Fencing a Vegetable Plot Against a Wall
The agriculture club has 40 m of fencing to enclose a rectangular vegetable plot. One side of the plot will run along an existing straight brick wall, which needs no fencing, so fencing is used on the other three sides only. Each of the two sides running out from the wall is x metres long, and the side parallel to the wall uses whatever fencing is left.
The club must fix the dimensions before the posts are ordered. Members want the largest possible area for a school garden competition. A teacher suggests a plot 5 m deep and 30 m long, arguing that a long plot uses the wall best. Another member prefers 15 m deep and 10 m long. Both are measured out from the wall and both use exactly 40 m of fencing, so the argument cannot be settled by checking the perimeter. The club has one afternoon to agree a design.
1. Write the area A of the plot as a function of x, state the values of x for which the design makes sense, and find the dimensions that give the greatest area. Test the two suggested designs against your answer.
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Model. The fencing covers two sides of length x and one side of length 40 − 2x, so A = x(40 − 2x) = 40x − 2x².
Domain. The side along the wall must be positive: 40 − 2x > 0 → x < 20. Also x > 0. So 0 < x < 20.
Maximum by completing the square. A = −2(x² − 20x) = −2[(x − 10)² − 100] = 200 − 2(x − 10)². Since 2(x − 10)² ≥ 0, the area never exceeds 200, and it equals 200 only when x = 10.
Best design. x = 10 m deep, side along the wall = 40 − 20 = 20 m, area 200 m². The graph is a downward parabola with vertex (10, 200) and intercepts at x = 0 and x = 20.
The two suggestions. 5 m × 30 m: 5(40 − 10) = 5 × 30 = 150 m². 15 m × 10 m: 15(40 − 30) = 15 × 10 = 150 m². Both are arithmetically correct and both use exactly 40 m of fence, yet both waste 50 m². They sit symmetrically either side of the vertex, which is why they give identical areas - matching perimeters does not imply matching areas.
2. The competition rules require a plot of at least 150 m². Solve an inequality to find the range of possible values of x, and explain what the answer means for the club's choice.
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Inequality. 40x − 2x² ≥ 150.
Rearrange. 2x² − 40x + 150 ≤ 0 (divide by −2 and reverse the sign, or move everything to the right): x² − 20x + 75 ≤ 0.
Factorise. (x − 5)(x − 15) ≤ 0. The product is negative or zero between the roots, so 5 ≤ x ≤ 15.
Check the boundaries. x = 5 gives 5 × 30 = 150 m²; x = 15 gives 15 × 10 = 150 m². x = 10 gives 200 m², comfortably inside. x = 4 gives 4 × 32 = 128 m², which fails; x = 16 gives 16 × 8 = 128 m², which also fails.
Interpretation. Any depth from 5 m to 15 m inclusive qualifies, so the club has a range of legal designs, not a single one. The two proposals in the debate sit exactly on the boundary - they qualify, but with zero margin, so any measuring error on site could put the plot below the 150 m² requirement. Choosing x = 10 m (200 m²) keeps a 50 m² safety margin while still using only the 40 m of fencing available. Reversing the inequality sign, or answering 'x ≤ 5 or x ≥ 15', is the standard error here and would approve plots of 4 m or 16 m depth that fail the rule.
3. The brick wall turns out to be only 12 m long. Assuming the plot must still use the wall for its whole length and the same 40 m of fencing, find the greatest area now possible and say how much the restriction costs the club.
Read the worked answer
New constraint. The side along the wall is 40 − 2x, and it cannot exceed 12 m: 40 − 2x ≤ 12 → −2x ≤ −28 → x ≥ 14. The domain is now 14 ≤ x < 20.
Where is the maximum? The unconstrained maximum is at x = 10, and A = 200 − 2(x − 10)² decreases steadily for x > 10. On the interval 14 ≤ x < 20 the function is therefore decreasing, so the largest allowed area occurs at the smallest allowed x, namely x = 14.
Evaluate. A = 14 × (40 − 28) = 14 × 12 = 168 m² (check the end of the interval: x = 15 gives 150 m², x = 19 gives 19 × 2 = 38 m²).
Cost of the restriction. 200 − 168 = 32 m² lost, about 16% of the best possible area. Note that 168 m² still beats both original proposals (150 m²), and it sits inside the competition's 5 ≤ x ≤ 15 band, so the club stays eligible. Substituting x = 10 into the new situation would be wrong - it needs 20 m of wall, which no longer exists.