Maximizing Area/Perimeter

Imagine a square box of side $100$ . then, imagine that there is a rubber band constraining a gas (suppose the rubber band doesn't let the gas pass through)

This is exactly the kind of setup we need, as the rubber band tries to minimize its length (perimeter), and the gass tries to maximize its volume (in this case, area).

If you start blowing more gas into the rubber band's confinments, it will initially be a circle, until the rubber band starts hitting the walls of the container, then it becomes 4 circular arcs of 90 degrees, connected by line segments

The ratio is $\large \frac{\text{Area}}{\text{Perimeter}}=\frac{100^2+\left(\pi-4\right)x^2}{400+\left(2\pi-8\right)x}$ .

The maximum value of this function is $26.508$ .

proof under construction still not complete!

The problem of "maximizing the ratio of area to perimeter" is that it is related to but not the same as the classic isoperimetric problem, which asks for the maximum possible area for a given perimeter. See 2D Isoperimetric Theorem. However, if the perimeter is not a given, and is allowed to vary without any constraints, then consider a circle of radius $r$ , which has an area of $\pi r^2$ and perimeter of $2\pi r$ , so that the ratio becomes $\frac{r}{2}$ , which means any ratio depending on choice of $r$ . Then such problems require constraints on the area and perimeter in order to have unique (or finite number?) of such ratio maxima, such as having a bounding closed exterior figure, such as a square.

Let there by a closed figure with an area $A$ and perimeter $P$ , so that the ratio is $\frac{A}{P}$ . Then we can do any of the following which all will improve that ratio. When speaking about segments of the perimeter, we will only consider the length of that segment of the perimeter and do not count the distance $d$ of that segment.

1) We can scale up the closed figure by any factor $s$ , so that this ratio is increased by that factor $s$

2) Let there be a segment of its perimeter which has length $q$ , and such that its end points are separated by some straight distance $d<q$ and the segment does not cross the line $d$ . Then if we are able to increase the area $A$ by $b$ by changing that segment of the perimeter while reducing its length to distance $d$ by making it a straight segment, we have a new ratio $\frac{A+b}{P-(q-d)}$ , which is greater than what it was before. Thus, any closed figure made convex by this means will have an improved $\frac{A}{P}$ ratio.

3) Let there be a segment of its perimeter which has length $q$ , its end points separated by some distance $d$ . Then if we are able to increase the area $A$ by $b$ by changing that segment of the perimeter without changing its length $q$ or distance $d$ , we have a new ratio $\frac{A+b}{P}$ , which is greater than what it was before. The maximum that this ratio can be improved is if this segment takes the shape of an arc of a circle. Or a whole circle if we maximize the area $A$ for its given perimeter $P$ .

4) Let there be a segment of its perimeter which has length $q$ , its end points separated by some distance $d$ . Then if we are able to reduce the perimeter $P$ by $c$ by changing that segment of the perimeter without changing the area $A$ or distance $d$ , we have a new ratio $\frac{A}{P-c}$ , which is greater than what it was before. The maximum that this ratio can be improved is if this segment takes the shape of an arc of a circle. Or a whole circle if we minimize the perimeter $P$ for a given area $A$ .

5) Now, we look at where an arbitrary segment of the perimeter is bounded by two straight sections . There exists a circular arc that is tangent to both straight sections that while the area is unchanged, the perimeter can be reduced. This new circular arc intersects the old perimeter such that the net of bounded areas can be made $0$ (Intermediate Value Theorem), and the perimeter is shortest (per Isoperimetric Theorem). This is not the same as 4) because the size and location of the circular arc tangent to the straight segments determines where line $d$ goes. For this reason, applying this step may not always be possible.

By applying 2), 3), 4) repeatedly, we can arrive at a convex figure which perimeter is composed of straight segments and circular arcs, which we'll call a straight+arc figure. And by applying 5) , we can arrive at a figure where the circular segments are tangent to the adjoining straight segments, but not always, if boundary conditions prevent this step.

If the closed figure has to be contained within a square, then provided if it's ultimately convered into a straight+arc figure without touching the sides, the new figure will have increased $\frac{A}{P}$ ratio, which we can then rotate and scale up until it touches the square at all four sides (using the Intermediate Value Theorem). If the closed figure is already or ends up touching at least 2 opposite sides of the square, then we can expand it until it touches the square on all four sides by adding a parallelogram section of area $sx$ , where $s$ is the length of the side of the square, and $x$ is the distance the sections to be separated by this addition. Hence we've improved the ratio further

$\dfrac{A}{P}<\dfrac{A+sx}{P+2x}$

which follows from the fact that

$\dfrac{A}{P} \le \dfrac{s}{2}$

since any closed figure with a "width" of $s$ cannot have a ratio $\frac{A}{P}$ that exceeds this. An rectangle of one finite width $s$ and one infinite length has ratio $\frac{sL}{2(s+L)} = \frac{s}{2}$ as $L\rightarrow \infty$ . $\;\;$ A circle has the same property.

So, finally, from this and by use of 2), 3), 4), 5) , we can have a figure inscribed inside the square which are characterized by quarter circles at the corners (or with some circular arcs which terminates at the vertices--see below). We can draw smaller squares at each corner, so that if the ratio $\frac{A}{P}$ (counting only perimeter length from side to side and not counting any other perimeter to make it closed) is greater for any one of them as compared to another square of the same size, then we can replace the one with the smaller ratio with the larger. If all the quarter circles have a radius equal to or less than half of the side length of the square, then they can immediately be made all alike, so that the figure is symmetric with respect to both axes. Otherwise, we have a figure with two pairs of different quarter circle radii.

Let $r_1, r_2$ be the two possible radii, in a square of side length $s$ . Then the area and perimeter are as follows

$A=\dfrac{1}{2}\pi(r_1^2+r_2^2)+s^2-2(r_1^2+r_2^2)$
$P=\pi(r_1+r_2)+4s-4(r_1+r_2)$

Optimizing the ratio $\frac{A}{P}$ gets us

$r_1=r_2=\dfrac{s}{2+\sqrt{\pi}}$ , so that for $s=100$ , the maximum ratio is $26.5079...$

In one special case of two quarter arcs meeting at corners (which has the maximum ratio $\frac{A}{P}$ of all variations of circular arcs meeting at the corners) so that none of the steps 1), 2), 3), 4), 5) can readily be used to improve on the ratio $\frac{A}{P}$ (we can use 4) converting the interior figure into a circle with same area, but this require computational work to ensure it fits within the square), we can still go ahead with the method used to determine $r1, r2$ by starting with $r1=0$ .

There are still other special cases where circular arcs end at vertices and are not necessarily tangent to both sides, such as

of which this one can be improved by application of 5) by replacing the two circular arcs with a semicircle tangent to the sides other than at the vertices, and then expanding until the figure once again meets all four sides of the square. Then we end up with the general case of corners truncated by quarter circles of two different radii.

In the general case of finding the maximum ratio $\frac{A}{P}$ of any interior closed figure bounded by a closed exterior figure, the optimum interior figure will always have its perimeter coinciding with that of the exterior figure and/or having circular arcs that are tangent to the exterior figure. Then, for a given closed exterior figure, all such tangent circular arcs will have to be balanced for maximum $\frac{A}{P}$ , as most of the time using 1), 2), 3), 4), 5) alone wil not be able to make this determination. For example, one could start out with an interior closed figure that has a longer perimeter than the perimeter of the exterior figure bounding it. Then by using 3) , we can expand the interior closed figure to match the exterior closed figure, which may not yield maximum $\frac{A}{P}$ , even though it is greater than what it was before.

Addendum: Here's a difficulty with assuming that expanding any circular arc into one that is tangent to the sides of the exterior boundary, while not changing where it touches it, will always improve the ratio $\frac{A}{P}$ . It doesn't always work. Have a look at this figure. The perimeter of the interior figure includes the blue circular arc $b$ and the sides of the large square, but without the quarter round cyan $c$ cutouts.

Here, the blue arc $a$ the entrance to the small square is not tangent to the sides of that small square. Expanding it to the red semicircle $b$ will result in a drop in $\frac{A}{B}$ , not an increase. The red semicircle $b$ has a total angle of $180$ degrees, while the blue arc $a$ has a total angle of about $154$ degrees. The drop in the ratio is very tiny but very real, from $0.5127$ to $0.5125$ . But in fact, the optimum figure has that red semicircle $b$ right where it is, not touching the top of the small square, but with 3 identical cyan $c$ quarter rounds in each of the 3 corners of the large square. The maximum ratio possible here is $0.5339$ . This is an example of where intuition fails , thinking that "the balloon inside expands to touch all sides of the exterior boundary", assuming that all the corners are truncated by quarter rounds will lead to the optimum ratio. It does not. An interesting consequence of this is that the exterior boundary doesn't necessarily have to be a closed figure, for an unique solution with maximum ratio to exist.

[This is not a complete solution. It suggests that the answer isn't 25, to starve off reports claiming the answer is 25.]

Most people think because "for a given perimeter, the circle encloses the most area", hence we must use a circle to maximize the ratio.
Using a circle of radius 50, we calculate the ratio $\frac { \pi r^2 } { 2 \pi r } = \frac{r}{2} = 25$ .
The fact that we get the same ratio by using the large square should suggest that this reasoning isn't completely correct and that the ratio can be optimized further.
What is wrong with this reasoning?

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Note: I do not have a rigorous proof of the actual answer, and I might be incorrect (though I believe this is indeed the correct answer).
If you can get a higher ratio than 26.508, please let me know! (Go ahead and report the problem!)

Note: If you're feeling particularly serious, skip to the last line of my solution

By the 2D Isoperimetric Inequality, we have $4\pi A \leq P^2$ so by rearranging, $\frac{A}{P} \leq \sqrt{\frac{A}{4\pi}}.$ Since $A$ obviously is maximized when we enclose the whole square, we have $\frac{A}{P} \leq \sqrt{\frac{100^2}{4\pi}} \approx 28.21$ and finally since we know the square itself gives $A/P = 25$ , the maximum must satisfy $25.00 \leq \max_{A,P} \frac{A}{P} \leq 28.21$

Therefore all we have to do is test all $2821-2500+1 = 322$ possible answers to two decimal places to find the answer is $26.51$ !

Yes, while the math is completely correct, this solution is my idea of a joke (Hopefully that's okay, Brilliant staff?). Feel free to comment, but please don't upvote it. Go read one of the other solutions and upvote them--they put a lot of effort into their solutions.

My solution is similar to Pedro Cardoso's solution. Instead of a rubber band, I imagined the behavior a soap bubble. It could be summarized for any regular polygon. For example, it can be done for an equilateral triangle. As the soap bubble becomes bigger and bigger and reaches the sides of the triangle, its base( initially a circle) will try to cover maximum area, and minimize its perimeter. The circle(base) will flatten along the three sides of the triangle, and at the corners will form arcs, each one a third of a circle, as shown in the drawing bellow. Surprisingly, the ratio this time is maximized by the triangle itself.

Started with a triangle inscribed inside the 100x100 square and found ratio of A to P to be about 15.5.

The square, octagon, circle all have ratios of A to P of 25. Intuition says it must be a combination of shapes that would yield ratio higher than 25. It would most likely be a square with rounded out corners. It is not likely to be a rounded out larger n-gon because this shape would have a ratio close to that of a circle. Wouldn’t make sense to have any concave polygons either since area would be small.

Rounding out the corner of a square with varying radii from 1 to 50 and checking the ratio of A to P in between we find corner radii 26 and 27 both yield a ratio of 26.51. As we round out the square to make it change from a square to a circle, the ratio follows a parabolic path.