# Maximizing the Area of Rectangles with Fixed perimeter



## vcuber13 (Jan 6, 2011)

I think most people already know that a square has the largest area for a fixed perimeter of a quadrilateral.
This thread is about a relation I noticed yesterday in math, and I was hoping someone could tell me if anyone has ever noticed it before.

Basically if you take rectangle like shapes (I'll explain it better in a moment) and assign the "vertical" lines x and the "horizontal" lines y, when the sum of the lengths of x's equal the sum of the lengths of the y's you will yield the largest area.
If we take the square (or a "normal" rectangle) as an example we will have 2y's and 2x's, so 2y=2x therefor x=y and it would be a square.

This is intended for "rectangles" where there is something like a river as one of the sides of the rectangle, or if there is another line (or fence if you are closing an area) through the rectangle.

Proof



Spoiler



\( A=xy \)

\( P=ax+by \)

\( y=\frac{P-ax}{b} \)

\( A=x \bigg(\frac{P-ax}{b}\bigg) \)

\( A=\frac{Px-ax^2}{b} \)

\( \frac{d}{dx}\:\:\frac{Px-ax^2}{b}=\frac{(P-2ax)\cdot b-(Px-ax^2)\cdot0}{b^2} \)

\( \frac{P-2ax}{b} \)

solve for 0

\( 0=P-2ax \)

\( P=2ax \)

\( P=ax+ax \)

but

\( P=ax+by \)

so

\( ax+ax=ax+by \)

\( ax=by \)



Examples



Spoiler










In this example the top side is supposed to be a river or stream, and we are maximizing area. A=xy and P=2x+y according to what I found if we need to maximize the area 2x will always equal y (in this example its 2x=y but others will have different co-efficients). If we have 600m of fencing to go around the 3 sides 2x+y=600 but since 2x=y 2x+2x=600 and so 4x=600 or x=150. 2(150)=y, y=300 and A=150*300.






In this there is 3 x's and 2 y's, and we have 600m of fencing again. P=3x+2y, but 3x=2y so P=3x+3x. Therefor 600=6x x=100 and y would equal 150.






Lastly, there are 5 x's and 3 y's and we have 1000 m of fencing. 1000=5x+3y, 5x=3y, 1000=5x+5x, x=100, 5x=3y, 500=3y, y=500/3.



Has anyone ever known about this?


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## TiLiMayor (Jan 6, 2011)

I guess the ilustrations doesn't help much but each of the inner quadrilaterals (idk if its the way to write this in english) are squares, isnt this called proportionality anyway?


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## Forte (Jan 6, 2011)

People normally use calculus for this:
So your problem would be like P = perimeter, A = area, a and b are constants such that P=ax+by and A=xy?
So likeeeee
Since P=ax+by, y=(P-ax)/b and A=xy=x(P-ax)/b=(Px-ax^2)/b
So you'd go all PSHEW and differentiate A
dA/dx = (P-2ax)/b and set to 0
so (P-2ax)/b=0
P=2ax
but since P= ax+by you'd get 2ax = ax+by
or ax=by

It turns out that you're right
COOLIO

i hadn't noticed that before ):


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## Whyusosrs? (Jan 6, 2011)

PSHEW PSHEW


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## vcuber13 (Jan 6, 2011)

TiLiMayor said:


> I guess the ilustrations doesn't help much but each of the inner quadrilaterals (idk if its the way to write this in english) are squares, isnt this called proportionality anyway?


 
If you read it properly you would notice in the secon example the ratio of the sides isnt 2:1 its 2:3, thats what the whole post was about.


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## abctoshiro (Jan 6, 2011)

I understood it a bit. A square has the maximum area possible. 
In your example with the stream, you have 600m of fencing and is trying to maximize the area. The way I go around the problem is this:

x=600/3 (divide the fence into three equal parts)
x=200

and since x=y, then,

fence=2x+y
fence=600
P=800m

and the area is 200(200) = 40000m^2.

We learned about this kind of problems in class and I discovered by myself that a square always gives the largest area. The "formula" I use for this kind of problems is A=(P/n)^2 where P is the perimeter or a fence and n is the no. of sides to be enclosed. 

That only applies to the first problem though (I think so). Correct me if there are any mistakes.


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## Stefan (Jan 6, 2011)

I stopped reading as soon as you started to use explicit numbers.


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## TiLiMayor (Jan 6, 2011)

vcuber13 said:


> If you read it properly you would notice in the secon example the ratio of the sides isnt 2:1 its 2:3, thats what the whole post was about.


I didn't meant to refute, Yuu are still maximizing area for a grid by building up squares/rectagles inside with the most high factors possible, thus making it as similar to a square as possible, and with this array the sumatory of the lineal fencing equals the higest posible area.


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## vcuber13 (Jan 6, 2011)

Forte said:


> People normally use calculus for this


another way of doing it is to create a quadratic equation and either do -b/2a or derive it and get 2ax+b then solve for 0.
i agree that there is a better way of solving it but it grade 10 they teach us to make a quadratic equation out of it, then to factor it into vertex form and solve for 0.



TiLiMayor said:


> I didn't meant to refute, Yuu are still maximizing area for a grid by building up squares/rectagles inside with the most high factors possible, thus making it as similar to a square as possible, and with this array the sumatory of the lineal fencing equals the higest posible area.


 
I'm not sure i understand you.


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## steph1389 (Jan 6, 2011)

Isn't this type of stuff in the linear programming topic in decision maths?

Maximizing profit while keeping spending small (for a business plan).


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## TiLiMayor (Jan 7, 2011)

vcuber13 said:


> I'm not sure i understand you.


If yuu were to build a grid of 20 slots with some fencing the maximum area you could achive would be in a 4x5 array rather than a 2x10 or 1x20.


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## vcuber13 (Jan 7, 2011)

I see what you mean now and it makes sense, but how does that apply to all of the examples?


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## daniel0731ex (Jan 7, 2011)

cool, never realized this before.

So I derived this equation:

x= F^2/4a^2

where x is the length of the side you are solving for, F is the fencing length available, _a_ is the # of x sides

Could somebody check this equation?

EDIT: oops, forgot to factor out in the end. the correct equation should be x=F/2a lol


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## TiLiMayor (Jan 7, 2011)

vcuber13 said:


> I see what you mean now and it makes sense, but how does that apply to all of the examples?


You can't apply this to the first example but arrange your second example as 1x8 area is less but fencing as the same, still 1x=8y.



daniel0731ex said:


> Could somebody check this equation?


Its somehow wrong, and why did yuu raised to the square anyway?. x=(F/2)/a


> where x is the length of the side you are solving for, F is the fencing length available, a is the # of x sides


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## daniel0731ex (Jan 7, 2011)

TiLiMayor said:


> You can't apply this to the first example but arrange your second example as 1x8 area is less but fencing as the same, still 1x=8y.
> 
> 
> Its somehow wrong, and why did yuu raised to the square anyway?. x=(F/2)/a


 
Oops, I forgot to factor as the last step of a completing the square process :fp

so yes, now I got x=f/2a lol


full equation i got:

A=-a/b(x^2-Fx/a+F^2/4a^2)+F^2/4ab
A=-a/b(x-F/2a)^2+F^2/4ab

therefore Area is a the maximum when x=F/2a.


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## LewisJ (Jan 7, 2011)

HEY GUYS I SAW THE COOL MATHEMATICS MARATHON THREAD BUT I THOUGHT MY PROBLEM WAS MORE IMPORTANT SO I MADE IT ITS OWN THREAD

isn't that cool guys? guys?


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## vcuber13 (Jan 7, 2011)

TiLiMayor said:


> You can't apply this to the first example but arrange your second example as 1x8 area is less but fencing as the same, still 1x=8y.


 
the coefficients arent the lengths in units the are the number of x lengths, so if you do have a rectangle it is 2x and 2y. the equation is basically ax=by where a is the number of xs and b is the number of ys. also this equation is for only when you are maximizing the area. also if you did have x+8y=600 x=300 y=150/4.
and like daniel said, only when you need the area maximized, it would be y=ax/b and x=P/2a or y=P/2b and x=by/a


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## TiLiMayor (Jan 7, 2011)

Oh srry I typed wrong, if yuu were to have a 1x8 grid it would meant the fencing as 2x+9y.


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## vcuber13 (Jan 7, 2011)

Stefan said:


> I stopped reading as soon as you started to use explicit numbers.


 
I added a prove using only algebra


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## einstein00 (Jan 8, 2011)

There's a simple method to solve these problems without any calculus or quadratics. Honestly, I'm kind of surprised no one has posted this before. It's known as AM-GM and you can read about it here.

In short, AM-GM states that the arithmetic mean of two real numbers is always greater than or equal to the geometric mean, with equality if and only if the two numbers are equal. Algebraically, this means that \( \frac{a+b}{2} \ge \sqrt{ab} \).

anyway. Here's an example. For the rectangle/square problem, let \( x \) and \( y \) be the sides of the rectangle with a fixed perimeter \( P \). Of course, \( P=2(x+y) \). AM-GM states that \( \frac{x+y}{2} \ge \sqrt{xy} \), or \( \frac{P}{4} \ge \sqrt{xy} \). Since both sides are positive, we can square both sides: \( xy \le \frac{P^2}{16} \). \( xy \) is maximized when \( xy = \frac{P^2}{16} \), and AM-GM states that this equality is only true when \( x = y \). Therefore, the maximum area of a rectangle with a fixed perimeter occurs when the rectangle is a square.


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## flan (Jan 8, 2011)

steph1389 said:


> Isn't this type of stuff in the linear programming topic in decision maths?
> 
> Maximizing profit while keeping spending small (for a business plan).


 
I <3 decision. We weren't allowed to do Decision 2, supposed to be too easy, Have to do fp2 instead 

Edit: off topic, sorry.


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## vcuber13 (Jan 8, 2011)

einstein00 said:


> anyway. Here's an example. For the rectangle/square problem, let \( x \) and \( y \) be the sides of the rectangle with a fixed perimeter \( P \). Of course, \( P=2(x+y) \). AM-GM states that \( \frac{x+y}{2} \ge \sqrt{xy} \), or \( \frac{P}{4} \ge \sqrt{xy} \). Since both sides are positive, we can square both sides: \( xy \le \frac{P^2}{16} \). \( xy \) is maximized when \( xy = \frac{P^2}{16} \), and AM-GM states that this equality is only true when \( x = y \). Therefore, the maximum area of a rectangle with a fixed perimeter occurs when the rectangle is a square.


 
thats quite interesting, but will that only work if it is an actual rectangle? like if p=3x+2y will it still work?


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## einstein00 (Jan 8, 2011)

vcuber13 said:


> thats quite interesting, but will that only work if it is an actual rectangle? like if p=3x+2y will it still work?


 
Sure. I'm guessing some people might want to try this on their own, so it's in the spoiler.



Spoiler



\( P = 3x+2y \)
\( A = xy \)


\( \frac{3x+2y}{2} \ge \sqrt{3x\times2y} \) (here, X (reference to my above post) = 3x and Y = 2y.)
\( \frac{P}{2} \ge \sqrt{6xy} \)
\( \frac{P^2}{4} \ge 6xy \)
\( A = xy \le \frac{P^2}{24} \)

We wish to maximize A, which happens at equality, meaning X = Y. (or 3x = 2y)

\( A = \frac{(600 \text{ meters})^2}{24} = \boxed{15000 \text{ meters}^2} \)

\( P = 3x+3x = 6x = 600 \text{ meters} \)
\( x = \boxed{100 \text{ meters}} \)
\( y = \boxed{150 \text{ meters}} \)


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## vcuber13 (Jan 8, 2011)

thats very interesting


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