Fencing A Region Against A Wall

Introduction

A wall can serve as one side of a rectangular enclosure, so fencing is needed along only the other three sides. This changes the constraint from the fixed-perimeter rectangle problem: two fenced sides have the same length, while the third runs parallel to the wall. The key is to model those lengths carefully before maximizing the area.

We will use one variable for the distance from the wall and use the fencing constraint to express the remaining side. Then we can find the largest area and check every feasible boundary case.

Set Up the Model

Suppose the available length of fencing is \(F\), where \(F>0\). Let \(x\) be the distance perpendicular to the wall, and let \(y\) be the side parallel to the wall, opposite it. The two perpendicular sides both require fencing, as does the side of length \(y\). The wall itself supplies the fourth side, so it uses no fencing.

Solving the constraint for \(y\) gives \(y=F-2x\). Since both dimensions must be nonnegative, \(x\ge 0\) and \(F-2x\ge 0\). Therefore the feasible interval is \(0\le x\le F/2\). Substituting into the area formula gives the one-variable objective

The endpoints have a direct geometric meaning. If \(x=0\), the perpendicular sides have zero length and the area is zero. If \(x=F/2\), then \(y=0\), so the side parallel to the wall has zero length and the area is again zero. These are degenerate limiting cases rather than usable enclosures, but including them makes the feasible interval closed and lets us compare them directly with the interior candidate.

Find and Check the Maximum

On the interior of its feasible interval, the area function is differentiable. Its derivative is \(A'(x)=F-4x\). A critical point occurs when \(A'(x)=0\), so \(F-4x=0\) and \(x=F/4\). This is feasible because \(0<F/4<F/2\) when \(F>0\). The corresponding side parallel to the wall is \(y=F-2(F/4)=F/2\).

A critical point alone does not settle a constrained optimization problem: feasible boundary cases must also be checked. Here, \(A(0)=0\) and \(A(F/2)=0\), whereas the area at the interior point is positive. The result can be established even more directly by rewriting the entire objective.

Why These Dimensions Give the Global Maximum

The derivative identifies the only interior critical point, but completing the square proves that no other feasible choice can have a greater area. Starting from \(A(x)=Fx-2x^2\), factor out \(-2\) from the quadratic terms and complete the square:

For every feasible \(x\), the square \(\left(x-\frac{F}{4}\right)^2\) is nonnegative. Therefore \(A(x)\le F^2/8\). Equality holds exactly when the square is zero, which requires \(x=F/4\). The constraint then gives \(y=F/2\). Since this \(x\) is feasible, the bound is attained, proving that it is the global maximum, not merely a local candidate. At either endpoint, the area is zero, strictly less than \(F^2/8\) because \(F>0\).

This proof also explains the shape of the objective: moving away from \(x=F/4\) makes the subtracted square larger and the area smaller. The dimensions at the maximum are not equal. The side along the wall is twice the depth, since \(y=F/2=2(F/4)=2x\).

Worked Examples

Boundary Cases and Physical Feasibility

There are two related ways to describe the feasible set. If zero lengths are allowed for the purpose of checking limiting cases, the set is the closed interval \(0\le x\le F/2\), and both endpoints are feasible degenerate configurations with zero area. If an actual enclosure is required to have positive side lengths, then \(x>0\) and \(y>0\), so the domain is the open interval \(0<x<F/2\). The maximum is still attained in that physical domain because \(x=F/4\) lies strictly inside it.

It is important not to treat an endpoint as a usable enclosure: a zero side gives no enclosed area. It is equally important not to ignore endpoints when using the closed feasible interval. Here their area values confirm that they cannot beat the interior candidate. In other problems, endpoints can be important candidates, which is why they are checked rather than assumed irrelevant.

Common Mistake: Fencing the Wall Side

A common modeling error is to count all four sides as requiring fencing. That changes the constraint to \(2x+2y=F\), which describes a rectangle fenced on every side, not an enclosure that uses a wall.

For example, with \(120\) metres of fencing, the incorrect four-sided model would suggest a square with sides \(30\) metres, because \(4(30)=120\). Its area would be \(30\cdot30=900\) square metres. But for the wall enclosure, fencing three sides of length \(30\) metres uses only \(2(30)+30=90\) metres. It does not use all \(120\) metres, so it is not the solution to the stated fixed-fencing problem. The correct three-sided model gives dimensions \(30\) metres by \(60\) metres, uses \(2(30)+60=120\) metres, and encloses \(1800\) square metres.

Why the Model Matters

The optimization is short once the constraint is right. The wall determines which side needs no fencing; the two equal depths contribute \(2x\); and the remaining fenced side is whatever length is left, \(F-2x\). Reversing these roles or counting a wall side as fenced produces a different objective and a different answer.

The result gives a practical design rule: with a fixed amount of fencing and a sufficiently long straight wall, use one quarter of the fencing length for each perpendicular side and one half for the side opposite the wall. The rule assumes the wall can supply the entire side of length \(F/2\), the ground is suitable for a rectangle, and no additional materials or geometric restrictions change the constraint. Under those assumptions, the square-completion argument verifies the best possible area for every feasible design.