maximum area

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Onimekh
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Joined: Mon Apr 27, 2015 8:30 pm

maximum area

Unread post by Onimekh » Mon Apr 27, 2015 8:39 pm

for N meter length, which area will be maximum ? and why ?
this area can be a triangle, rectangle, circle & etc...

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seemanta001
Posts: 13
Joined: Sat Jun 06, 2015 9:31 am
Location: Chittagong

Re: maximum area

Unread post by seemanta001 » Sun Jun 28, 2015 2:29 am

When the perimeters are equal, the area of circle will be maximum.
In this case while $N$ is the perimeter,for the triangles, the area of the equilateral triangle will be greater than the others.If $A$ is the area and $$s=(a+b+c)/2$$ of a triangle we can write it as $$A^2=s(s-a)(s-b)(s-c)$$.From the AM-GM Inequality we get $$s^4/27 \geq A^2$$.
Here equality is only possible when $$s-a=s-b=s-c$$ or $$a=b=c$$.
So, it's clear that among triangles with same perimeters, the area of the equilateral triangle is the greatest.
Now, if $N$ is the perimeter of an equilateral triangle,it's side will be $N/3$.
Thus,it's area is $$A_1=N^2/12\sqrt{3}$$.
Now, the area of a square is greater than the areas of other quadrangles with equal perimeters can be proved using the AM-GM Inequality.
If $N$ is the perimeter of a square, it;s is side will be $N/4$.
Then it's area will be $$A_2=N^2/16$$.
Now if the perimeter of a circle is $N$, it's radius will be $N/2\pi$.
And, the area will be $$A_3=N^2/4\pi$$.
We can generate the areas of the other polygons through this procedure and can easily observe that the areas of the other polygons are less than them.Because,for $n$ sided regular polygons with sides $a$,the area is $$A_4=na^2cot(180^o/n)/4$$.Here of course, $$a=N/n$$.
We can easily observe that $A_4<A_2<A_1<A_3$.
Hence,we get our answer. :mrgreen:
"Sometimes it's the very people who no one imagines anything of who do the things no one can imagine"

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