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Homework Help: Optimization help!

  1. Apr 26, 2010 #1
    The problem states: The trough in the figure is to be made to the dimensions shown. Only the angle theta can be varied. What value of theta will maximize the troughs volume?

    http://img81.imageshack.us/img81/5963/24ni3.jpg [Broken] (There is an image of the problem)

    I know the height in terms of theta is h/1 = h

    So the cross sectional area is

    1/2(h)(1)sin(theta) = 1/2 cos(theta)sin(theta)

    This is as far as I have gotten. I don't even know if this is right. Please help!
    Last edited by a moderator: May 4, 2017
  2. jcsd
  3. Apr 27, 2010 #2
    I dont know the procedure but i can say this:
    Area of a triangle = (1/2)h(x). You have two triangles so Area = h(x). 'x' is the base of triangle and h is height. Now you have some portion in-between the two triangles and its area = h.
    So final area of the cross section is = [tex]h(x+1)[/tex].
    Consider one triangle and we have by Pythagoras law:
    [tex]x^2+h^2=1[/tex] and [tex]\sin\theta=x[/tex] so [tex]h^2=1-\sin^2\theta[/tex].
    Now Area becomes = [tex]h(1+\sin\theta)=(1+\sin\theta)\sqrt{(1-\sin^2\theta)}[/tex]. [Note: in this relation you have only angle].
    Finally you get:
    Area = [tex]\cos\theta(1+\sin\theta)[/tex].
    This is all what i got. I think may be you have to solve for the area such that area becomes maximum.
    Actually i don't know how to solve using formula..but by plotting i guess it is << answer deleted by berkeman >>
    hope this helps.
    Last edited by a moderator: Apr 27, 2010
  4. Apr 27, 2010 #3


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    Staff: Mentor

    Rajini, please take care not to do too much of the OP's work for them, and please do not post answers. Thanks.
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