensitivity

A model is sensitive to a parameter if a small change in that parameter causes a large change in the output.

A model is not very sensitive to a parameter if changing it slightly only changes the output a little.

Suppose a model for cost is

$$C = 50n + 200$$

where \(n\) is the number of items made. The parameter 50 represents cost per item, and 200 represents fixed cost.

If the cost per item changes from 50 to 52, then the new model is

$$C = 52n + 200$$

For \(n = 10\), the cost changes from

$$50(10) + 200 = 700$$

to

$$52(10) + 200 = 720$$

That is a change of only 20.

But for \(n = 500\), the cost changes from

$$50(500) + 200 = 25200$$

to

$$52(500) + 200 = 26200$$

Now the difference is 1000. This shows that sensitivity can also depend on the input values, not just on the parameter itself.

3. Why sensitivity matters in modeling

When we build a model, we often estimate parameter values from incomplete data. For example:

• a growth rate may be estimated from last year’s trend,
• a cost per unit may change because of inflation,
• a speed or time estimate may vary in real life.

If the model is very sensitive to one parameter, then we need to estimate that parameter carefully. Even a small mistake could make the prediction unreliable.

If the model is not very sensitive to a parameter, then small estimation errors may not matter much.

This helps us decide:

• which values need the most attention,
• which assumptions are safe,
• whether the model is strong enough for decision-making.

4. How to test parameter sensitivity

A simple way to test sensitivity is to change one parameter at a time while keeping the others fixed.

This process often follows these steps:

1. Choose a model and identify its parameters.
2. Find the original prediction.
3. Change one parameter slightly, such as by \(+5\%\) or \(-5\%\).
4. Recalculate the prediction.
5. Compare the new result to the original result.

You can measure the change using:

• difference: new output minus original output,
• percent change:

$$\text{Percent change} = \frac{\text{new output} - \text{original output}}{\text{original output}} \times 100\%$$

If a small parameter change causes a large percent change in output, the model is sensitive to that parameter.

5. Worked Example 1: Linear model sensitivity

A school fundraiser model is

$$R = 8s + 150$$

where \(R\) is total revenue in dollars and \(s\) is the number of shirts sold.

The parameter 8 means each shirt sells for \(\$8\), and 150 is an initial donation.

Suppose \(s = 40\).

Step 1: Find the original prediction.

$$R = 8(40) + 150 = 320 + 150 = 470$$

Step 2: Change the price parameter slightly.

If the price was actually \(\$8.50\) per shirt, the new model is

$$R = 8.5s + 150$$

Step 3: Find the new prediction.

$$R = 8.5(40) + 150 = 340 + 150 = 490$$

Step 4: Compare the results.

The revenue changed by

$$490 - 470 = 20$$

The percent change in revenue is

$$\frac{20}{470} \times 100\% \approx 4.26\%$$

Conclusion: A \(\$0.50\) change in shirt price caused about a \(4.26\%\) change in predicted revenue when 40 shirts are sold.

6. Worked Example 2: Exponential model sensitivity

A bacteria culture is modeled by

$$P(t) = 500(1.12)^t$$

where \(t\) is time in hours.

The growth rate parameter is \(12\%\), or \(r = 0.12\).

Find how sensitive the prediction at \(t = 6\) is to a small change in growth rate.

Step 1: Original prediction

$$P(6) = 500(1.12)^6$$ $$P(6) \approx 500(1.9738) \approx 986.9$$

Step 2: Increase the growth rate from \(12\%\) to \(13\%\)

New model:

$$P(t) = 500(1.13)^t$$

New prediction:

$$P(6) = 500(1.13)^6$$ $$P(6) \approx 500(2.0800) \approx 1040.0$$

Step 3: Compare

Difference:

$$1040.0 - 986.9 = 53.1$$

Percent change in output:

$$\frac{53.1}{986.9} \times 100\% \approx 5.38\%$$

Conclusion: Increasing the growth rate by only 1 percentage point changed the prediction after 6 hours by about \(5.38\%\). Exponential models are often quite sensitive to growth rate.

7. Validation: checking whether the model makes sense

Even if a model is written correctly, it still may not fit reality well. That is why we validate it.

To validate a model means to compare its predictions with actual data or with what should reasonably happen in the real world.

Common validation questions include:

• Does the model match known data points?
• Are the predictions close to observed results?
• Do the outputs make sense in context?
• Does the model work only for a short time, or for a wide range of inputs?

A model can fail validation for several reasons:

• important variables were left out,
• parameter values were estimated poorly,
• the chosen type of model was not appropriate,
• the model is being used outside the range where it makes sense.

8. Worked Example 3: Validating a model with data

A student models the height of a plant with

$$h(t) = 2.5t + 8$$

where \(h\) is height in centimeters and \(t\) is time in weeks.

The student records actual data:

• At \(t = 2\), actual height is 13 cm
• At \(t = 4\), actual height is 18 cm
• At \(t = 6\), actual height is 22 cm

Step 1: Use the model to predict each value.

For \(t = 2\):

$$h(2) = 2.5(2) + 8 = 13$$

For \(t = 4\):

$$h(4) = 2.5(4) + 8 = 18$$

For \(t = 6\):

$$h(6) = 2.5(6) + 8 = 23$$

Step 2: Compare predicted and actual values.

• Week 2: predicted 13, actual 13
• Week 4: predicted 18, actual 18
• Week 6: predicted 23, actual 22

Step 3: Decide whether the model is valid.

The model matches the first two data points exactly and is only off by 1 cm at week 6. This suggests the model is reasonably valid for this time period.

However, we should still be careful. A plant cannot keep growing linearly forever. Later, the growth may slow down. So the model may be valid only for the first several weeks.

9. Sensitivity and validation together

These two ideas work best when used together.

A model may fit past data well, but still be risky if it is extremely sensitive to one parameter. In that case, small errors in measuring that parameter can make future predictions less trustworthy.

On the other hand, a model may not be very sensitive, but still be a poor model if it does not match actual data.

A strong model should:

• fit the data reasonably well,
• use realistic assumptions,
• not change wildly from tiny parameter errors unless that behavior is expected,
• be used only in a sensible domain.

10. Worked Example 4: Sensitivity and validation in one problem

A theater models ticket income with

$$I = pn$$

where \(p\) is ticket price and \(n\) is number of tickets sold.

They expect to sell 300 tickets at \(\$12\) each, so the predicted income is

$$I = 12(300) = 3600$$

Now suppose the estimated number of tickets sold could be off by 10 tickets.

Sensitivity check:

If \(n = 290\), then

$$I = 12(290) = 3480$$

If \(n = 310\), then

$$I = 12(310) = 3720$$

So the income could vary from 3480 to 3720.

Compared to 3600, this is a change of 120 dollars.

Percent change:

$$\frac{120}{3600} \times 100\% = 3.33\%$$

So the model is somewhat sensitive to the ticket estimate, but not extremely so.

Validation check:

Suppose actual sales for a similar event last month were 295 tickets at the same price, giving actual income

$$12(295) = 3540$$

The model predicted 3600, which is off by

$$3600 - 3540 = 60$$

This is a small difference, so the model seems reasonably valid for planning.

11. Questions to ask when evaluating a model

When checking sensitivity and validation, ask yourself:

• What are the important parameters in this model?
• Which parameters are estimated, and how accurate are those estimates?
• What happens if one parameter increases or decreases slightly?
• Does the model still match real data?
• Are the predictions realistic?
• Is the model being used outside the interval where it was built?

12. Common mistakes

• Changing more than one parameter at once when trying to test sensitivity. This makes it harder to tell which parameter caused the output change.
• Assuming a model is correct just because it looks nice. A model must be checked against data.
• Ignoring context. A negative population or impossible height does not make sense, even if the algebra is correct.
• Using a model too far into the future. Many models work only over a limited range.
• Looking only at one data point. Validation is stronger when several data points are checked.

13. Practical strategy for students

When you are asked to analyze a model, use this simple method:

1. Write down the model clearly.
2. Identify the parameters.
3. Calculate the original prediction.
4. Change one parameter slightly.
5. Recalculate and compare the results.
6. Use data or context to decide whether the model is valid.
7. State a conclusion in words.

Brief Summary

Parameter sensitivity tells us how much a model’s prediction changes when a parameter changes a little. Validation tells us whether the model matches real data and makes sense in context. A good mathematical model is not just easy to compute—it is also tested, checked, and used carefully.

Linear Optimization and Constraints is about making the best choice when you have limits.

In real life, people often want to get the most or least of something. For example, a business may want to make the most profit, or a family may want to spend the least money while staying within a budget.

In math, this is called optimization. When the limits are written as math rules, they are called constraints.

In this lesson, you will learn how to:

• understand what optimization means,
• identify constraints,
• write simple linear expressions and inequalities,
• test possible answers,
• find the maximum profit or minimum cost.

1. What does linear optimization mean?

The word linear means the relationships are made with variables that have a power of 1, such as \(x\) or \(y\). A linear expression looks like \(5x + 3y\).

The word optimization means finding the best value. Usually this means:

• maximum: the greatest amount, such as the highest profit,
• minimum: the smallest amount, such as the lowest cost.

So, linear optimization means finding the greatest or smallest value of a linear expression while following certain rules.

2. What are constraints?

Constraints are the limits or restrictions in a problem. They tell you what is allowed and what is not allowed.

Constraints are often written as inequalities, such as:

• \(x + y \leq 10\)
• \(2x + 3y \leq 18\)
• \(x \geq 0\)
• \(y \geq 0\)

These tell us that the values of \(x\) and \(y\) cannot be just anything. They must stay within the limits.

For example, if \(x\) is the number of sandwiches and \(y\) is the number of drinks sold, then:

• \(x \geq 0\) means you cannot sell a negative number of sandwiches,
• \(y \geq 0\) means you cannot sell a negative number of drinks,
• \(x + y \leq 10\) could mean you can sell at most 10 total items.

3. Important parts of an optimization problem

Most linear optimization problems have these parts:

1. Variables: what the numbers stand for.
2. Objective: what you want to maximize or minimize.
3. Constraints: the limits you must follow.

For example:

Let \(x\) = number of fruit cups
Let \(y\) = number of snack bags

If each fruit cup earns \(\$3\) profit and each snack bag earns \(\$2\) profit, then the profit is:

$$P = 3x + 2y$$

This is called the objective function because it is what we want to make as large as possible.

If there are only enough supplies to make 12 items total, then one constraint is:

$$x + y \leq 12$$

We also know:

$$x \geq 0 \quad \text{and} \quad y \geq 0$$

4. How to solve a simple linear optimization problem

At this level, one helpful way is to:

1. Define the variables.
2. Write the objective function.
3. Write the constraints.
4. Find possible solutions that follow all constraints.
5. Test those solutions in the objective function.
6. Choose the greatest or smallest value, depending on the question.

This is often done by making a table of possible values.

Worked Example 1: Maximum profit with one constraint

A student club is selling notebooks and pens.

• A notebook earns \(\$4\) profit.
• A pen earns \(\$2\) profit.
• They can sell at most 10 items total.

Find the maximum profit.

Step 1: Define the variables

Let \(n\) = number of notebooks
Let \(p\) = number of pens

Step 2: Write the objective function

$$\text{Profit} = 4n + 2p$$

Step 3: Write the constraint

$$n + p \leq 10$$$$n \geq 0, \quad p \geq 0$$

Step 4: Test possible solutions

Since notebooks earn more profit than pens, we should check what happens if we sell more notebooks.

• If \(n = 10\) and \(p = 0\), profit is \(4(10) + 2(0) = 40\).
• If \(n = 8\) and \(p = 2\), profit is \(4(8) + 2(2) = 32 + 4 = 36\).
• If \(n = 5\) and \(p = 5\), profit is \(4(5) + 2(5) = 20 + 10 = 30\).
• If \(n = 0\) and \(p = 10\), profit is \(0 + 20 = 20\).

Answer: The maximum profit is \(\$40\) by selling 10 notebooks and 0 pens.

This makes sense because notebooks give more profit per item.

Worked Example 2: Maximum profit with two constraints

A snack stand sells smoothies and granola cups.

• Each smoothie earns \(\$5\).
• Each granola cup earns \(\$3\).
• They can make at most 8 items total.
• They have enough fruit for at most 5 smoothies.

Find the maximum profit.

Step 1: Define the variables

Let \(s\) = number of smoothies
Let \(g\) = number of granola cups

Step 2: Objective function

$$P = 5s + 3g$$

Step 3: Constraints

$$s + g \leq 8$$$$s \leq 5$$$$s \geq 0, \quad g \geq 0$$

Step 4: Find possible solutions

Since smoothies earn more profit, try the largest possible number of smoothies first.

• \(s = 5\), then \(g\) can be at most 3. Profit: \(5(5) + 3(3) = 25 + 9 = 34\).
• \(s = 4\), then \(g\) can be at most 4. Profit: \(5(4) + 3(4) = 20 + 12 = 32\).
• \(s = 3\), then \(g\) can be at most 5. Profit: \(15 + 15 = 30\).
• \(s = 2\), then \(g\) can be at most 6. Profit: \(10 + 18 = 28\).

Answer: The maximum profit is \(\$34\) when the stand sells 5 smoothies and 3 granola cups.

5. Minimum cost problems

Optimization does not always mean getting the biggest number. Sometimes we want the smallest number.

For example, a family may want to buy supplies for the lowest cost while still getting enough of each item.

Worked Example 3: Minimum cost

A teacher needs to buy folders and notebooks for a class.

• Folders cost \(\$2\) each.
• Notebooks cost \(\$3\) each.
• She needs at least 12 total items.
• She must buy at least 4 notebooks.

Find the minimum cost.

Step 1: Define the variables

Let \(f\) = number of folders
Let \(n\) = number of notebooks

Step 2: Objective function

$$C = 2f + 3n$$

We want to minimize \(C\).

Step 3: Constraints

$$f + n \geq 12$$$$n \geq 4$$$$f \geq 0$$

Step 4: Reason through the choices

Folders cost less than notebooks, so to keep cost low, we should use the smallest allowed number of notebooks.

If \(n = 4\), then we need at least 8 folders to make 12 items total.

Cost:

$$C = 2(8) + 3(4) = 16 + 12 = 28$$

Check another choice:

• If \(n = 5\), then \(f = 7\). Cost: \(2(7) + 3(5) = 14 + 15 = 29\).
• If \(n = 6\), then \(f = 6\). Cost: \(12 + 18 = 30\).

Answer: The minimum cost is \(\$28\), with 8 folders and 4 notebooks.

6. How constraints help us eliminate impossible answers

Suppose a problem says:

• \(x + y \leq 6\)
• \(x \geq 0\)
• \(y \geq 0\)

Then:

• \((3,2)\) works because \(3 + 2 = 5\), which is less than or equal to 6.
• \((4,2)\) works because \(4 + 2 = 6\).
• \((5,3)\) does not work because \(5 + 3 = 8\), which is too large.
• \((-1,4)\) does not work because \(x\) cannot be negative.

This is important because only values that follow all constraints are allowed.

7. Thinking about the best choice

When you solve these problems, ask yourself:

• What do the variables mean?
• What am I trying to maximize or minimize?
• What limits must I follow?
• Which choices are possible?
• Which possible choice gives the best result?

Sometimes the item with the larger profit should be chosen more often. But not always. A second constraint may limit how many you can use.

That is why we must check the constraints carefully.

Worked Example 4: A fuller optimization problem

A school store sells water bottles and keychains.

• Each water bottle earns \(\$6\) profit.
• Each keychain earns \(\$4\) profit.
• The store can stock at most 9 items total.
• Because of limited shelf space, it can stock at most 4 water bottles.

Find the maximum profit.

Step 1: Define variables

Let \(w\) = number of water bottles
Let \(k\) = number of keychains

Step 2: Write the objective function

$$P = 6w + 4k$$

Step 3: Write the constraints

$$w + k \leq 9$$$$w \leq 4$$$$w \geq 0, \quad k \geq 0$$

Step 4: Test the most important possible values

Use as many water bottles as possible first, because they earn more profit.

• If \(w = 4\), then \(k\) can be at most 5. Profit: \(6(4) + 4(5) = 24 + 20 = 44\).
• If \(w = 3\), then \(k\) can be at most 6. Profit: \(18 + 24 = 42\).
• If \(w = 2\), then \(k\) can be at most 7. Profit: \(12 + 28 = 40\).
• If \(w = 1\), then \(k\) can be at most 8. Profit: \(6 + 32 = 38\).
• If \(w = 0\), then \(k = 9\). Profit: \(36\).

Answer: The maximum profit is \(\$44\) by stocking 4 water bottles and 5 keychains.

8. Common mistakes to avoid

• Forgetting what the variables mean. Always define them clearly.
• Ignoring a constraint. A solution must follow every rule.
• Mixing up maximize and minimize. Check whether you want the greatest or smallest value.
• Using negative numbers. In many real-life problems, the number of items cannot be negative.
• Not testing enough possibilities. Make sure you compare valid choices before deciding.

9. Quick step-by-step guide

1. Read the problem carefully.
2. Choose variables for the unknown amounts.
3. Write the objective function.
4. Write all constraints as inequalities.
5. List or test possible solutions that satisfy the constraints.
6. Compute the value of the objective function for those solutions.
7. Pick the solution with the maximum or minimum value.

10. Summary

Linear optimization helps us make the best decision when there are limits.

The objective function tells us what we want to maximize or minimize. The constraints tell us what values are allowed.

To solve a problem, define the variables, write the objective function, write the constraints, test valid solutions, and choose the best one.

This kind of math is useful in business, budgeting, shopping, planning, and many other real-life situations.