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Subject: "Rectangle areas"     Previous Topic | Next Topic
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Oskar
guest
Mar-24-03, 01:13 PM (EST)
 
"Rectangle areas"
 
   Hi,

A friend of mine gave me a problem that has been bugging me for quite some time now, and i was hoping to get som help (excuse me for me bad english)

If the edge of a certain rectangle how long should the sides be?

I know that the answer should be that all sides should be equal (so its really a square). But what is the proof.


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Vladimir
Member since Jun-22-03
Jul-28-03, 06:46 PM (EST)
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1. "RE: Rectangle areas"
In response to message #0
 
   LAST EDITED ON Jul-28-03 AT 07:10 PM (EST)
 
As I had to learn English at an adult age, you have my sympathy. Your question does not make sense. Perhaps what you meant was

"If the circumference (not edge!) of a certain rectangle is given, how long should the sides be to get the maximum area?"

Denote the rectangle sides x, y (x <= y). The circumference s and the area A are

s = 2x + 2y
A = xy

Since y = s/2 - x, the area is

A = x·(s/2 - x) = x·s/2 - x2

The area will be maximum, if the derivative dA/dx equals to zero.

dA/dx = d/dx{x·s/2 - x2} = s/2 - 2x = 0
x = s/4
y = s/2 - x = s/2 - s/4 = s/4

So that rectangle sides x = y = s/4 are equal and the rectangle is a square. But perhaps you are not familiar with calculus and the derivatives yet. Then look at the equation for the area A as a quadratic equation for x:

A = x·(s/2 - x) = x·s/2 - x2
x2 - x·s/2 + A = 0

The equation will have a real solution if its discriminant D is greater than or equal to zero:

D = (s/2)2 - 4A >= 0
4A <= (s/2)2
A <= (s/4)2

So the area A is always less than or equal to (s/4)2 and it will be maximum if it equals to (s/4)2. In that case the quadratic equation for x is

x2 - x·s/2 + (s/4)2 = 0
(x - s/4)2 = 0
x = s/4
y = s/2 - x = s/2 - s/4 = s/4

Again, the rectangle sides x = y = s/4 are equal and the rectangle is a square.

Regards, Vladimir


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Vladimir
Member since Jun-22-03
Jul-28-03, 08:05 PM (EST)
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2. "RE: Rectangle areas"
In response to message #0
 
   LAST EDITED ON Sep-01-03 AT 10:03 AM (EST)
 
You can also find out the answer without knowing about the quadratic equations. For a given circumference s, we can always make a square with sides a = s/4 and area

Asquare = (s/4)2

Suppose you deform this square by making the horizontal side longer by a distance Da > 0. Since the cirumference s is fixed, you have to shorten the vertical side by the same distance:

s = 4a = 2a + 2a = 2(a + Da) + 2(a - Da) = 2x + 2y

Now you have a rectangle with different horizontal and vertical sides x and y

x = a + Da = s/4 + Da
y = a - Da = s/4 - Da

Area of this rectangle is

Arectangle = xy = (s/4 + Da)·(s/4 - Da) = (s/4)2 - (Da)2 < (s/4)2 = Asquare

simply because the square of any real number greater than zero (actually, any real number not equal to zero), incuding Da, is positive. Therefore, whatever rectangle you make by deforming the square while keeping the circumference fixed, the rectangle area is always less than the area of the square.


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Joe
guest
Sep-03-03, 04:46 PM (EST)
 
3. "RE: Rectangle areas"
In response to message #0
 
   Let the length, width, and perimeter of a rectangle be l, w, and p respectively, then l+w=p/2. By the Arithmetic Mean - Geometric Mean Inequality, p^2/16=((l+w)/2)^2>=lw, with equality if and only if l=w, or l=w=p/4. Maximum area is the left hand side, p^2/16.


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