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I've been resolving some problems of maxima and minima and I don't understand why my first intent doesn't work. Here is the problem.

A wire of length 100 cm it's going to divide in two pieces, one of the pieces it will fold to form a circumference, and the other in an equilateral triangle. How to cut the wire so that the sum of the areas of the circumference and the triangle is maximum?

I already solve the problem:

One piece will have length $x$ and the other $L-x$. My first intent was: the piece of length $x$ it's the triangle and the other pice, length $L-x$, the circumference. If we name $A_T$:the area of the triangle and $A_C$:the area of the circumference then the function: $$A=A_T+A_C$$ It's the function I have to maximize.

So for the pice of length $x$, I have to divided in three segments of the same length to form the equilateral triangle. Then I get $$A_T=\frac{\sqrt{3}}{36} \cdot x^2$$ For the other piece I have that $L-x=2 \pi r$ where $r$ it's the radious of the circumference. Then $$A_C=\frac{(L-x)^2}{4 \pi}$$ Then $$A=\frac{\sqrt{3}}{36} \cdot x^2 + \frac{(L-x)^2}{4 \pi}$$ So I start the process and I don't get the results the book said, but if I try the other way; $x$ the circumference, $L-x$ the triangle, it works perfectly.

Can anyone explain to me why?

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    $\begingroup$ Just a detail : isosceles is not equilateral in the general case (at least to me). $\endgroup$ – Claude Leibovici Dec 14 '15 at 10:39
  • $\begingroup$ yes, I mean equilateral. Thanks for the observation $\endgroup$ – Luis Victoria Dec 14 '15 at 10:50
  • $\begingroup$ The area of the triangle seems wrong. $\endgroup$ – Emilio Novati Dec 14 '15 at 10:57
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    $\begingroup$ Anyway, among the (isosceles) triangles with same perimeter the one with maximal area is the equilateral (which in your case is $\frac{\sqrt 3}{36}x^2$, and not $\frac{\sqrt 3}{25}x^2$. $\endgroup$ – AndreasT Dec 14 '15 at 10:57
  • $\begingroup$ After all the mistakes I made; I get the inverse value, in the firts one, of the second one. The second one it's what I want and the question is why. $\endgroup$ – Luis Victoria Dec 14 '15 at 11:21
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Hint: the area of an equilateral triangle with side length $a$ is $A = \dfrac{\sqrt{3}}{4} a^2$, so if your side length is $\dfrac{x}{3}$ the area is $\dfrac{\sqrt{3}}{4}\left(\dfrac{x}{3}\right)^2 = \dfrac{\sqrt{3}}{36}x^2$.

The other shape you're describing is a circle - the circumference is the length around the circle.

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  • $\begingroup$ Yes, thank you and sorry for not paying attention. And for the concepts I'm not a native english speaker so I get lost in the translation. $\endgroup$ – Luis Victoria Dec 14 '15 at 11:18
  • $\begingroup$ You're doing well - the readers on the page understood what you intended :-). $\endgroup$ – Frentos Dec 14 '15 at 11:29

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