Garden Plot with a Fixed Perimeter and Diagonal
A community garden committee has a rectangular plot with a perimeter of 34 m and a diagonal of 13 m. They need the exact length and width before ordering timber edging, which is sold in 1 m lengths and must be rounded up. Separately, the committee has permission to use an existing stone wall as one long side of a new rectangular plot. For this second plot they have 34 m of new fencing for the other three sides and want the largest possible area. A member suggests making the second plot 12 m by 5 m, the same as the first plot, because 'the perimeter is the same'. The chair is not convinced and asks for calculations before the next meeting. The committee also wants the first plot to keep its 34 m perimeter but have an area of at least 60 m², and needs to know the possible widths of the shorter side.
1. Set up and solve a pair of equations to find the length and width of the first plot. Explain how you assign the two roots.
Read the worked answer
Let the length be x metres and the width be y metres. The perimeter gives 2x + 2y = 34, so x + y = 17. The diagonal gives Pythagoras: x² + y² = 13² = 169. Square x + y: (x + y)² = 289, so x² + 2xy + y² = 289. Substituting x² + y² = 169 gives 169 + 2xy = 289, so 2xy = 120 and xy = 60. Therefore x and y are the roots of t² − 17t + 60 = 0. This factorises as (t − 12)(t − 5) = 0, so the roots are 12 and 5. Both are positive, so neither is rejected; the longer side is the length, 12 m, and the shorter side is the width, 5 m. Check: 12 + 5 = 17 and 12² + 5² = 144 + 25 = 169.
2. For the second plot, with 34 m of fencing for three sides and the wall as the fourth side, find the dimensions that maximise the area. Compare with the member's 12 m by 5 m suggestion.
Read the worked answer
Let y be the depth away from the wall and x the length along the wall. The fencing covers x + 2y = 34, so x = 34 − 2y. Area A = xy = y(34 − 2y) = 34y − 2y². Completing the square: A = −2(y² − 17y) = −2[(y − 8.5)² − 72.25] = 144.5 − 2(y − 8.5)². So the maximum area is 144.5 m² when y = 8.5 m and x = 34 − 17 = 17 m. The suggested 12 m by 5 m plot has area 60 m² and would use only 12 + 10 = 22 m of the 34 m of fencing, leaving 12 m unused. It is much smaller than the maximum and does not use the available fencing efficiently. The 'same perimeter' argument is misleading because the wall replaces one side of fencing.
3. For the first plot, with perimeter fixed at 34 m, find the possible widths of the shorter side if the area must be at least 60 m².
Read the worked answer
Let the shorter side be w metres. The other side is 17 − w metres, because the perimeter is 34 m. The area condition is w(17 − w) ≥ 60. Rearranging: 17w − w² ≥ 60, so w² − 17w + 60 ≤ 0. Factorising gives (w − 5)(w − 12) ≤ 0. This holds between the roots, so 5 ≤ w ≤ 12. But w is defined as the shorter side, so w ≤ half the perimeter, w ≤ 8.5. Combining: 5 ≤ w ≤ 8.5. At w = 5 the plot is 5 m by 12 m, with area 60 m²; at w = 8.5 the plot is a square of area 72.25 m². Both endpoints satisfy the condition, so the shorter side can be any value from 5 m to 8.5 m inclusive.