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## Main Question or Discussion Point

Hello,

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

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 [tex]\frac{x^2}{16}[/tex]area.

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

Since, [tex]\frac{x^2}{16}[/tex] is more than [tex]\frac{\sqrt{3}x^2}{36}[/tex], 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

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!

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 [tex]\frac{x^2}{16}[/tex]area.

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

Since, [tex]\frac{x^2}{16}[/tex] is more than [tex]\frac{\sqrt{3}x^2}{36}[/tex], 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?"*
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