Herb Garden with Surrounding Path
The eco-club is laying a rectangular herb garden that is 4 m longer than it is wide. A uniform 1 m wide gravel path will surround it on all four sides. The club has enough gravel for exactly 36 m² of path and wants the total outer rectangle to enclose 96 m². The school also has 45 m of edging fence for the outer boundary. Before digging, the club must confirm the inner dimensions, the outer dimensions, and whether the fence and gravel quantities are sufficient. A member suggests making the garden 5 m longer than wide instead, but the gravel and outer-area figures were calculated for the 4 m difference. The caretaker is worried that a negative root might appear in the equation and wants the final layout checked against the physical constraints.
1. Form a second-degree equation for the width of the garden and solve it, rejecting any impossible value.
Read the worked answer
Let the inner width be $x$ metres. Then the inner length is $x+4$ metres.
With a 1 m path on every side, the outer width is $x+2$ and the outer length is $x+4+2=x+6$.
The outer area is 96 m²:
$$(x+2)(x+6)=96.$$
Expanding:
$$x^2+8x+12=96 \Rightarrow x^2+8x-84=0.$$
The discriminant is $8^2-4(1)(-84)=64+336=400$, so
$$x=\frac{-8\pm20}{2}.$$
This gives $x=6$ or $x=-14$. A negative width is impossible, so the inner width is 6 m and the inner length is 10 m.
2. Check the path area and the fencing requirement.
Read the worked answer
The inner garden area is $6\times10=60$ m².
The outer rectangle is $8$ m by $12$ m, with area $96$ m².
Path area $=96-60=36$ m², exactly the gravel available.
The outer perimeter is
$$2(8+12)=40\text{ m}.$$
The school has 45 m of edging fence, so 40 m is sufficient, with 5 m spare. The 1 m path is uniform because the outer dimensions exceed the inner dimensions by 2 m in both directions, i.e. 1 m on each side.
3. A member suggests changing the inner length to 5 m more than the width. Explain whether the given gravel and outer-area figures still work.
Read the worked answer
If the inner width is $x$ and the inner length is $x+5$, the outer dimensions become $x+2$ and $x+7$. The outer-area condition would give
$$(x+2)(x+7)=96 \Rightarrow x^2+9x+14=96 \Rightarrow x^2+9x-82=0.$$
The discriminant is $81+328=409$, which is not a perfect square, so the width would not be a neat measurable value. More importantly, the path area would no longer be guaranteed to be 36 m². The original data were built for the 4 m difference: $6$ m by $10$ m inner, $8$ m by $12$ m outer. Changing the difference without recalculating gravel and outer area is not valid.