Difficulty with Intuitional Understanding of an Area Problem

[DocZaius]

Hello,

Perhaps you could help me out. I am having trouble intuitively understanding the result of the following problem:

A piece of wire is cut into two pieces. One piece is bent into an equilateral triangle, the other is bent into a square. Where should the wire be cut to yield the largest total enclosed area? Also, where should the wire be cut to yield the smallest total enclosed area? Note: The wire can be "not cut" with your pick of the shape it all goes to.

Although the way to go about the answer is straightforward (create the total area equation, find the derivative and thus the critical points) I wanted to intuitively step through it.

My (wrong) logic was that I would find out which of the two shapes, given an equal length of wire would yield the most area out of it, and which would give the least. My answer would then be to give all the wire to the biggest area-yielder to maximize total area, and conversely give all the wire to the smallest-area yielder to minimize total area.

Given a length x to be bent into a square, that square gives \frac{x^2}{16}area.

Given a length x to be bent into an equilateral triangle, that triangle gives \frac{\sqrt{3}x^2}{36}area.

Since, \frac{x^2}{16} is more than \frac{\sqrt{3}x^2}{36}, give it all to the square for maximum area and all to the triangle for minimum area. Simple right? Wrong.

Turns out that giving it all to the square does maximize total area but that giving it all to the equilateral triangle doesn't minimize total area. (You should give 0.43495 of a wire of length 1 to the square and the rest to the triangle to get the smallest total area.)

Can anyone give a reason why splitting the wire between a high area yielding shape and low area yielding shape gives a smallest total area? Thanks!

PS: If you skimmed this post, please understand that I do know how to solve this problem through simple calculus. My question isn't "How do I solve this problem?", but rather "What mental approaches would be helpful in gaining an intuitional understanding of the answer?"

 

[Mentallic]

Well it's a simple problem really, just take the derivative of...
haha just kidding


I myself also don't understand it very well intuitively, but this isn't the only problem in mathematics that had an answer which contradicted my intuitions
I know this won't be giving much insight, but what the hell...

It has to do with the fact that the relationship between perimeter and area isn't quite as simple. Let's say that the length of the wire is 10 units, and the triangle is taking up 9 units of wiring. If we add 1 more unit to the triangle, this is added to the perimeter of an already large triangle. You can imagine that adding 1//3 length to each side adds quite a bit to the total area since you're approximately multiplying this small length by a much bigger length on the adjacent side of the triangle.

You'll have \frac{\sqrt{3}}{2}(3)^2 become \frac{\sqrt{3}}{2}(3+\frac{1}{3})^2=\frac{\sqrt{3}}{2}\left((3)^2+2(3)(\frac{1}{3})+(\frac{1}{3})^2\right)

It's this middle term 2(3)(1/3) that adds to a lot of the area (relative to the square).

If we instead added the 1 unit to the square, we're creating a new shape from scratch, and all we'll have is adding (\frac{1}{4})^2 to the total area.

So conclusively (and poorly at that), the minimum area is not by adding all the wiring to the triangle, but instead a nice blend of a smaller size of both square and triangle - by adding a bit more than half to the triangle than the square, which you've already solved.

 

[Hurkyl]

Area grows faster than length. You want to minimize how much you let area grow -- which means roughly the same contribution of length to each shape. Since square area grows faster than triangle area, you want to allocate a little less to the square than the triangle.

But really, the algebra told you that -- you just have to learn how to read it.

 

[DocZaius]

Thank you both for the answers. It really helped out! Yes of course, starting a shape from scratch is always the best decision if your goal is to minimize area with a given perimeter length.