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.

 

[CellCoree]

Dear fellow scientist,

Thank you for bringing up this interesting problem. It seems that you have already come up with a logical solution using calculus, but are struggling with understanding it intuitively.

In this case, it may be helpful to think about the shapes in terms of their perimeter and how that affects their area. The perimeter of a square is 4x, while the perimeter of an equilateral triangle is 3x. This means that for a given length of wire, the square will have a larger perimeter, leaving less wire to be used for area. On the other hand, the equilateral triangle has a smaller perimeter, leaving more wire to be used for area.

So, when we give all the wire to the square, we are maximizing the perimeter and minimizing the area. This results in the largest total area. On the other hand, when we give all the wire to the triangle, we are minimizing the perimeter and maximizing the area. This results in the smallest total area.

In essence, it is a balance between perimeter and area that determines the optimal solution. I hope this helps in gaining a better intuitional understanding of the problem. Keep exploring and questioning, that's what science is all about!

Best regards,

 

[pmacias]

Dear reader,

Thank you for sharing your difficulty with understanding this problem intuitively. It is understandable that you would want to approach it in this way, as intuition can often provide valuable insights in problem-solving. However, in this case, it seems that your initial logic was flawed.

To gain an intuitive understanding of this problem, it is important to consider the properties of the shapes involved. The square has four equal sides, while the equilateral triangle has three equal sides. This means that for a given length of wire, the square will have a larger perimeter than the equilateral triangle. This also means that the triangle will have a larger area than the square, since it has a greater number of sides to distribute the wire on.

When considering the maximum and minimum total area, it is important to remember that the wire must be cut into two pieces. Giving all the wire to the shape with the highest area yield (the square) will indeed yield the maximum total area. However, giving all the wire to the shape with the lowest area yield (the triangle) will not necessarily yield the minimum total area. This is because the wire is limited by its length and can only be cut into two pieces.

To achieve the minimum total area, it is necessary to distribute the wire between the two shapes in a way that takes into account their respective perimeters and areas. This is where calculus comes into play, as it allows us to find the optimum point where the total area is minimized. It is important to note that this is not always the case in all problems, and sometimes intuition alone may not provide the correct solution.

In summary, to gain an intuitive understanding of this problem, it is important to consider the properties of the shapes involved and how they affect the distribution of the wire. However, in order to find the optimum solution, it is necessary to use mathematical tools such as calculus. I hope this helps in your understanding of this problem. Good luck!