The Playground Fence: A Correct Derivative With an Impossible Answer
A secondary school in a governorate town has received 60 m of new fencing wire. Along one edge of the sports yard runs an existing straight brick wall, 25 m long, which the administration wants to use as one complete side of a rectangular playground, so that only three sides need new fence. A student in the mathematics club models the problem as follows: let x be the length of each of the two sides perpendicular to the wall and let y be the side built along the wall. Then 2x + y = 60 and the area is A = xy, so A(x) = x(60 − 2x) = 60x − 2x². Setting A′(x) = 60 − 4x = 0 gives x = 15 m, y = 30 m, and A = 450 m². The student submits 450 m² as the maximum area, and the head teacher is about to order the posts cut to these lengths. The workshop teacher then measures the wall and reports its length.
1. The differentiation is correct, but the conclusion is not. Identify exactly where the reasoning breaks down and write down the set of x values that the 25 m wall actually permits.
Read the worked answer
The calculus is faultless: A(x) = 60x − 2x², A′(x) = 60 − 4x, so x = 15 and A(15) = 15 × 30 = 450 m². What is missing is the domain of the model.
The side along the wall is y = 60 − 2x, and that side must fit inside a 25 m wall, so y ≤ 25:
60 − 2x ≤ 25 → x ≥ 17.5 m
Lengths must also be positive: x > 0 and 60 − 2x > 0 → x < 30.
So the admissible values are only x ∈ [17.5, 30). The stationary point x = 15 lies outside this interval: the student has silently optimized as if the wall were infinitely long. A derivative test locates the vertex of the parabola; it says nothing about whether the vertex is a point the builder is allowed to use. Every optimization answer must be checked against the physical constraints before it is reported.
2. Find the largest area that can actually be fenced, with dimensions, and justify why no further calculus is needed.
Read the worked answer
On the admissible interval [17.5, 30) we have A′(x) = 60 − 4x < 0 for every x > 15, so A is strictly decreasing throughout the interval. A decreasing function on a closed-left interval takes its maximum at the left endpoint:
x = 17.5 m, y = 60 − 2(17.5) = 25 m, A = 17.5 × 25 = 437.5 m²
No second derivative and no new equation solving are required: the sign of A′ already fixes the behaviour, and the only candidate left is the boundary imposed by the wall. The loss relative to the student's figure is 450 − 437.5 = 12.5 m², about 2.8 %. That is small as a percentage but decisive as an instruction to the workshop: the posts must be cut at 17.5 m, not 15 m, or the fence will not close.
3. Before the posts are cut, the municipality offers to extend the wall to 40 m at no cost to the school. Without redoing the whole optimization, state the new maximum and explain in one line why no new calculation is needed.
Read the worked answer
The new constraint is y ≤ 40, i.e. 60 − 2x ≤ 40, so x ≥ 10. The interval becomes [10, 30), and the stationary point x = 15 now lies inside it. The parabola's vertex is therefore attainable: x = 15 m, y = 30 m, A = 450 m², exactly the student's original answer.
The general lesson is that an optimization problem has two separate answers depending on whether the constraint is binding. When the wall is 25 m the constraint binds and the answer sits on the boundary; when the wall is 40 m the constraint is slack and the answer returns to the interior stationary point. Recognising which case applies saves a full re-derivation and explains why the same calculation can be right in one situation and wrong in another.