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Solving for perimeter and dimensions of a rectangle

  1. Apr 26, 2009 #1
    I understand calculus, I just don't understand how it is applied to solve these sorts of problems.

    1. The problem statement, all variables and given/known data
    What is the smallest perimeter for a rectangle with an area of 16, and what are its dimensions?


    2. Relevant equations
    A=L*W
    P=2W+2L

    3. The attempt at a solution

    I managed to get the perimeter equation down to one variable like so:

    Area is 16.

    A=LW
    16=LW
    16/L=W

    P=2L+2(16/L)
    P=(2L2+32)/L

    and this is where I stop. I just don't see how this can be applied to the situation stated.
     
  2. jcsd
  3. Apr 26, 2009 #2

    danago

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    Gold Member

    Well you now have a function, P, of just one variable, L. You want to find when this function is a minimum? Have you been shown how to find the minimum points of a function using calculus?
     
  4. Apr 26, 2009 #3
    Yes I have.

    I need to look for locations where the first derivative is zero and the second derivative is positive on both sides of the critical point.

    I'm just having a tricky time actually applying calculus to these situations.
     
  5. Apr 26, 2009 #4

    danago

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    Gold Member

    Ok. The easiest way to do these types of problems is probably to first find all of the critical point of the function (i.e. where the derivative is zero or undefined). From there, you can then check whether the critical point(s) you find are local maximums/minimums or inflections by evaluating the second derivative at those points.
     
  6. Apr 26, 2009 #5

    HallsofIvy

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    Staff Emeritus
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    Perhaps because you are not trying! What is the derivative of P= 2L+ 32/L? Where is that derivative 0?
     
  7. Feb 4, 2012 #6
    P=(2L^2+32)/L

    Where do you get the power of 2?

    "P=2L+2(16/L)
    P=(2L2+32)/L" ??

    P=2L+2(16/L)
    P=2(L+16/L)

    or

    2L+32/L
    L+16/L
     
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