1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Local max

  1. Jun 13, 2014 #1
    A long piece of sheet metal w inches wide is to be bent into a SYMMETRIC form with three straight sides to make a rain gutter. A cross section is shown below.

    \_____/

    The base is w-2x and the angled side lengths are both x with a theta between top horizontal.

    A. Determine dimensions that allow maximum possible flow or maximum cross sectional area.

    B. Would it be better to bend the material into a gutter with a semicircular cross section than a three sided cross section?
     
  2. jcsd
  3. Jun 13, 2014 #2
    I can't seem to find a constraint. Without constraint, I cannot use lagrangian method.

    So, I take the partials of the cross sectional area function with respect to x and then with respect to theta.

    I get nowhere afterwards.
     
  4. Jun 13, 2014 #3

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    First this is clearly a home-work type question and should have been posted there.

    Second the "constraint" is x= w/4.

    Frankly I wouldn't consider this a "Lagrange multiplier" problem to begin with. I would write the volume of the trapezoid as a function of x and [itex]\theta[/itex], replace x by w/4, and treat it as a "Calculus" I problem of finding the value of the single variable [itex]\theta[/itex] that maximizes the area.
     
    Last edited: Jun 13, 2014
  5. Jun 17, 2014 #4
    Why would constraint be w/4
     
  6. Jun 17, 2014 #5
    Also, volume of trapezoid is not important here. I only need the cross sectional area of this trapezoid.
     
  7. Jun 18, 2014 #6
  8. Jun 18, 2014 #7

    verty

    User Avatar
    Homework Helper

    Show some working out please. Give us the formula for the cross-sectional area, it'll have 3 variables: w, x, Θ. If you can do that, we can start to help.
     
  9. Jun 18, 2014 #8
    [A(x,b)] = (x^2-b^2)^.5 (w + b - 2x)

    Where b is the leg of each of the right triangles in the corners.


    [A(x,@)] = (w)(x)[sin(@)] + (x^2)[sin(@)][cos(@)] - 2x^2[sin(@)]

    Above are the cross sectional surface area equations.
     
  10. Jun 18, 2014 #9

    verty

    User Avatar
    Homework Helper

    I get the same formula although you should try to simplify it a little. Okay, you want to find the maximum. Can you see how to do it? It will of course have something to do with the gradient. Refer to your textbook if you need to.
     
  11. Jun 18, 2014 #10
    Yes. No problems with gradient. It's the CRITICAL POINTS that are hard to find.
     
  12. Jun 18, 2014 #11

    verty

    User Avatar
    Homework Helper

    I've helped all I can, sorry.
     
  13. Jun 18, 2014 #12
    Lol. How have you helped?

    You've merely agreed with me... No more
     
  14. Jun 18, 2014 #13

    micromass

    User Avatar
    Staff Emeritus
    Science Advisor
    Education Advisor
    2016 Award

    Please post an attempt. Also, this thread belongs in homework, so please post there in the future.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Local max
  1. Local max/min (Replies: 11)

  2. Local Max & Min (Replies: 4)

Loading...