# Thread: Showing that a square has largest area

1. ## Showing that a square has largest area

This is the question:

"Let S be the set of all rectangles with perimeter P. Show that the square is the element of S with the largest area"

I'm having trouble proving this. Any assistance would be greatly appreciated.

2. Hint:

Try to find a function which is describes an area of rectangles with perimeter P.

remember that:

P=2x+2y (when a & y are sides or some rectangle)

3. I did that, but I don't know what to do with it.

I wrote down A = x^2 (for a square) and A = xy (for any rectangle). But I don't know how to prove the square is bigger.

4. let A=xy=x(p/2 -x)=f(x)

f'(x)=p/2-2x=0

p=4x

==> 2x=2y ==>x=y

or something like that...

5. Put $y$ in terms of $x$ then find $A'=0$

6. I don't think I'm supposed to use calculus. This is pre-calculus material.

7. Originally Posted by Glitch
I don't think I'm supposed to use calculus. This is pre-calculus material.
Well it should be posted in pre-calculus sub-forum not in the university level sub-forum.

If you need to solve it w/out calculus take a number to fix P i.e 10 then with trial and error for x and y you will find a maximum area when x=y

8. Yeah, I don't know where to post things any more. My last question was from the same problem set, and I was told it should be put here.

9. Originally Posted by Glitch
I don't think I'm supposed to use calculus. This is pre-calculus material.
ok....

f(x)=x(p/2 -x)=-x^2+xp/2 this is sad parabola, when she reach the maximum?