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Maximizing Area of Cross-Section of Isosceles Trapezoid

  1. Mar 7, 2009 #1
    1. The problem statement, all variables and given/known data

    attached. It is just part a i am attempting for now.

    2. Relevant equations

    area of trapezoid = h(b1+b2 / 2)
    where h is height, b1 is base one and b2 is base two.

    3. The attempt at a solution

    i really tried, i didn't know where to start. the answer i got was 30cm for base, while the correct answer was 20cm for base and sides. I am not sure where to go with this problem my first try was a guess and obviously didn't work properly. I think maybe i need someone to better explain this problem to me.
     

    Attached Files:

  2. jcsd
  3. Mar 8, 2009 #2

    tiny-tim

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    Hi ha9981! :wink:

    Show us your full calculations, so we can see where the problem is, and we'll know how to help! :smile:
     
  4. Mar 8, 2009 #3
    I am stuck on understanding exactly what the problem is asking.
    I know i have to maximize the area, which i have the equation for, so ill find the derivative and set it to zero (due to turning points which are max and mins).

    But I need a related equation as well, is it going to be perimeter?


    here is my attempt:

    Let x be the length folded.
    Let y be the width folded.

    total length = 5 - 2x

    total width = AD + BC + AB
    0.60 = 2y + AB

    also i know:

    x = BCsin60
    x = ysin60

    area of trapezoid = ysin60 ((0.60+(0.60-2y)) / 2)
    area of trapezoid = ysin60 ((1.20 - 2y)/2)

    Now i know there is something fatally wrong here since we didn't learn derivatives of sin function so i can't progress!
     
  5. Mar 8, 2009 #4

    tiny-tim

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    Hi ha9981! :smile:

    (i haven't checked your formula, but …)

    sin60 is a constant

    differentiation not needed! :wink:
     
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