Sixty Metres of Wire: Whose Plan Gives the Biggest Vegetable Plot?
Chidi has 60 m of fencing wire and wants to enclose a rectangular vegetable plot against the straight wall of his compound, using the wall as one whole side so that only three sides need wire. His brother, who is preparing for the same examination, claims that the largest possible area comes from a plot whose side parallel to the wall is 30 m and whose two equal sides are 15 m each. Chidi's father insists that the plot must be at least 9 m deep, measured perpendicular to the wall, so that a wheelbarrow can turn, and that no single side may be longer than 25 m, because the wire is supplied in pre-cut 25 m lengths. The planting season starts in two weeks. Chidi wants to know whether his brother is right, what the best dimensions are once his father's two rules are applied, and how much area he gives up by obeying them.
1. Test the brother's 30 m by 15 m claim by expressing the area as a function of one variable, and state whether it is truly a maximum rather than merely one possible layout.
Read the worked answer
Set up one variable. Let x be the depth perpendicular to the wall. The two equal sides each measure x, so the side parallel to the wall measures 60 − 2x.
Write the area function. A(x) = x(60 − 2x) = 60x − 2x², with 0 < x < 30.
Complete the square. A = −2(x² − 30x) = −2[(x − 15)² − 225] = 450 − 2(x − 15)².
Because (x − 15)² is never negative, A ≤ 450 m², and the value 450 is reached only when x = 15. Then the parallel side is 60 − 30 = 30 m.
Check by calculus. A′(x) = 60 − 4x = 0 gives x = 15; A″(x) = −4 < 0, so this is a maximum.
Judgement. The brother is right for the unconstrained problem: 30 m by 15 m gives 450 m² and is the largest rectangle the 60 m will enclose against a wall. But the claim has only been proved for the ideal case; the father's rules have not yet been applied, and they may exclude this layout.
2. Apply both of the father's rules and find the best permitted dimensions and the area surrendered.
Read the worked answer
Translate the rules into inequalities.
- Depth at least 9 m: x ≥ 9.
- No side longer than 25 m: this applies to the depth (x ≤ 25) and to the side parallel to the wall (60 − 2x ≤ 25, so 2x ≥ 35, so x ≥ 17.5).
- The area function requires x < 30.
Feasible range. 17.5 ≤ x ≤ 25.
Optimise on that range. A(x) = 450 − 2(x − 15)². For every x ≥ 17.5 the bracket (x − 15)² grows as x grows, so A decreases across the whole feasible range. The maximum therefore sits at the smallest permitted depth, x = 17.5.
Dimensions. Depth 17.5 m; parallel side 60 − 35 = 25 m; the other two sides are 17.5 m each. All sides ≤ 25 m and the depth exceeds 9 m, so the layout is legal.
Area and loss. A = 17.5 × 25 = 437.5 m². Loss = 450 − 437.5 = 12.5 m², about 2.8% of the unconstrained maximum.
Judgement. The cost of the rules is small, and the practical answer is a 25 m by 17.5 m plot. Note that the 25 m limit on the long side, not the 9 m depth, is what forces the change.
3. The father later relaxes the minimum depth from 9 m to 12 m. Does Chidi's answer change, and what does this reveal about which rule is actually binding?
Read the worked answer
Re-solve. The depth condition becomes x ≥ 12. Combined with x ≥ 17.5 from the 25 m limit on the parallel side, the feasible range is still 17.5 ≤ x ≤ 25.
Result. Nothing changes: same dimensions, same 437.5 m². The relaxation has no effect because the depth rule was never the constraint that mattered.
General lesson. A constraint only changes an optimum if it is binding - that is, if it cuts off the unconstrained optimum (x = 15) or pushes the feasible region away from it. Here the 25 m maximum on the parallel side forces x ≥ 17.5, which already satisfies any depth requirement up to 17.5 m.
When would it bite? If the father demanded a depth of at least 20 m, the feasible range would start at 20 and A = 450 − 2(20 − 15)² = 450 − 50 = 400 m², a much bigger loss. That is the case in which the depth rule dominates.
Exam habit. Always identify which constraint is active before optimising; students who simply set the derivative to zero and stop would have missed the whole point of this question.