Fencing a Plot Against a Wall of Limited Length
A neighbourhood association has 30 metres of fencing and wants to enclose a rectangular vegetable plot against an existing straight stone wall, using the wall as one full side so that only three sides need fencing. The section of wall the council will allow them to use is 7 metres long, and the plot's side along the wall may not exceed that length. The two sides perpendicular to the wall each have length x metres and the side parallel to the wall has length y metres, so 2x + y = 30 with y ≤ 7, x > 0 and y > 0. One member drafts a plan assuming the best plot is always given by the vertex of the area function, obtaining x = 7.5 and an area of 112.5 m². The chair is doubtful, because 7.5 m of wall is not available. Separately, a second proposal requires the plot to have an area of at least 72 m² while keeping the same 30 m of fencing and the same 7 m wall limit.
1. Find the dimensions giving the largest possible area under the 7 m wall limit.
Read the worked answer
From 2x + y = 30, y = 30 − 2x, so
S(x) = x(30 − 2x) = −2x² + 30x.
The constraints: y > 0 gives x < 15; y ≤ 7 gives 30 − 2x ≤ 7, that is x ≥ 11.5. So the true domain is 11.5 ≤ x < 15.
The parabola opens downward with vertex at x = −30/(2·(−2)) = 7.5. That value is outside the domain, so the vertex is not admissible. On the domain the derivative S′(x) = −4x + 30 is negative throughout (since x ≥ 11.5 > 7.5), so S is strictly decreasing and the maximum is at the left endpoint x = 11.5.
Then y = 30 − 23 = 7, matching the wall limit exactly, and S = 11.5 × 7 = 80.5 m². The key judgement is that the vertex had to be tested against the feasible interval before being used; here the constraint, not the vertex, decides the answer.
2. For what x does the plot have area at least 72 m²?
Read the worked answer
Solve −2x² + 30x ≥ 72. Rearranging and dividing by −2 (which reverses the inequality):
x² − 15x + 36 ≤ 0, i.e. (x − 3)(x − 12) ≤ 0,
so 3 ≤ x ≤ 12.
This is only the condition from the area requirement. The fencing and wall constraints still impose 11.5 ≤ x < 15. Intersecting the two sets gives
11.5 ≤ x ≤ 12.
Check the ends: x = 11.5 gives y = 7 and area 80.5 ≥ 72; x = 12 gives y = 6 and area exactly 72; x = 12.5 would give y = 5 and area 62.5, which fails. So the answer is a short interval, and reporting 3 ≤ x ≤ 12 would be wrong because it ignores the wall.
3. Explain why the 112.5 m² plan cannot be built and state when it would be valid.
Read the worked answer
The value 112.5 m² comes from substituting the vertex x = 7.5 into the area function, which forces y = 30 − 2(7.5) = 15. A plot 7.5 m by 15 m would need 15 m of useable wall, but only 7 m is permitted, so the point (7.5, 15) violates y ≤ 7 and is not in the feasible set.
Geometrically, the member optimised an unconstrained problem and then treated the constraints as an afterthought. The vertex is the correct answer only if the wall offers at least 15 m; in that case x = 7.5, y = 15, and the area is 112.5 m². With a wall of exactly 15 m the vertex sits on the boundary, and with more than 15 m it is interior and the derivative test applies directly.
The general lesson for this type of problem: write the domain first, decide whether the stationary point lies inside it, and compare with the endpoint values before quoting a maximum.