Fencing a Kitchen Garden: Two Roots, One Wall
A farmer in a drought-prone block has exactly 40 m of galvanised wire fencing, which she will use on three sides of a rectangular kitchen garden; the fourth side is an existing brick wall that is 20 m long and cannot be extended. She measures x m along each of the two sides perpendicular to the wall and y m along the side parallel to it. Her first target is an area of 192 m². If that works she wants to try for 210 m². Her neighbour, who has just finished a chapter on quadratic equations, says: 'With 40 m of wire you cannot beat 192 m²; the equation gives two answers and they are both the same area anyway.' The block development officer will release subsidised seed only for plots between 150 m² and 200 m².
1. Set up the equations for the 192 m² target, solve them, and decide which solution the farmer can actually use.
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Wire used: 2x + y = 40, so y = 40 − 2x. Area: xy = 192 → x(40 − 2x) = 192 → 2x² − 40x + 192 = 0 → x² − 20x + 96 = 0. Factorising: (x − 8)(x − 12) = 0 → x = 8 or x = 12.
- x = 8 → y = 40 − 16 = 24 m
- x = 12 → y = 40 − 24 = 16 m
Both give area 8 × 24 = 12 × 16 = 192 m², and both use exactly 40 m of wire. But the brick wall is only 20 m long, and y is the side lying along that wall, so y ≤ 20.
The 8 m × 24 m plot is rejected (24 m of fence would stand along a 20 m wall). The usable plot is 12 m × 16 m, which also falls inside the 150–200 m² subsidy band.
2. Test the 210 m² ambition. Is it attainable with the same 40 m of wire? Justify your answer without guessing.
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Write the area as a function of x: A(x) = x(40 − 2x) = 40x − 2x² = 200 − 2(x − 10)².
Since (x − 10)² ≥ 0, A(x) ≤ 200, and the maximum 200 m² is reached at x = 10 m, y = 20 m - a plot that exactly fits the 20 m wall.
Discriminant check: x(40 − 2x) = 210 → x² − 20x + 105 = 0 → D = 400 − 420 = −20 < 0, so there is no real value of x. The algebra confirms the geometry.
So 210 m² is impossible; the farmer's best plot is 10 m × 20 m, which still qualifies for subsidised seed since 200 m² is within the 150–200 m² limit.
3. Evaluate the neighbour's statement carefully. Which part is right and which part is wrong?
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Right: the equation for 192 m² does have two real roots, and both genuinely give 192 m² (8 × 24 and 12 × 16). The two solutions are not a mistake in the solving.
Wrong on two counts:
- The roots are not interchangeable here. The 24 m side would have to lie along a 20 m wall, so only one root is physically admissible - a constraint, not the algebra, decides.
- 'You cannot beat 192 m²' is false. Completing the square shows the maximum is 200 m² at 10 m × 20 m, so 192 m² is not a ceiling.
The correct conclusion: the pair of roots is a set of candidate answers, and each candidate must be tested against the real-world constraint before it becomes a solution.