# Application of derivative: maximum porblems.

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?

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?

• Just a detail : isosceles is not equilateral in the general case (at least to me). – Claude Leibovici Dec 14 '15 at 10:39
• yes, I mean equilateral. Thanks for the observation – Luis Victoria Dec 14 '15 at 10:50
• The area of the triangle seems wrong. – Emilio Novati Dec 14 '15 at 10:57
• 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 $\frac{\sqrt 3}{25}x^2$. – AndreasT Dec 14 '15 at 10:57
• 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. – Luis Victoria Dec 14 '15 at 11:21

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 $\dfrac{\sqrt{3}}{4}\left(\dfrac{x}{3}\right)^2 = \dfrac{\sqrt{3}}{36}x^2$.