Using the Extreme Value Theorem on rectangles?

[LilTaru]

Homework Statement

Fix a positive number P. Let R denote the set of all rectangles with perimeter P. Prove that there is a member of R that has maximum area. What are the dimensions of the rectangle of maximum area? HINT: Express the area of an arbitrary element of R as a function of the length of one of the sides.

Homework Equations

Perimeter = P = 2(l + w)
Area = (P/2)w - w^2 --> as a function of the length of one side

The Attempt at a Solution

I don't know how to prove there is a rectangle with a maximum area using the Extreme Value Theorem? Help?!

 

[micromass]

If x is the length of one, side, then the other length is (P-2x)/2. The area is then x(P-2x)/2. We can put this information in a function:

f(x)=\frac{x(P-2x)}{2}

All you'll have to do now is proving that f reaches a maximum.

 

[LilTaru]

Yeah. That's the part I'm having trouble with because in the chapter, it doesn't show how to find the maximum, only that there exists one according to the theorem...

 

[micromass]

Yes, but you don't need to find the maximum do you? You just need to show a maximum exists.

If you really want to find what the maximum is. Then you will want to find the deravitive f'. The maximum then has f'(x)=0.

 

[LilTaru]

Two questions/problems:

1) How do you prove in general a maximum exists?

2) When I differentiate the function I get

f¹(x) = (-Px + 2x² + 2P - 8x)/4
What am I doing wrong because this seems so incorrect to me!

 

[micromass]

1) Proving in general that a maximum exists is done by the extreme value theorem. It states that any function f:[a,v]\rightarrow \mathbb{R} has a maximum and a minimum. But it doesn't state explicitly how to find that maximum.

2) hmm, your derivative seems wrong... I get f^\prime(x)=-2x+\frac{P}{2}. This is zero if x=P/4. Thus the rectangle's area is maximal if it is a square...

 

[LilTaru]

Yes I finally solved it! Thank you to everyone who helped!