Write an Area Model from a Perimeter
Translate a fixed perimeter into a quadratic rule and domain.
Supplied materials
- Paper or a note app.
- Fictional rectangle perimeter: 24 units. Let width be w and length be L; 2w + 2L = 24.
Learn the idea
The perimeter counts two widths and two lengths. Dividing the perimeter equation by two gives w + L = 12, so L = 12 − w. Multiplying width by length gives A(w) = w(12 − w).
A genuine rectangle needs positive width and positive length, so 0 < w < 12. The mathematical boundary values 0 and 12 give area zero and collapsed shapes. Stating the domain keeps algebraic inputs tied to the design rather than allowing a negative length.
- Divide 2w + 2L = 24 by two and solve for L.
- Multiply w by 12 − w to form the area rule.
- State the positive-length domain, then compute area at w = 3.
See the method in action
At w = 3, L = 12 − 3 = 9 and A = 3 × 9 = 27 square units. The perimeter check is 2(3 + 9) = 24 units.
Check your reasoning
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See the answer and explanation
A = w(14 − w)
Dividing 2(w + L) = 28 by two gives w + L = 14. Thus L = 14 − w and A = w(14 − w), with 0 < w < 14 for positive sides.
Take it further
Change perimeter to 32. Then L = 16 − w, A = w(16 − w), and 0 < w < 16. At w = 4, length is 12 and area is 48 square units.
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Check the subject sources
- Common Core: creating equations and interpreting constraints
- Common Core: interpreting function models
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