Find a rectangle’s largest modelled area
Connect a quadratic expression with a table and a constrained maximum.
Supplied materials
- Fictional paper rectangle has width w and length 10 − w, measured in units. For this model, 0 ≤ w ≤ 10; zero gives a collapsed boundary shape.
- Area rule: A(w) = w(10 − w) = 10w − w². You need multiplication and squares.
Learn the idea
The area is quadratic because multiplying w by 10 − w produces a squared term. Unlike a linear rule, equal increases in width do not produce a constant area increase.
The expression can also be written A(w) = 25 − (w − 5)². Since a square cannot be negative, the area cannot exceed 25 in this model, and that value occurs when w = 5.
- Calculate area at widths 2, 5 and 8.
- Compare widths equally far from 5.
- Use the completed-square form to explain the maximum.
- Calculate the changed width in the question.
See the method in action
A(2) = 2 × 8 = 16; A(5) = 5 × 5 = 25; A(8) = 8 × 2 = 16 square units. The equal areas at 2 and 8 reflect symmetry around w = 5.
Check your reasoning
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See the answer and explanation
21 square units
A(3) = 3(10 − 3) = 3 × 7 = 21. In the other form, 25 − (3 − 5)² = 25 − 4 = 21, providing an independent check.
Take it further
Change the fixed width-plus-length total from 10 to 12. Show that A(w) = 36 − (w − 6)² and the maximum is 36 at w = 6.
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