How do you turn a design goal into a one-variable function?
You will be able to: Separate the quantity to optimize from the constraint and state the feasible domain.
How do you turn a design goal into a one-variable function?
A club has 100 meters of fencing for three sides of a rectangular garden beside a straight river. Making the garden wider uses more fence on two sides, leaving less for the third.
A useful starting point: What can derivative tables prove—and what can they only suggest? →
Words and symbols before equations
- Objective
- The quantity to maximize or minimize, here area.
- Constraint
- The fixed relationship designs must satisfy, here 2x+y=100.
- Feasible domain
- Inputs representing allowed designs.
- One-variable model
- The objective after using the constraint to eliminate another variable.
What this picture assumes
Original model; numerical readouts are rounded. Exactly 100 m of fence on three sides; straight river supplies the fourth boundary. y=100−2x, A=xy, 0<x<50. No gates or waste. Equal geometric length scales.
Read the picture in three steps
- Read the axes and labels first. Identify what each symbol and line represents. Read the units and fixed conditions before comparing quantities.
- x=25 m, y=50 m; 2x+y=100 m. A=1250 m², A′=0 m²/m. Maximum-area design: 25 m by 50 m.
- Check what the picture assumes below. Use the Explore task to predict one change before moving a control.
Connect the picture to the mathematics
Let x be each side perpendicular to the river and y the side parallel to it. Area A=xy is the objective; the fence constraint is 2x+y=100, not the area formula.
Solve y=100−2x and substitute to get A(x)=x(100−2x)=100x−2x². Both side lengths must be positive, so 0<x<50.
The objective has units m², while A′ has units m² per meter of x. A derivative of zero is a candidate for the best area, but a sign test or candidate comparison must justify a maximum.
The model assumes a straight usable river edge, no fence along the river, and no gates, terrain or material waste. State assumptions because a changed constraint creates a different problem.
A worked example, step by step
A rectangle has perimeter 40 m on all four sides. Write its one-variable area and feasible domain.
- Let sides be x and y; objective A=xy.
- Constraint 2x+2y=40 gives y=20−x.
- Then A(x)=20x−x².
- A nondegenerate rectangle requires 0<x<20; the endpoints would have zero area and are excluded as actual designs.
Differentiate the objective after imposing the constraint. Optimizing the constraint itself or treating both dimensions as independent answers the wrong problem.
Why is x<50?
Compare with an explanation
The remaining side y=100−2x must stay positive.
Predict. Change one thing. Explain.
Change x. Check that 2x+y always equals 100 and observe the area trade-off. Predict where the largest area occurs before checking the derivative.
On narrow screens, swipe or scroll diagrams sideways to read all labels.
x=25 m, y=50 m; 2x+y=100 m. A=1250 m², A′=0 m²/m. Maximum-area design: 25 m by 50 m.
Original model; numerical readouts are rounded. Exactly 100 m of fence on three sides; straight river supplies the fourth boundary. y=100−2x, A=xy, 0<x<50. No gates or waste. Equal geometric length scales.
Explain what you noticed: Answer the investigation prompt above. State one observation and explain it using the relevant derivative signs, theorem conditions, domain or geometric constraint. Identify what the representation cannot tell you.
Apply the idea to a fresh problem Practice →Show what you understand.
Two original questions are a starting check, not proof of mastery. Explain your choice before revealing the answer.
Original written challenge
4 points · self-check · not an official AP questionA garden uses 60 m of fence on three sides beside a wall. Define the variables, constraint, area function and nondegenerate domain.
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Compare with the answer and four-point rubric
- 1 point: Let x be each wall-perpendicular side and y the parallel fenced side.
- 1 point: 2x+y=60.
- 1 point: A(x)=x(60−2x)=60x−2x².
- 1 point: Require 0<x<30 so both lengths remain positive.
Accept equivalent correct methods and explanations. This is a Refresh Kid teaching rubric, not an official AP scoring guideline.
Retrieve it before you reveal it.
RECALL 1What is an objective?
The quantity to optimize.
RECALL 2What does a constraint do?
Restricts which designs are allowed and relates variables.
RECALL 3Why state a domain?
An algebraic answer may not represent a physical design.
Revisit these tomorrow and a week later. Try a fresh problem and explain why the method applies.
How do you turn a design goal into a one-variable function?
- Define variables and units → objective → constraint → substitute → domain.
- A critical point is a candidate, not yet a proven optimum.
Remember: Differentiate the objective after imposing the constraint. Optimizing the constraint itself or treating both dimensions as independent answers the wrong problem.
Conditions: Original model; numerical readouts are rounded. Exactly 100 m of fence on three sides; straight river supplies the fourth boundary. y=100−2x, A=xy, 0<x<50. No gates or waste. Equal geometric length scales.
Refresh Kid · AP Calculus AB Unit 5 · Objectives FUN-4.B · Review edition
Framework, scope and review status
Mapped to College Board CED, Topic 5.10, FUN-4.B. CED effective Fall 2020, with Fall 2026 clarifications, checked September 17, 2026. Unit 5 has twelve official topics. Focused lesson titles and questions are original Refresh Kid teaching material.
Existence theorems require their stated hypotheses. Critical numbers must belong to the function’s domain; interior local extrema and included endpoints are handled explicitly. Extrema and inflection candidates need justification, not just a zero derivative. Sparse derivative samples do not establish interval-wide signs. Optimization includes the constraint, feasible domain and global comparison. Implicit-curve conclusions specify the branch and distinguish finite slopes from vertical tangents.
The Organic Chemistry Tutor Mean Value Theorem and Optimization Problems video titles, creator and relevant descriptions were checked; full videos were not reviewed. The free optimization video mentions additional paid material, which is not required here. Khan Academy’s unit destination was checked; its JavaScript lesson contents were not fully readable by the research tool. OpenStax Sections 4.3, 4.4, 4.5 and 4.7 were consulted for conceptual cross-checking. No creator scripts, questions, diagrams or artwork were copied. Refresh Kid is not affiliated with or endorsed by these providers.
GitHub’s 3D website collection and its camera-control example informed optional spatial inspection. The open-box model is original geometry using existing self-hosted Three.js with its MIT license. A 12 cm square sheet loses four equal corner squares. The remaining net folds into an open box; the geometry uses the same lengths as the labeled 2D net. Fold angle is a construction view, not an independent design variable; displayed volume refers to the fully upright box. Camera rotation changes no mathematical values. Complete 2D diagrams and explanations remain available without WebGL.
Independent teacher review and observation of students remain pending. Technical checks do not certify mathematical accuracy, accessibility or learning effectiveness. This is a review edition.
Optional official resource: Released AP Calculus AB questions and scoring guides. The archive spans multiple units; no entire exam question is assigned here.
Learn → Explore → Practice → Review is informed by the IES learning guide; this exact implementation has not been evaluated with learners.
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