for N meter length, which area will be maximum ? and why ?

this area can be a triangle, rectangle, circle & etc...

## maximum area

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

### Re: maximum area

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

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

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.

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.

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