AP Syllabus focus:
‘To begin an optimization problem, translate the verbal description into an objective function, expressing the quantity to be optimized in terms of a single variable.’
Building an objective function requires interpreting a real-world situation, identifying the quantity to optimize, and rewriting all relevant relationships in terms of one variable to prepare for differentiation.
Building an Objective Function from Context
Constructing an objective function is the foundational step in solving any optimization problem. The objective function is the mathematical expression that represents the quantity you want to maximize or minimize, such as area, cost, time, or distance. Because optimization relies on derivative techniques, the function must be expressed using a single independent variable. Translating a verbal description into a usable mathematical form requires careful reading, recognition of constraints, and selection of appropriate relationships among variables.
When reading a problem, students must distinguish between given information, relationships among quantities, and the target expression they ultimately want to optimize. This translation from words to mathematical structure is what allows the problem to be solved using calculus tools.
Identifying the Target Quantity
Determining What Must Be Optimized
In any context, the first task is to locate the explicit or implied quantity that the question asks to make as large or as small as possible. Typical quantities include area, volume, surface area, cost, and time, all of which are appropriate candidates because they depend on measurable variables.
Objective Function: An expression written in terms of one variable that represents the quantity to be maximized or minimized.
Once the target quantity is identified, it becomes the function you intend to differentiate later, after forming it in terms of a single variable.
Distinguishing the Objective from Constraints
The objective is not the only component in the problem. Most contexts include constraints, which are restrictions on the variables, often expressed through geometric or algebraic relationships. These constraints enable the reduction from multiple variables to one.
Using Constraints to Reduce Variables
Expressing All Variables in Terms of One
Real-world descriptions typically involve several variables. For optimization, however, the goal is to express the objective function using only one.
Typical sources of these relationships include:
Geometric formulas such as perimeter, area, or volume
Physical formulas like distance–rate–time
Descriptions linking quantities (e.g., “the length is twice the width”)
Fixed resources, such as a maximum amount of material
These relationships allow you to rewrite the objective function so that all variables except one are replaced using constraint equations.
This figure shows how a verbal description about cutting squares from a rectangular sheet becomes a labeled diagram with variables on each edge. From the labels, students can write the volume of the box as an objective function V(x)V(x)V(x) in a single variable. The diagram also hints at later calculus steps (finding a maximum volume) but the essential focus here is on constructing V(x)V(x)V(x) from the geometric context. Source.
= First variable in the context
= Second variable related to through the constraint
After rewriting using the constraint, the objective function will depend only on the chosen independent variable.
A constraint must always be interpreted within the physical or contextual boundaries of the problem. These boundaries ensure that the chosen domain for the variable is reasonable and meaningful.
Choosing an Appropriate Independent Variable
Why One Variable Matters
Differentiation methods used later in optimization require a single-variable function. Choosing the independent variable should be guided by convenience, clarity, and the directness with which other variables can be expressed in terms of it. Often a variable representing a length, radius, or time interval is easiest to manipulate algebraically.
Ensuring the Domain Makes Sense
Every meaningful objective function must be accompanied by a domain reflecting the possible real-world values the variable can take. Quantities like length, area, and time must be nonnegative, and many problems further restrict these values based on physical limits, resource constraints, or the structure of the relationship.
Translating Contextual Details into Mathematical Expressions
Interpreting Words into Structure
A critical step in building the objective function is transforming descriptive language into mathematical form. Common translation strategies include:
Rewriting proportional relationships (“twice as long,” “half the distance”)
Recognizing total quantities (“a fixed perimeter of 20 meters”)
Converting physical descriptions into formulas (“water drains at a constant rate”)
Identifying dependent quantities (“the height changes as the radius increases”)
These interpretations ensure that each component of the verbal description is accounted for algebraically.
Organizing Relationships Before Forming the Objective
Before writing the final expression, it is helpful to gather and list all relevant mathematical relationships. These may include:
The target expression to maximize or minimize
All constraint equations
Auxiliary relationships derived from geometry or physical laws
Domain restrictions arising from context
Such organization provides a clear structure for constructing the single-variable objective function.
Assembling the Objective Function
Substituting to Create a Single-Variable Expression
After identifying the target quantity and all relationships, substitution is used to eliminate extra variables. The final form must express the desired quantity solely in terms of the chosen independent variable. This expression becomes the objective function to be differentiated in later steps of the optimization process.
The graph shows an objective function A(x)A(x)A(x) of a single variable, obtained from a rectangular-garden context, with its maximum clearly marked. It illustrates how, once the verbal description has been translated into A(x)A(x)A(x), calculus tools can be used to locate the value of xxx that maximizes the quantity of interest. The numerical maximum extends slightly beyond the “building the objective function” step but supports understanding of the purpose of creating a single-variable expression. Source.
Ensuring the Function Reflects the Context
Once the objective function is formed, it should be checked for consistency with the original problem. The structure of the expression, the domain, and the relationships used must all align with the contextual description to ensure that subsequent calculus techniques produce meaningful and valid solutions.
FAQ
Choose the variable that allows the other quantities to be expressed most simply using the given constraints. In many problems, this is the variable directly tied to the dimension or measure that changes freely within the scenario.
You should avoid selecting a variable that produces complicated algebraic expressions if another simpler option exists. Simpler expressions reduce errors and make later calculus steps more manageable.
Sometimes a context appears to contain several quantities that could be optimised. In such cases:
• Identify which quantity the question explicitly asks you to maximise or minimise.
• If nothing is stated, determine which quantity makes physical or practical sense.
Focus solely on the objective tied to the problem’s purpose. Extra quantities may help form constraints but should not become the objective function themselves.
The objective function itself is not altered by domain restrictions, but its valid inputs are.
For example, lengths, radii, and areas must be strictly non-negative. If a substitution produces values outside the physical limits of the situation, those values must be excluded, even if they satisfy the algebraic relationship.
Domain awareness prevents misinterpretation when the function is later analysed for extrema.
You may need to infer a constraint from the relationships described verbally.
You can do this by:
• Identifying fixed resources, totals, or constant quantities.
• Converting geometric or physical descriptions into formulae.
• Noting any proportional relationships.
These inferred constraints serve the same purpose as explicit equations: they allow reduction to a single-variable objective function.
Differentiation for optimisation requires a single-variable function so that turning points correspond to critical points in one dimension.
With multiple variables, you cannot apply standard single-variable optimisation rules. Reducing the expression to one variable ensures that the derivative describes how the objective changes with respect to exactly one independent factor, making the optimisation process valid and interpretable
Practice Questions
Question 1 (1–3 marks)
A rectangular garden is to be built using 30 metres of fencing for the perimeter. Let the width of the garden be w metres and the length be l metres.
(a) State the perimeter constraint relating l and w. (1 mark)
(b) Write the area A of the garden as an objective function in terms of w only. (2 marks)
Question 1
(a)
• Correct perimeter equation: 2l + 2w = 30.
Award 1 mark for the correct relationship.
(b)
• Substituting l = 15 − w into A = lw. (1 mark)
• Correct objective function: A(w) = w(15 − w). (1 mark)
Question 2 (4–6 marks)
A manufacturer designs an open-top cylindrical container using exactly 800 square centimetres of material for the curved surface and base. The radius of the container is r cm and the height is h cm.
(a) Write a constraint relating r and h based on the total material used. (2 marks)
(b) Using your constraint, express the volume V of the container entirely in terms of r. (2 marks)
(c) Explain why the function obtained in part (b) is suitable as an objective function for an optimisation problem. (1–2 marks)
Question 2
(a)
• Recognising the total surface area for an open-top cylinder: base area plus curved surface area.
• Correct constraint: pi r^2 + 2 pi r h = 800.
Award 2 marks for a correct equation; 1 mark if partially correct.
(b)
• Rearranging the constraint to express h in terms of r, e.g. h = (800 − pi r^2) / (2 pi r). (1 mark)
• Substituting correctly into the volume formula V = pi r^2 h. (1 mark)
• Final expression for V in terms of r only. Credit is given for an algebraically equivalent form. (1 mark if needed for total)
(c)
• Explanation that V is written in a single variable and therefore can be differentiated. (1 mark)
• Explanation that it reflects the physical context and satisfies the constraint. (1 mark)
