The Fence That Doubled Back
A landscape architect is fencing a rectangular garden against a straight wall. The wall forms one full side, so fencing is needed on only three sides: two equal ends of length x metres and one side parallel to the wall. The client has 40 metres of fencing and wants the enclosed area to be at least 150 square metres. The architect writes the area as A(x) = x(40 - 2x) and proposes x = 10, giving A = 200. A colleague objects that x = 10 is not the largest possible area and that the feasible set of x values is not all positive reals: the parallel side 40 - 2x must be positive, and the area constraint A(x) >= 150 must also hold. The architect must decide which x values are admissible, whether the proposed design meets the client's requirement, and what the true maximum area is under the fencing constraint.
1. Determine the full set of admissible values of x, and state whether x = 10 is admissible. Justify each inequality used.
Read the worked answer
Admissibility requires three conditions.
- Positive ends: x > 0.
- Positive parallel side: 40 - 2x > 0, so x < 20.
- Area requirement: A(x) = x(40 - 2x) >= 150.
Solve the area inequality:
40x - 2x^2 >= 150 => 2x^2 - 40x + 150 <= 0 => x^2 - 20x + 75 <= 0 => (x - 5)(x - 15) <= 0 => 5 <= x <= 15.
Intersecting with 0 < x < 20 gives the admissible set 5 <= x <= 15.
At x = 10: A = 10(40 - 20) = 200 >= 150, and 10 lies in [5, 15], so x = 10 is admissible. The colleague is right that x = 10 is not the maximiser, but wrong to suggest it is inadmissible.
2. Find the maximum area achievable and the x that produces it, then explain why the maximum is not at an endpoint of the admissible interval.
Read the worked answer
Complete the square or differentiate.
A(x) = 40x - 2x^2 = -2(x^2 - 20x) = -2[(x - 10)^2 - 100] = 200 - 2(x - 10)^2.
This is a downward parabola with vertex at x = 10, so the maximum on all reals is 200 at x = 10. Since 10 lies inside the admissible interval [5, 15], the constrained maximum is also 200 square metres at x = 10.
Why not an endpoint? On [5, 15] the function is a concave quadratic whose vertex is interior. Endpoint values are A(5) = 5(30) = 150 and A(15) = 15(10) = 150, both smaller than 200. A concave function on a closed interval attains its maximum at the stationary point when that point lies inside the interval; endpoints only win when the vertex is outside. Here the vertex is interior, so the boundary values are minima of the admissible range, not maxima. The colleague's objection about x = 10 being non-maximal is therefore incorrect: x = 10 is exactly the maximiser.