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Global Extrema/Area of a Rectangle

  1. Apr 18, 2009 #1
    1. The problem statement, all variables and given/known data
    Find the dimensions of of the rectangle with perimeter 200 meters that has the largest area.

    2. Relevant equations

    3. The attempt at a solution
    This is in the section on Global Maxima/Minima so I know it has to be something with graphing a formula and finding the maxima, but I cannot figure where to start on this. The book is no help (Applied Calculus - Hughes-Hallett, et.al) and this is not the first time it has presented problems that it does not even began to describe in the text. :grumpy:

    Can someone tell me where to start?
  2. jcsd
  3. Apr 18, 2009 #2
    What is the equation for the perimeter of a rectangle with length x and height y (and what do you have to set this equal to)? What is the equation for the area of this rectangle?
  4. Apr 19, 2009 #3
    Ok, so I have 200=2(x+y), and the area is A=xy, right? I must be missing something terribly obvious, because I am still stuck on where to go from here, except that maybe 100=x+y. How does this fit into a Global Extrema problem?
  5. Apr 19, 2009 #4
    Okay, good! Now, you need to write the area in terms of just one variable so that you can find the maximum of the area function. Any ideas?
  6. Apr 19, 2009 #5
    y=A/x? Man, I can almost get this, I feel like I'm so close, but where does the perimeter function come in, y=100-x?

    Is this problem an an easy one? Because it seems like it, and it's making me feel so dumb that I can't figure it out!
  7. Apr 19, 2009 #6
    You are so close! You are trying to maximize the area, so you need to write the area as a function of x (or y) only. You can solve for x in terms of y (and vice versa) using the perimeter constraint. 2x+2y = 200 => y = 100-x => A = x(100-x). Now how do we find the value of x when this function is at its maximum? Once you find this x, you can solve for y using the relationship we found between x and y using the perimeter constraint. Is this clear?
  8. Apr 19, 2009 #7
    50x50! Dude, thanks for your patience on that one, it was driving me crazy! It makes sense now.
  9. Apr 19, 2009 #8
    You're welcome! Also, note that this answer may have been what you expected from the beginning: a square!
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