Fencing a rectangle
One constraint expressed three ways: as a perimeter, as a function, and as a maximum.
Scenario
A gardener has 60 metres of fencing and wants to enclose a rectangular bed against a long straight wall. The wall forms one of the long sides, so fencing is needed on only three sides.
Let w be the length of each of the two sides perpendicular to the wall.
Answers in order, explanations withheld, and a report on where the chain broke.
- 9.1Equivalent expressionsStandard
Which expression gives the length of the side parallel to the wall?
- A60 - 2w
- B60 - w
- C30 - w
- D(60 - w) / 2
Show the answer and reasoningHide the reasoning
60 - 2w
Three sides are fenced: two of length w and one parallel to the wall. The parallel side takes whatever fencing is left, which is 60 - 2w.
- 9.2Quadratic functionsChallenge
Builds on: The expression 60 - 2w from question 1. A wrong answer there produces a plausible wrong answer here.
Which expression gives the enclosed area in terms of w?
- A60w - 2w^2
- B60w - w^2
- C30w - w^2
- D60 - 2w^2
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60w - 2w^2
Area is the two dimensions multiplied: w(60 - 2w) = 60w - 2w^2. Using 60 - w for the parallel side, which is the error available from question 1, produces 60w - w^2 instead.
- 9.3Quadratic functionsChallenge
Builds on: The area expression from question 2 — the roots of that expression are what locate the maximum.
What value of w encloses the greatest area?
- A15 metres
- B20 metres
- C30 metres
- D10 metres
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15 metres
The area function 60w - 2w^2 factors as 2w(30 - w), so it is zero at w = 0 and w = 30. A parabola peaks halfway between its roots, at w = 15.
- 9.4Interpreting modelsStandard
Builds on: The optimal w = 15 from question 3, substituted back into the expression from question 2.
What is the greatest area the gardener can enclose?
- A450 square metres
- B900 square metres
- C225 square metres
- D400 square metres
Show the answer and reasoningHide the reasoning
450 square metres
Substituting w = 15 gives 60(15) - 2(225) = 900 - 450 = 450 square metres. The 900 figure is the first half of that calculation, before the subtraction.