Fencing a Rectangular Vegetable Plot With a Fixed Length of Wire
A farmer in Barabanki has 70 m of fencing wire left over from an earlier job and wants to enclose a rectangular vegetable plot behind his house, using the wire for all four sides with no gate. He wants a plot of exactly 300 square metres because his drip-line has 300 emitters, one per square metre of bed. He drives four corner posts, stretches the wire, and then checks that the corners are right angles by measuring the diagonal with a 30 m tape. A neighbour tells him that with the same 70 m of wire he could simply make the plot twice as large, 600 square metres, by choosing a different length and breadth. The farmer is willing to re-set the posts but not to buy more wire, and he wants the plot to remain rectangular. Work in metres and keep answers exact where possible.
1. Find the length and breadth of the plot. Explain what happens to both roots of the quadratic and why neither is discarded.
Read the worked answer
Let the length be x m and the breadth y m, with x ≥ y.
Perimeter condition: 2(x + y) = 70 → x + y = 35. Area condition: xy = 300.
Substitute y = 35 − x: x(35 − x) = 300 → 35x − x² = 300 → x² − 35x + 300 = 0.
Discriminant: (−35)² − 4(1)(300) = 1225 − 1200 = 25 = 5². A positive perfect square means two distinct rational roots: x = (35 ± 5)/2 → x = 20 or x = 15.
If x = 20 then y = 15; if x = 15 then y = 20. Neither root is rejected, because the two roots are simply the two dimensions of the same rectangle. The convention x ≥ y picks 20 m × 15 m, not because 15 is invalid but because it is the breadth in that labelling.
Check: 2(20 + 15) = 70 ✓ and 20 × 15 = 300 ✓ - so 300 emitters fit exactly.
Judgement note: had the discriminant come out negative, no such rectangle would exist and the farmer would have to change either the wire length or the target area. A negative root, on the other hand, would have to be discarded as a length.
2. What diagonal should the farmer measure to confirm the corners are right angles, and is his 30 m tape long enough?
Read the worked answer
For a rectangle, the diagonal follows from Pythagoras: d = √(20² + 15²) = √(400 + 225) = √625 = 25 m.
The 30 m tape is long enough, with 5 m to spare.
How to use the number: if all four sides are correct (20, 15, 20, 15) and both diagonals measure 25 m, the plot is a true rectangle - this is the converse of the Pythagoras theorem. If a diagonal comes out longer, say 26 m, then 20² + 15² ≠ 26², the angle between the 20 m and 15 m sides is not 90°, and the shape is a parallelogram with the same area formula but skewed corners, so the emitter spacing will not line up.
Checking only one diagonal is not enough on a four-sided plot; both must match, since a shape can have equal sides and one correct diagonal and still be distorted at the other pair of corners.
3. The neighbour claims 600 m² is achievable with the same 70 m of wire. Test this claim and state the largest area possible.
Read the worked answer
Keep the perimeter: x + y = 35. Demand the new area: xy = 600.
Then x and y are roots of t² − 35t + 600 = 0. Discriminant = 1225 − 4(600) = 1225 − 2400 = −1175 < 0.
A negative discriminant means no real roots: no rectangle with perimeter 70 m has an area of 600 m². The neighbour's plan is impossible, not merely awkward.
Largest possible area: for a fixed sum x + y = 35, the product xy is greatest when x = y = 17.5 m, giving 17.5² = 306.25 m². So even the very best rectangle with this wire beats the farmer's 300 m² by only 6.25 m². Doubling to 600 m² is nowhere near possible - it would need about 4√600 ≈ 98 m of wire (a square of side ≈ 24.5 m).
Judgement: the farmer should keep 20 m × 15 m, which is already within 2% of the theoretical maximum for his wire, and tell the neighbour that area cannot be increased without either more wire or giving up the rectangular shape.