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Math Help - Optimization of a Trough

  1. #1
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    Optimization of a Trough

    pts)
    A covered metal triangular trough is constructed as follows:

    A square shaped sheet of metal which is 200 centimeters wide and long square is folded along the center. Next, two pieces of metal in the shape of isosceles triangles are are welded to the ends. (See picture above). Finally, a metal cover is attached to the top (not shown).
    We want to find the smallest and largest surfaces area that a so constructed trough can have, and at what opening angle of the triangular pieces it is attained. Proceed as follows:
    Find the surfaces area as a function of the angle. (Be sure to include all five sides of the trough).

    The natural domain of is an interval with left endpoint and right endpoint .
    On its domain has one stationary point . Although there is no explicit formula for the value of itself, it is possible to derive the exact value of the cosine of . Find it:

    Find from this the surfaces area at the stationary point:
    Find also the global minimal value of on its natural domain:


    Ok I have tried doing the first part and was pretty confident that I had the right answer but the computer was telling me that I was wrong. I know that 0 is the endpoint of the domain of w. Please Help
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  2. #2
    MHF Contributor Unknown008's Avatar
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    Could you show us what you've done?
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    ok i just now figured out the first part
    but i know surface area is the area of all of the sides added to each other
    i know that the square folded into a v has side lengths of 200 so A=lw so that Area=40,000
    the area of a triangle is A=.5bh
    h=100cos(w) and b=100sin(w)
    there are 2 b for each triangle so the area of one of the triangles is A=.5*100cos(w)*2*100sin(w)=10000cos(w)sin(w)
    there are two triangles so their area is A=20000sin(w)cos(w)
    for the rectangle on top A=lw
    the length is 200 and the width is 2b so A=200*2*100sin(w)=40000sin(w)
    Add all these areas up and you get
    Surface Area=40000+20000sin(w)cos(w)+40000sin(w)

    i also know that the domain of w is from (0,pi/2) because it can't have an angle greater than 90 degrees
    I am stuck on the next part where it is talking about a stationary value (which is where the first derivative=0) and cos(c)
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  4. #4
    MHF Contributor Unknown008's Avatar
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    Yes, find the derivative of S(w).

    At w = c, S'(c) = 0.
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    i found the derivative to of S(w) to be S'(w)=20000(cos^2(w)-sin^2(w))+40000cos(w)
    i set s'(w)=0 and got w=1.196062
    i'm not sure what it wants when it asks for cos(c)=?
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  6. #6
    MHF Contributor Unknown008's Avatar
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    What about using the identity;

    \cos2A = \cos^2A - sin^2A = 2cos^2A - 1 = 1 - 2sin^2A

    Choose the one which you think will lead you to the answer
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  7. #7
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    ok im confused which one to use
    i have
    cos^2(w)-sin^2(w)+2cos(w)=0
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  8. #8
    MHF Contributor Unknown008's Avatar
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    Quote Originally Posted by rawkstar View Post
    ok im confused which one to use
    i have
    cos^2(w)-sin^2(w)+2cos(w)=0
    w is now c.

    And use 2cos^2A - 1 to get everything in terms of cos.
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  9. #9
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    so correct me if im wrong, you're telling me to change
    cos^2(c)-sin^2(c)+2cos(c)=0
    to
    2cos^2(c)-1+2cos(c)=0 ?
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  10. #10
    MHF Contributor Unknown008's Avatar
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    Yes.

    However, your problem seems you give you all sorts of 'odd' numbers. I'm afraid you'll have to express it into radicals, using the equadratic formula.
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