Extreme Value Theorem: Maximum Area of Rectangle

  • Thread starter tmlrlz
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  • #1
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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[itex]\in[/itex] [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!

Homework Statement





Homework Equations





The Attempt at a Solution

 

Answers and Replies

  • #2
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You did use the EVT, although you didn't realize it. When you write, "Find where A'(w) = 0 which (i guess) means either a max or a min," you are implicitly using EVT.

The Extreme Value Theorem says, simply, that a continuous function on a closed, bounded interval attains both its maximum and its minimum at least once each inside of the interval.

Consider what values w can realistically take in the equation P = 2l + 2w. Can w ever be greater than P/2? If l = 0, then what is w? If l = P/2, then what is w? That gives you the possible interval for your Area as a function of w.
 
  • #3
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I looked at the problem again and realized that if you were to graph the function then you would realize that it was a parabola facing downwards. Then you realize as well that the area of a rectangle and the perimeter of a rectangle cannot be less than zero, they must be positive, which means that there is a maximum in the interval where A(x) = 0 which is at 0 and P/2 so between those two values for w there will be a maximum which occurs at P/4 and resulting in a length of P/4. Thank You!
 

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