# Extreme Value Theorem: Maximum Area of Rectangle

## Homework Statement

This problem had been previously posted, however i have specific questions about it and don't feel that it was completely answered for it did not explain how you use the extreme value theorem in the problem and specifically, HOW DO YOU GET THE INTERVAL [a,b], The answer has the interval [0,P/2] but i don't understand why.
"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."
*Also i didn't think that replying on the original thread would be productive because i realized that it had been posted well over 2 years ago and i didn't think i would get a response lol

## Homework Equations

P = 2l + 2w → l = (P-2w)/2
A = lw → A(w) = w((P-2w)/2)

## The Attempt at a Solution

Let w rep the width
Let l rep the length
A(w) = w((P-2w)/2) = (Pw - 2w2)/2
First Derivative: A'(w) = (P - 4w)/2
Find where A'(w) = 0 which (i guess) means either a max or a min
0 = (P - 4w)/2 → P/2 -2w → 2w = P/2 → w = P/4
Using w we can find l :
l = (P - 2(P/4))/2 = P/4
Therefore the maximum area the rectangle can be is a square of dimensions P/4 x P/4.
However i did not use the Extreme Value Theorem anywhere in this answer! For some reason the answer page says that the interval is w$\in$ [0,P/2], did not take the derivative of A, had A = -w2 + Pw/2 then completing the square to get:
A = -(w - P/4)2 + P2/16
From that they determined that w = P/4 (I don't know how they did that)

Please help me answer this question using Extreme Value Theorem and explaining how you obtain the interval as being [0, P/2] (Also can it be other values or does it have to be this interval) THANK YOU!