The School Garden Path
The Eco Club of a school in Mangaluru has a rectangular kitchen garden 20 m long and 12 m wide. The club wants a gravel path of uniform width all around the outside of the garden. The total area of the garden plus path is to be 560 m². The garden sits in a corner of the campus, and the available clearance beyond the garden is never more than 3.2 m on any side because of a boundary wall and a drain. The club president solves for the width and gets a value that looks neat, but the teacher asks the students to check it against the site constraint before ordering gravel. The club also has a fallback proposal: a 3 m wide path instead. Students must decide whether the original 560 m² plan is physically possible, what maximum total area the site allows at 3.2 m width, and what area the fallback gives.
1. Form a quadratic equation for the path width and solve it.
Read the worked answer
Let the path width be x metres. The total dimensions are (20 + 2x) and (12 + 2x). (20 + 2x)(12 + 2x) = 560 240 + 40x + 24x + 4x² = 560 4x² + 64x − 320 = 0 x² + 16x − 80 = 0 (x + 20)(x − 4) = 0 x = 4 or x = −20.
Reject x = −20 because width cannot be negative. So x = 4 m. Check: (20 + 8)(12 + 8) = 28 × 20 = 560 m².
2. Check whether the 4 m width is physically possible. If not, find the maximum total area allowed by the 3.2 m clearance.
Read the worked answer
The site allows at most 3.2 m clearance, but the solution requires 4 m. Therefore the 560 m² plan is not physically possible without moving the wall or drain.
At the maximum x = 3.2 m, total area = (20 + 6.4)(12 + 6.4) = 26.4 × 18.4 = 485.76 m².
So the site can provide at most 485.76 m², which is 560 − 485.76 = 74.24 m² short of the target.
3. Evaluate the fallback proposal of a 3 m wide path.
Read the worked answer
For x = 3 m, total area = (20 + 6)(12 + 6) = 26 × 18 = 468 m².
This is feasible within the 3.2 m clearance but gives only 468 m², not 560 m². The club must either accept a smaller total area or change the garden dimensions or site.