Build the largest enclosure
Which rectangle gives the most space for a fixed fence?
Entering midway? Revisit Turn a production plan into resource needs for the preceding worked example, or start here and use the explanation below.
Start with a concrete question
Design a rectangular garden using a fixed amount of fencing for all four sides. Allocate part of the half-perimeter to its width.
Try the initial inputs, then write a prediction before changing one setting. All calculations run locally on your CPU.
What do you expect?
Predict whether a long thin rectangle or a square gives more area.
Change one thing. Inspect what happens.
- At 40 m of fence, compare width shares 20%, 50% and 80%.
- Explain why 20% and 80% give equal area.
- Double the fence while keeping the shape. Predict the area multiplier.
Your settingsMechanism
Calculation belowEvidence
Compare outcomes
Compare the current results
Every bar starts at zero and uses the same numeric scale in this chart. The upper bound automatically adjusts to the largest current value (at least 1), so compare the numbers when changing inputs. Units and quantities come from each label; different kinds of quantity are not interchangeable. Displayed values are rounded to at most six significant digits.Follow the calculation
- Four sides use 20 + 20 = 40 m of fence.
- Half-perimeter is 20 m, so length = 20 − 10 = 10.
- Area = 10 × 10 = 100 m².
- At equal sides: 10 × 10 = 100 m².
With a fixed rectangular perimeter, balancing the side lengths increases area. Doubling the fence can quadruple area.
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Explain the mechanism
Build the largest enclosure
Design a rectangular garden using a fixed amount of fencing for all four sides. Allocate part of the half-perimeter to its width.
Before you change anything
Predict whether a long thin rectangle or a square gives more area.
Follow the mechanism
Perimeter counts the boundary length; area counts the surface inside. They have different units.
With 40 m of fence, width plus length must be 20 m. Width 4 m gives length 16 m and area 64 m²; width 10 m gives 100 m².
Moving away from equal sides by d metres gives (10 − d)(10 + d) = 100 − d². Every imbalance loses area.
Calculate it yourself
Length = fence/2 − width; area = width × length. Square optimum = fence²/16.
Try a counterexample
At 40 m of fence, compare width shares 20%, 50% and 80%.
Explain why 20% and 80% give equal area.
Double the fence while keeping the shape. Predict the area multiplier.
What this model leaves out
All four sides need fence; using an existing wall changes the optimum.
Terrain, gates, minimum dimensions and construction costs are omitted.
Take it further
High school: Sketch three rectangles with the same perimeter and label their areas.
College: Optimize with calculus, then change the constraint to a three-sided fence.
Working professionals: Write a constrained-design recommendation that includes a minimum access width.
Try a fresh case
A new rectangle uses 28 m of fence on all four sides and has width 5 m. What is its area in square metres?
This is a public practice check. Retries support learning; matching a number does not certify independent proficiency.
Take the investigation further
Choose any extension. These are project prompts for you to carry out; they are not automatically assessed.
School investigation
Sketch three rectangles with the same perimeter and label their areas.
College investigation
Optimize with calculus, then change the constraint to a three-sided fence.
Workplace investigation
Write a constrained-design recommendation that includes a minimum access width.
Keep the evidence, choose your next step
Completion records your own decision that you have worked through this investigation. Practice checks and independent project evidence are separate.