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**RauLiTo****Member**- From: Bahrain
- Registered: 2006-01-11
- Posts: 142

we have a retangle , a square and a circle ... they are all have got the same Perimeter ... the eustion is : which one has got the biggest area ?!

by putting numbers it like this :

1 - circle

2 - square

3 - rectangle

i proved that the circle has got an area bigger than the square like this :

4 X = 2 Pi radius >>> radius = 2X / Pi

therefore the area of the circle is : 4 X ² / Pi which is nearly = 1.27 X ²

the area of the square is = x ²

1.27 X ² > X ²

but now i want to prove that the rectangle has got smaller area than the quare ... who can help me ?

ImPo$$!BLe = NoTH!nG

Go DowN DeeP iNTo aNyTHinG U WiLL FinD MaTHeMaTiCs ...

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**MathsIsFun****Administrator**- Registered: 2005-01-21
- Posts: 7,713

I "prove" it this way: the circle is the solution to largest area for given perimeter. So it wins top spot. And a square is the closest a rectangle gets to being a circle, so it gets 2nd spot. Crude but effective

To be a little more helpful:

A = wh

P = 2(w+h)

rearrange P formula: h = P/2-w

substitute in A formula: A = w(P/2-w) = Pw/2 - w²

Now, you can find the minima/maxima by finding where A'=0 (derivative of A = 0):

A' = P/2 - 2w = 0

then: P/2 = 2w

then P = 4w

So there is a minimum/maximum at w = P/4

And at that point h = P/2 - P/4 = P/4 also

Tidy it up and you are done!

"The physicists defer only to mathematicians, and the mathematicians defer only to God ..." - Leon M. Lederman

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**RauLiTo****Member**- From: Bahrain
- Registered: 2006-01-11
- Posts: 142

sorry dear MathsIsFun but i didnt understand that about minima and maxima

ImPo$$!BLe = NoTH!nG

Go DowN DeeP iNTo aNyTHinG U WiLL FinD MaTHeMaTiCs ...

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**MathsIsFun****Administrator**- Registered: 2005-01-21
- Posts: 7,713

The derivative of a function gives you the slope at any point.

The function is: A = Pw/2 - w²

It's derivative is: A' = P/2 - 2w

(I am not sure if you know about derivatives, but it is part of calculus.)

Now, for the sake of making a graph I set P=10, and plotted the functions together: Plot of 10*x/2-x^2 vs 10/2-2*x (x means w)

Notice that the "10/2 - 2w" line shows you the slope of the "10*w/2 - w²" line?

Also notice that where the "10/2 - 2w" crosses the zero line is exactly where the "10*w/2 - w²" changes direction?

And, as it turns out, that is the maximum value for the "A = 10*w/2 - w²" function.

That is because a curve flattens out (slope=0) on the top (or bottom) when it reaches a maximum (or minimum) value.

So I used this idea to find the largest value of A. That is, A will be at its peak when A'=0.

There can be pitfalls with this - you don't actually know if it is a maximum or minimum, or the curve may later come back and go even further, or it may stay flattened out, etc. But with a simple function like this it works nicely.

"The physicists defer only to mathematicians, and the mathematicians defer only to God ..." - Leon M. Lederman

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**RauLiTo****Member**- From: Bahrain
- Registered: 2006-01-11
- Posts: 142

hehehe maybe i didnt learn that !

is there any other way to prove it ?

ImPo$$!BLe = NoTH!nG

Go DowN DeeP iNTo aNyTHinG U WiLL FinD MaTHeMaTiCs ...

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**Ricky****Moderator**- Registered: 2005-12-04
- Posts: 3,791

So we have a square of side s, such that 4s = p, the perimiter. Now we have the sides of a rectangle, x, y such that x≠y and 2x + 2y = p. Without a loss of generallity, we can state that x > y. So 4s = 2x + 2y and 2s = x + y.

Area of square: s²

Area of rectangle: xy

What we wish to conclude is xy < s².

Thus, xy < s²

"In the real world, this would be a problem. But in mathematics, we can just define a place where this problem doesn't exist. So we'll go ahead and do that now..."

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**MathsIsFun****Administrator**- Registered: 2005-01-21
- Posts: 7,713

I think that pretty much proves it.

I like your "read from the middle outwards" comparison.

Another approach would be to compare the areas for a square and for a rectangle that is just "Δx" from being a square.

"The physicists defer only to mathematicians, and the mathematicians defer only to God ..." - Leon M. Lederman

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**RauLiTo****Member**- From: Bahrain
- Registered: 2006-01-11
- Posts: 142

guys sorry but still didnt get it exactly

( XY + 4 (( X + y ) / 2 )) / 2 = ( X + y ) ² / 4 ... how is that ?

( X + Y ) ² = X² + 2 XY + Y ²

ImPo$$!BLe = NoTH!nG

Go DowN DeeP iNTo aNyTHinG U WiLL FinD MaTHeMaTiCs ...

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**Ricky****Moderator**- Registered: 2005-12-04
- Posts: 3,791

Bah. Good call. I made a mistake. I'll see if I can save my proof.

"In the real world, this would be a problem. But in mathematics, we can just define a place where this problem doesn't exist. So we'll go ahead and do that now..."

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**Ricky****Moderator**- Registered: 2005-12-04
- Posts: 3,791

Alright, let's have another go. Maybe it's simplier than I thought.

Since x and y are both positive, it must be that:

-xy < x² + y²

So:

-xy + 2xy < x² + 2xy + y²

And:

xy < x² + 2xy + y² = s²

"In the real world, this would be a problem. But in mathematics, we can just define a place where this problem doesn't exist. So we'll go ahead and do that now..."

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**RauLiTo****Member**- From: Bahrain
- Registered: 2006-01-11
- Posts: 142

s² = ( x² + 2xy + y² ) / 4 !

ImPo$$!BLe = NoTH!nG

Go DowN DeeP iNTo aNyTHinG U WiLL FinD MaTHeMaTiCs ...

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**Ricky****Moderator**- Registered: 2005-12-04
- Posts: 3,791

Arggggg!!! Stupid mistakes! Alright, we'll have another go at it:

Since x > y, it must be that 0 < x - y and 0 < (x - y)(x - y). So...

So

Edit: Ok, I checked it over like 500 times. That *has* to be right.

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**Ricky****Moderator**- Registered: 2005-12-04
- Posts: 3,791

Oh, and just as a note, to do a problem like this, you start off with a hunch. That is, assume your conclusions. Then for your conclusion to be true, find out what else has to be true and try to work your way backwards till you get to a statement you can prove. Then just write it in reverse.

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**RauLiTo****Member**- From: Bahrain
- Registered: 2006-01-11
- Posts: 142

woow brilliant ! ... just one thing :

xy < \frac{x^2 - y^2}{2}

how did it become :

\frac{xy}{2} < \frac{x^2 + y^2}{4}

i meant the - how did it become + ?!

sorry really sorry i bothered u alot !

ImPo$$!BLe = NoTH!nG

Go DowN DeeP iNTo aNyTHinG U WiLL FinD MaTHeMaTiCs ...

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**RauLiTo****Member**- From: Bahrain
- Registered: 2006-01-11
- Posts: 142

thaaaaaaaaaaaanks thaaaaaaaaaanks ...∞ thanks alot darling !!! you are amazing !!! thanks again

ImPo$$!BLe = NoTH!nG

Go DowN DeeP iNTo aNyTHinG U WiLL FinD MaTHeMaTiCs ...

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**luca-deltodesco****Member**- Registered: 2006-05-05
- Posts: 1,470

RauLiTo wrote:

woow brilliant ! ... just one thing :

i meant the - how did it become + ?!

following what else he wrote, it seems to be a typo

and off topic? who likes my avatar, i think it looks rather cool

*Last edited by luca-deltodesco (2006-06-26 03:18:24)*

The Beginning Of All Things To End.

The End Of All Things To Come.

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**Ricky****Moderator**- Registered: 2005-12-04
- Posts: 3,791

What typo?

As for you avatar, what exactly is it?

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**luca****Guest**

Ricky wrote:

What typo?

As for you avatar, what exactly is it?

well... the intricate design was created by joining lots of circles together in a chain, then deleting the alternate segments and filling in the middle, as for what it IS, i have no idea, but i think it looks quite cool :X

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