AP Syllabus focus:
‘We verify that optimization solutions satisfy all given constraints and lie in the allowed domain, rejecting extraneous mathematical solutions that do not make sense in context.’
Optimization results are only meaningful when they obey every condition in the problem. Ensuring solutions satisfy constraints prevents unrealistic outcomes and confirms the mathematics truly represents the scenario.
Checking That Solutions Satisfy Constraints
Optimization in AP Calculus AB often produces mathematically correct critical points that may not be valid in the real-world context of a problem. Because optimization problems are grounded in physical, geometric, or practical scenarios, every proposed solution must be checked against constraints—the limits, conditions, and allowable values stated or implied in the problem. Ignoring constraints can lead to impossible dimensions, negative measurements, or values outside the domain, all of which must be rejected.
Understanding Constraints in Optimization
Constraints describe the allowable domain of the variables involved. They may be explicitly stated or inferred from the situation. Evaluating constraints ensures that mathematical results produce meaningful values consistent with the context.
Constraint: A required condition or limit that a solution must satisfy in order to be valid within the physical or contextual setting of an optimization problem.
Realistic optimization always respects the situation’s limitations, which may arise from physical boundaries, domain restrictions, or contextual requirements.
Types of Constraints Common in AP Calculus AB Contexts
Students should recognize constraints from wording, diagram descriptions, or contextual clues. Typical constraints include:
Domain restrictions
Lengths, widths, heights, distances, and areas must be nonnegative.
Variables representing dimensions must usually be positive, not zero.
Interval constraints on variables
Problems may specify that a variable lies within a closed interval.
These intervals define the feasible region for the objective function.
Contextual constraints
Physical realities, such as not exceeding available material or space.
Practical requirements, including staying within speed limits or time limits.
Geometric constraints
Fixed perimeters, areas, or volumes.
Relationships among variables established by the problem’s geometry.
One essential part of validating solutions is recognizing which constraints are explicit and which must be logically inferred from the context.
Ensuring Solutions Lie Within the Allowed Domain
After finding possible extrema using derivatives, students must determine whether these candidate solutions fall within the allowed domain. This domain is shaped by constraints and is often a closed interval, meaning endpoints must also be checked. The process of confirmation involves analyzing both the structure of the objective function and the contextual limits.
After we have a list of candidates, we compare their function values, but only for those candidates that actually lie in the allowed domain or interval.
This graph shows a function with labeled local and global maxima and minima on its domain. It visually reinforces that extreme values can occur at interior critical points or at boundary points. Although the specific function is more complex than typical AP problems, the labels highlight the same classification process used when checking which candidate points satisfy given constraints. Source.
Why Checking Constraints Matters
Mathematical optimization sometimes produces extraneous solutions, which are values that satisfy equations used in the derivation but do not satisfy the original problem. These must be rejected. Reasons a solution might be extraneous include:
The variable value does not satisfy domain restrictions.
The dimensions violate a contextual condition.
The solution requires impossible or undefined behavior.
The value leads to a geometry inconsistent with the problem description.
A mathematically correct derivative does not guarantee a contextually valid solution.
Process for Verifying Solutions Against Constraints
To remain consistent with AP expectations and mathematical precision, a clear, structured process helps ensure thorough verification. Students should apply the following steps:
Identify all constraints before beginning mathematical work.
Rewrite any explicit conditions clearly.
Infer contextual limits where appropriate.
Construct the objective function and express it in terms of a single variable while keeping track of domain implications.
Differentiate and find critical points, but treat these points as candidates, not final answers.
Test candidate points by:
Confirming they fall within the permissible domain.
Ensuring they satisfy all given conditions.
Rejecting any that violate physical or contextual requirements.
Include endpoints of the interval if the domain is closed, since extrema might appear there even when not found by derivative methods.
Interpret the resulting values only after confirming they adhere to constraints.
This process ensures that the final result genuinely optimizes the quantity in question under realistic boundaries.
Rejecting Solutions That Do Not Make Sense in Context
A solution that violates constraints must be discarded regardless of the appeal of its algebraic form. Situations requiring rejection include:
Negative dimensions, which cannot represent physical lengths.
Undefined expressions, such as division by zero or square roots of negative numbers.
Values outside allowed intervals, even if they are mathematically derived.
Violations of real-world feasibility, such as speeds exceeding imposed limits or materials requiring more length than available.
When discarding a solution, it is essential to identify the exact constraint it violates. This reinforces conceptual understanding and ensures students can justify decisions when communicating solutions.
Interpreting Feasible Solutions in Context
A validated solution must be interpreted within the scenario and must reflect realistic, meaningful values for the variables involved. Once constraints have been checked, students can confidently determine whether a solution corresponds to the minimum or maximum described in the problem, ensuring the final answer is both mathematically correct and contextually sound.
The feasible set is the collection of all input values that satisfy every constraint simultaneously; only values in this set can represent acceptable optimization solutions.
This diagram shows a shaded feasible region formed by the intersection of multiple constraints. Any point inside the shaded region satisfies all the conditions at once, while points outside fail at least one constraint. Although displayed in a two-variable context, it provides a clear geometric analogy for understanding what it means for a solution to lie within the allowed domain in one-variable optimization. Source.
FAQ
A constraint is a restriction imposed by the context of the problem, such as dimensions that must remain positive or totals that cannot be exceeded.
A condition derived during optimisation, such as setting a derivative equal to zero, arises from the mathematics used to locate potential extrema. It does not itself determine whether the value is contextually acceptable.
Constraints must always be checked after solving these mathematical conditions to ensure the result is realistically valid.
Some objective functions have several points where the derivative equals zero, but not all of these lie within the physically meaningful domain.
Situations leading to multiple stationary points include:
• Objective functions with curved or oscillating shapes
• Scenarios where an algebraic model applies beyond the real context
• Problems with wide domains, allowing several mathematically valid turning points
Only stationary points that satisfy every contextual and domain constraint can be considered genuine candidates.
Students often miss constraints that are implied rather than explicitly stated. These include:
• Dimensions that cannot be zero, such as heights, widths, or radii
• Domains restricted by square roots, denominators, or physical limitations
• Situations where an endpoint cannot be used because it breaks the problem’s intended geometry
These overlooked constraints often lead to extraneous solutions if not checked carefully.
You must check an endpoint whenever the domain is closed or when the context specifies a fixed minimum or maximum allowable value.
Endpoints should be tested when:
• The variable is restricted to an interval
• The problem describes a maximum capacity or a smallest possible measurement
• The feasible region includes boundary cases that the derivative cannot detect
Endpoints may yield the true optimum even if they do not correspond to stationary points.
Yes. A solution can maximise or minimise an expression algebraically while still violating essential constraints.
This happens when:
• The variable falls outside the domain of the model
• The value creates negative or impossible dimensions
• It contradicts the physical or contextual restrictions of the scenario
A result is only valid if it both optimises the quantity and satisfies every constraint imposed by the context.
Practice Questions
Question 1 (1–3 marks)
A company uses a rectangular piece of cardboard to create an open-top box by cutting equal squares from each corner and folding up the sides. The cardboard measures 20 cm by 30 cm, and the side length of each cut-out square is labelled x.
After finding a critical point for the volume function, a student obtains x = 12.
Explain why x = 12 is not a valid solution for the problem.
Question 1
• 1 mark: States that x = 12 is greater than half of at least one side length of the cardboard.
• 1 mark: Explains that cutting out squares of side 12 cm would remove more cardboard than available, making the construction impossible.
• 1 mark: Concludes that x = 12 does not satisfy the geometric constraints and is therefore invalid.
Question 2 (4–6 marks)
A farmer has 40 metres of fencing to create a rectangular pen along an existing straight wall, which acts as one side of the pen. Only the remaining three sides require fencing.
The width of the pen is x metres, and the length (perpendicular to the wall) is y metres.
The area is given by A = xy.
The fencing constraint is 2y + x = 40.
(a) Show that the area A can be expressed in terms of x only.
(b) The derivative reveals a stationary point at x = 20. Verify whether x = 20 is valid by checking the constraints.
(c) Determine whether the stationary point gives a maximum area, and justify your answer using the context of the problem and the constraints.
Question 2
(a)
• 1 mark: Uses the fencing constraint to express y in terms of x (y = (40 − x)/2).
• 1 mark: Substitutes into A = xy to obtain A expressed solely in terms of x (A = x(40 − x)/2).
(b)
• 1 mark: States that if x = 20, then y = (40 − 20)/2 = 10.
• 1 mark: Notes that both x and y are positive and satisfy the constraint, so x = 20 is valid.
(c)
• 1 mark: States that the stationary point corresponds to a maximum because the feasible domain is 0 < x < 40 and the quadratic expression for A opens downwards.
• 1 mark: Gives a contextual justification (e.g., increasing x beyond 20 reduces y and total area due to the fixed amount of fencing).
