Netting a chicken run against the yard wall
Mrs Amutenya keeps laying hens and wants a rectangular run built against the straight back wall of her yard, so only three sides need netting. She has one 24 m roll of netting and intends to use all of it. She labels the two equal sides perpendicular to the wall x metres each, so the side parallel to the wall is (24 − 2x) metres.
Her neighbour, who sells eggs to the same market, insists that the run must enclose at least 64 m² or it is not worth keeping the hens, and also claims that if she buys a second 24 m roll and uses 48 m of netting in the same wall-to-wall design, the maximum possible area will simply double. Mrs Amutenya wants the largest possible run but is not sure whether the neighbour's two statements are consistent.
1. Express the enclosed area in terms of x, find the value of x that maximises it, and state the dimensions of the largest run.
Read the worked answer
A(x) = x(24 − 2x) = 24x − 2x² = −2(x² − 12x) = −2[(x − 6)² − 36] = 72 − 2(x − 6)².
Since 2(x − 6)² ≥ 0 for every x, A(x) ≤ 72, with equality only when x = 6. So the maximum area is 72 m², with perpendicular sides of 6 m and 6 m and a wall side of 24 − 2(6) = 12 m.
Domain: the wall side must be positive, so 24 − 2x > 0, giving 0 < x < 12. At x = 0 or x = 12 the area is zero, so the optimum is strictly inside the domain.
Check: netting used = 6 + 12 + 6 = 24 m, and enclosed area = 6 × 12 = 72 m².
Judgement: completing the square is what proves this is a maximum (the squared term is subtracted, so the parabola opens downwards); reading a maximum off a table of values alone does not.
2. Solve the inequality x(24 − 2x) ≥ 64 and explain what the solution means for Mrs Amutenya.
Read the worked answer
x(24 − 2x) ≥ 64
24x − 2x² ≥ 64
2x² − 24x + 64 ≤ 0 (multiplying by −1 reverses the inequality)
x² − 12x + 32 ≤ 0
(x − 4)(x − 8) ≤ 0
So 4 ≤ x ≤ 8.
Interpretation: the perpendicular sides must be between 4 m and 8 m, which makes the wall side between 16 m and 8 m. Both boundary values give exactly 64 m² (4 × 16 = 64 and 8 × 8 = 64), and they are included because the requirement is 'at least 64 m²'. Inside the interval the area is larger, for example x = 6 gives 72 m²; outside it, for example x = 3 gives 3 × 18 = 54 m², which fails the neighbour's requirement.
The interval lies wholly inside the valid domain 0 < x < 12, so it is a genuine solution. Note the profit of this method: the maximum area found earlier (72 m² at x = 6) sits inside the feasible interval, so she can satisfy the neighbour and still be at the best point.
3. Test the neighbour's claim that doubling the netting to 48 m doubles the maximum possible area, using the same wall-to-wall design.
Read the worked answer
With 48 m: A(x) = x(48 − 2x) = 48x − 2x² = −2(x² − 24x) = −2[(x − 12)² − 144] = 288 − 2(x − 12)².
Maximum = 288 m² at x = 12 m, giving sides 12 m, 12 m and 24 m.
Comparison: 288 ÷ 72 = 4, so the maximum area is four times as large, not twice.
Reason: the optimum always has the same proportions - perpendicular sides L/4 each and wall side L/2 - so doubling L doubles both dimensions, and area scales as the square of the linear dimension (2² = 4). Area is proportional to L².
Judgement: the claim is false. This is a scale-factor argument, not an arithmetic slip; any design where doubling the material doubles every dimension will quadruple the area. The neighbour's two requirements are consistent, but his doubling claim is not.