3

Consider a problem here :

There is a wall in your backyard. It is so long that you can’t see its endpoints. You want to build a fence of length L such that the area enclosed between the wall and the fence is maximized. The fence can be of arbitrary shape, but only its two endpoints may touch the wall.

what is the shape of fence ? answer is semi circle . But how does one arrive at such a conclusion without using high level mathematics ?? any simple answers ??

this problem is asked on : http://www.spoj.com/problems/FENCE1/

Number945
  • 151
  • 1
    The Isoperimetric Inequality (circle is optimal way to enclose an area) takes some effort to prove rigorously. If we assume it, the fence result is an easy consequence: reflect in the wall. – André Nicolas Mar 05 '14 at 17:33
  • I think we can confidently affirm that all efforts to “arrive at such a conclusion without using high level mathematics” will inevitably hit a brick wall. :-) – Lucian Mar 05 '14 at 17:58
  • What do you consider "high level mathematics"? The standard approach, using the Calculus of Variations, is indeed usually taught in college courses, but conceptually requires only elementary, 1D calculus. – user7530 Mar 14 '14 at 16:28

2 Answers2

2

Something very geometrical can be used to prove that one of the optimal solutions is a semi-circle.

"Optimal solution" means a solution that reaches the best result or that is very close to it.

$A$ and $B$ are the intersections of one of the optimal fences $F$ and the wall, and $O$ the middle.

Result 1 (not proved): any ray from $O$ intersect $F$ in exactly 1 point.

Result 2 : for every $n>0$, it exists an optimal solution where $OX_k=R_n$ for every $X_k$ intersection of the Fence and of the ray from $O$ defined by the angle $\frac{k*\pi}{2^n}$.

Proof Let $N$ be the normal to the wall from $O$. $N$ intersects $F$ in $C$/ $N$ separates the area in 2 quarters $Q_{AC}$ and $Q_{CB}$. They are both one of the best possible shapes for a quarter. So we can decide that $N$ is a symmetry axis. Using the bissectrix of $OA, OC$ that cuts $Q_{AC}$ into 2 subquarters and the argument of the best possible shapes, we can decide that $OA=OC$. The same construction can be applied to the quarters $n$ time.

For $X \in F$, $H_X$ is the half plane defined by the axis normal to $Ox$ in $X$ such as $O \in H_X$

Let $M_F$ be the set of points such as: $X \in M \implies \forall Y \in M, Y \in H_X $.

Result 3: $A \in M_F$

Proof: if not, there is point of F further on the left, and it is possible to build a better solution.

Imagine the neighbor does the same on the other side of the wall, and chooses the optimal solution that is the symmetry by $O$.

Let $W'$ be another wall that passes by $O$. It intersects each fence in 1 point

Result 4: If $W'$ intersects the fences in $A'$ and $B'$, the curve $A'B'$ is also an optimal solution.

Result 5: $F = M_F$

Result 6: the series $(R_n)$ converges to an $R$

Proof : using result 5, it is possible to bound the area with functions of $R_n$

Final Result: the series defined by result 2 converges to a semi circle of radius $R$

Sylvain Biehler
  • 478
1

The Wikipedia article on the isoperimetric inequality identifies this as the classical Dido's problem, and says the conclusion was reached in ancient Greece, but that the first convincing steps toward rigorous proof begin with Steiner in 1838. However this recent paper, The Isoperimetric Problem Revisited promotes the earlier contributions of Euler (1744).

"Our primary objective in this paper is, with the student in mind, to present a short, elementary, and teachable solution of the isoperimetric problem. A secondary objective is to give a brief, but reasonably complete, overview of the remarkable life of the isoperimetric problem."

So you will find there both a good history of the problem and a self-contained proof, requiring calculus and Lagrange multipliers.

hardmath
  • 37,015