SL 5.7 Solve contextual optimization problems | International Baccalaureate (IB) IB AA SL: Mathematics: Analysis and Approaches SL

International Baccalaureate (IB) IB AA SL: Mathematics: Analysis and Approaches SL

Calculus

Use a mathematical model to find the best possible value

Optimization means finding the greatest or least value of a quantity under given conditions. Examples include maximizing an area when the amount of fencing is fixed or minimizing the material needed for a container. The essential work is to identify what can vary, express the quantity of interest as a function of that variable, and find the best value within the allowed domain. You will use familiar algebra and the derivative rules for polynomials, then interpret the result in the original situation.

What you will learn

Translate a practical situation into a function to maximize or minimize.
Choose a suitable variable and state its allowed domain and units.
Use derivatives to locate candidate optimum values and check that they make sense in context.
Use a graphing calculator to check, rather than replace, mathematical reasoning.

1. From a situation to a function

Begin by identifying the quantity that must be optimized, such as area, volume, cost, or distance. Choose a variable for a dimension or other quantity that can change. Use the conditions in the problem to express the other quantities in terms of that variable. Substituting these relationships into the target quantity gives a function of one variable.

The domain is the set of values the variable is allowed to take. It comes from the physical situation: a length must be positive, and a box cut from a sheet cannot use a cut larger than half the sheet's shorter side. State the domain before optimizing; a mathematical result outside it cannot be a valid answer.

Keep track of units. If a length is measured in metres, an area is measured in square metres. If a function gives volume, its units are cubic units. Write down any assumptions that shape the model, such as a rectangular design or a fixed amount of fencing.

Define the variable and target quantity.
Use the context to reduce the target to a function of one variable.
State a realistic domain and include units in the final interpretation.

2. Finding and checking an optimum

The derivative describes how a function changes as its input changes. For a smooth function, an interior maximum or minimum can occur where the derivative is zero. Such an input is called a stationary point. For polynomial models, differentiate using the power rule: the derivative of

ax^n

is

anx^{n-1}

.

Solve the equation formed by setting the derivative equal to zero, then keep only solutions in the domain. A stationary point is a candidate, not automatically the required answer. Compare relevant candidate values of the original function, including domain endpoints when they are allowed. In many physical situations, a boundary may describe a degenerate object, such as a box with zero volume; explain whether it is included.

A graph offers a visual check. Near a maximum, the function rises and then falls; near a minimum, it falls and then rises. A graphing calculator can help estimate the location and value of a turning point. Use the derivative and the model to justify the result, and give the final answer at the requested accuracy.

\frac{d}{dx}(ax^n)=anx^{n-1}

Differentiate the target function with respect to its variable.
Solve $f'(x)=0$ and check the domain.
Check that the candidate gives the required maximum or minimum in context.

3. Representing and checking the model

The algebraic representation is the formula for the target quantity. The numerical representation is a table of values or a calculator estimate. The graph shows how the quantity changes across the allowed domain, while the contextual representation explains what an input and output mean in the real situation. These views should agree.

For a purposeful technology check, enter the function and restrict the viewing window to the domain. Use the graph's maximum or minimum feature to estimate the turning point, or inspect a table near the predicted value. A broad window can hide the important part of the graph, so set the horizontal range using the context. Retain enough calculator digits during working, then round only the final result.

Before presenting an answer, check the units, whether the dimensions are physically possible, and whether the value is genuinely a maximum or minimum. If the question asks for dimensions, report dimensions rather than only the optimized area or volume.

Use a graph or table as a check on the analytical result.
Set the calculator window to the allowed domain.
Interpret the optimum in words and with appropriate units.

Worked example

Maximum area with a fixed perimeter

A rectangular garden has a perimeter of 40 m. Find the dimensions that give the greatest area.

Choose a variable
Let the width be $x$ metres and the length be $y$ metres. Both dimensions must be positive.
$x>0,\quad y>0$
Build the model
The perimeter condition gives $2x+2y=40$ , so $y=20-x$ . The area is the product of the dimensions, and the physical domain is $0<x<20$ .
$A(x)=x(20-x)=20x-x^2$
Find the candidate
Differentiate the area and set the derivative to zero. The solution lies in the domain.
$A'(x)=20-2x=0\quad\Rightarrow\quad x=10$
Interpret and check
The corresponding length is $20-10=10$ m. The graph is a downward-opening quadratic, so its turning point is a maximum. The area at this point is $100$ square metres.
$A(10)=100$

Answer: The greatest area is

100\text{ m}^2

, achieved by a

10\text{ m}

by

10\text{ m}

garden.

Check: The dimensions have perimeter

2(10)+2(10)=40

m, as required.

Worked example

Maximum volume from a sheet

Squares of side

x

cm are cut from the corners of a

30

cm by

20

cm sheet. The sides are folded up to make an open box. Find the cut size that gives the greatest volume, to the nearest

0.01

cm.

State dimensions and domain
After folding, the height is $x$ cm, and the base dimensions are $30-2x$ cm and $20-2x$ cm. For all dimensions to be positive, $0<x<10$ .
$V(x)=x(30-2x)(20-2x)$
Differentiate
Expand the volume expression to make differentiation straightforward. Set the derivative equal to zero to locate stationary points.
$V(x)=600x-100x^2+4x^3,\quad V'(x)=600-200x+12x^2$
Select the feasible point
Solving $V'(x)=0$ gives $x=\frac{25\pm5\sqrt{7}}{3}$ . Only the smaller solution is in the domain $0<x<10$ .
$x=\frac{25-5\sqrt{7}}{3}\approx3.92$
Check the result
The volume is zero at the limiting cuts $x=0$ and $x=10$ . A graph restricted to the domain rises to a turning point near $3.92$ and then falls, confirming the interior maximum. Substitution gives a maximum volume of about $1056.2$ cubic centimetres.
$V(3.92)\approx1056.2$

Answer: Cut out squares with side length approximately

3.92

cm. The resulting maximum volume is approximately

1056.2\text{ cm}^3

.

Check: At this cut size, both base dimensions are positive: approximately

22.15

cm and

12.15

cm.

Worked example

Fencing beside a river

A farmer has

120

m of fencing to enclose a rectangular field beside a straight river. The river forms one side, so fencing is needed for only the other three sides. Find the dimensions that maximize the enclosed area.

Represent the sides
Let each of the two equal sides perpendicular to the river have length $x$ m. The fenced side parallel to the river then has length $120-2x$ m.
$0<x<60$
Form the area function
Area is the product of the perpendicular width and the fenced length parallel to the river.
$A(x)=x(120-2x)=120x-2x^2$
Optimize
Set the derivative to zero. The result lies within the domain, and the quadratic graph opens downwards, so its turning point is a maximum.
$A'(x)=120-4x=0\quad\Rightarrow\quad x=30$
Interpret the dimensions
The fenced side parallel to the river is $120-2(30)=60$ m. The maximum area is therefore $30$ times $60$ square metres.
$A(30)=1800$

Answer: Use two sides of

30

m perpendicular to the river and one fenced side of

60

m parallel to it. The maximum area is

1800\text{ m}^2

.

Check: The required fencing is

30+30+60=120

m.

Common mistakes and how to avoid them

Finding a stationary point but not checking whether it is a maximum or minimum.

Correction: Use the graph or compare function values at relevant candidates and domain boundaries. Explain why the selected candidate gives the requested extreme.

Keeping a solution that is outside the physical domain.

Correction: Write the domain from the context before solving, then reject any candidate that cannot represent the situation.

Optimizing the wrong expression or forgetting a dimension changes with the variable.

Correction: Use the constraints to express every changing dimension in terms of the chosen variable before forming the target function.

Rounding intermediate values too early or giving an answer without units.

Correction: Keep adequate calculator precision during the working and round the final contextual result to the requested accuracy, with units.

Lesson summary

Choose a variable, use the constraints to form a one-variable target function, and state its physical domain.
Differentiate, solve for stationary candidates, and keep only feasible values.
Check that the candidate gives the required maximum or minimum; use a domain-aware graph or table as a check.
State the result in context with suitable units and accuracy.

Check your understanding

Question 1

For

f(x)=18x-x^2

on

0<x<18

, which value of

x

gives the maximum?

$x=9$
$x=18$
$x=0$
$x=18$ and $x=0$ both give the maximum

Show answer and explanation

x=9

The derivative is

18-2x

, which is zero at

x=9

. The quadratic opens downwards, so this interior stationary point is the maximum.

Question 2

A square is cut from each corner of a

24

cm by

16

cm sheet to form an open box. Which domain for the cut size

x

keeps all dimensions positive?

$0<x<8$
$0<x<12$
$0<x<16$
$x>8$

Show answer and explanation

0<x<8

The base dimensions are

24-2x

and

16-2x

. The shorter side requires

16-2x>0

, so

x<8

, with

x>0

.

Question 3

A graph of a target function rises and then falls within the allowed domain. What does its turning point represent?

A local maximum candidate
A guaranteed minimum
A point outside the domain
A value that must be ignored

Show answer and explanation

A local maximum candidate

Rising and then falling indicates a local maximum. Check its domain and compare with any relevant boundary values before concluding it is the contextual maximum.

Key terms

Optimization: Finding the greatest or least value of a quantity subject to given conditions.
Domain: The allowed input values for a function, restricted in a contextual model by the situation.
Derivative: A measure of how a function changes as its input changes.
Stationary point: A point where the derivative is zero; it may be a maximum, a minimum, or neither.

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SL 5.7 · Solve contextual optimization problems

International Baccalaureate (IB) IB AA SL: Mathematics: Analysis and Approaches SL

Use a mathematical model to find the best possible value

What you will learn

1. From a situation to a function

2. Finding and checking an optimum

3. Representing and checking the model

Maximum area with a fixed perimeter

Maximum volume from a sheet

Fencing beside a river

Common mistakes and how to avoid them

Lesson summary

Check your understanding

Question 1

Question 2

Question 3

Key terms

Continue through IB AA SL

About this lesson and its review