Find Widths for a Target Area
Solve a quadratic area equation and retain both valid widths.
Supplied materials
- Paper or a note app.
- Supplied model A(w) = w(10 − w), 0 < w < 10; target area 16 square units.
Learn the idea
A quadratic can give the same output at two inputs. For this rectangle, exchanging width and length gives the same area. Set w(10 − w) = 16 to find every width that achieves the target rather than guessing only one.
Rearrange to w² − 10w + 16 = 0 and factor as (w − 2)(w − 8) = 0. A product is zero when at least one factor is zero, yielding w = 2 or w = 8. Check both in the original rule and check their positive lengths.
- Set the area expression equal to 16.
- Move all terms to one side and factor w² − 10w + 16.
- Find both zeros, substitute each, and compare their corresponding lengths.
See the method in action
Widths 2 and 8 both work: 2(10 − 2) = 16 and 8(10 − 8) = 16. The designs are 2 by 8 and 8 by 2, each with positive sides.
Check your reasoning
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See the answer and explanation
w = 3 and w = 7
w² − 10w + 21 = (w − 3)(w − 7), so w = 3 or 7. Both give 3 × 7 = 21 and lie in the positive-length domain.
Take it further
Change the target to 25. Then (w − 5)² = 0, so only width 5 works. A target of 26 has no real width in this model because 25 − (w − 5)² cannot exceed 25.
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