How can a fixed-volume container use the least material?
You will be able to: Minimize a one-variable surface-area model on an unbounded physical domain.
BC foundation: Unit 5 shares these analytical differentiation objectives with AB. Check theorem hypotheses, justify derivative signs over intervals, and compare feasible candidates before claiming an optimum. State the local branch when analyzing an implicit relation.
How can a fixed-volume container use the least material?
Two closed cylindrical cans can hold the same amount but use different amounts of metal. A very thin can needs tall walls; a very wide can needs large end caps.
A useful starting point: How large should the corner cuts be to maximize a box’s volume? →
Words and symbols before equations
- Closed cylinder
- A cylinder with both circular ends included.
- Fixed volume
- The quantity held constant, here 54π cm³.
- Surface area S
- Two cap areas plus curved side area.
- Unbounded domain
- The positive radius has no finite upper endpoint in this ideal model.
What this picture assumes
Original model; numerical readouts are rounded. Closed cylinder of volume 54π cm³. h=54/r², S=2πr²+108π/r for all r>0; the displayed window samples only part of that domain. Cross-section is schematic, explicitly not to scale. Algebra supplies the global proof.
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.
- r=3 cm, h=6 cm; volume=169.646 cm³=54π. Total closed surface=169.646 cm²; S′=0 cm²/cm. Minimum-area design: r=3,h=6.
- 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 radius r>0 and height h>0. Volume πr²h=54π gives h=54/r². The closed surface area is S=2πr²+2πrh, so S(r)=2πr²+108π/r.
Differentiate: S′=4πr−108π/r²=4π(r³−27)/r². Since the denominator is positive, its sign is negative for 0<r<3 and positive for r>3.
Therefore S decreases all the way to r=3 and increases afterward, proving the global minimum. The height is 54/9=6 cm and minimum area is 54π cm².
This open unbounded radius domain has no finite endpoint list for the closed-interval candidates test. The derivative signs settle the global comparison; additionally S tends to infinity as r→0+ or r→∞. Ignoring a lid would change the objective.
A worked example, step by step
For a closed cylinder with fixed volume 16π cm³, find the minimum-area dimensions.
- The constraint gives h=16/r².
- S=2πr²+32π/r, so S′=4π(r³−8)/r².
- The derivative changes negative to positive at r=2 over r>0.
- Thus h=4 cm and minimum area S=24π cm².
Include both lids for a closed can, and justify an optimum on the actual positive domain rather than inventing finite endpoints.
Would an open-top cylinder have the same area formula?
Compare with an explanation
No. It has only one circular cap, so the optimization model changes.
Predict. Change one thing. Explain.
Vary r while holding volume fixed. Watch height shrink as cap area grows; compare the total area and derivative sign on either side of r=3.
On narrow screens, swipe or scroll diagrams sideways to read all labels.
r=3 cm, h=6 cm; volume=169.646 cm³=54π. Total closed surface=169.646 cm²; S′=0 cm²/cm. Minimum-area design: r=3,h=6.
Original model; numerical readouts are rounded. Closed cylinder of volume 54π cm³. h=54/r², S=2πr²+108π/r for all r>0; the displayed window samples only part of that domain. Cross-section is schematic, explicitly not to scale. Algebra supplies the global proof.
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 closed cylinder has volume 2π cm³. Find and justify the minimum-area radius, height and surface area.
This response is not submitted or saved. Copy it before leaving.
Compare with the answer and four-point rubric
- 1 point: h=2/r² and S=2πr²+4π/r for r>0.
- 1 point: S′=4π(r³−1)/r².
- 1 point: The sign is negative below r=1 and positive above, proving a global minimum.
- 1 point: r=1 cm,h=2 cm, and S=6π cm².
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 1Why can the radius vary while volume stays fixed?
The height changes according to the constraint.
RECALL 2Which test proves the minimum here?
Derivative negative before and positive after across the positive domain.
RECALL 3Why distinguish closed from open-top?
They have different surface-area objectives.
Revisit these tomorrow and a week later. Try a fresh problem and explain why the method applies.
How can a fixed-volume container use the least material?
- Closed cylinder: S=2πr²+2πrh; volume=πr²h.
- For volume 54π, minimum at r=3,h=6,S=54π.
Remember: Include both lids for a closed can, and justify an optimum on the actual positive domain rather than inventing finite endpoints.
Conditions: Original model; numerical readouts are rounded. Closed cylinder of volume 54π cm³. h=54/r², S=2πr²+108π/r for all r>0; the displayed window samples only part of that domain. Cross-section is schematic, explicitly not to scale. Algebra supplies the global proof.
Refresh Kid · AP Calculus BC Unit 5 · Objectives FUN-4.C · Review edition
Framework, scope and review status
Mapped to College Board CED, Topic 5.11, FUN-4.C. 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 BC 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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