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Math Help - Word problem?

  1. #1
    Member maxpancho's Avatar
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    Word problem?

    The strength of a rectangular beam is proportional to the product of its width w times the square of its depth d. Find the dimensions of the strongest beam that can be cut from a cylindrical log of radius r



    Answers: $w=\dfrac{2r}{\sqrt{3}}, h=\dfrac{2\sqrt{2}r}{\sqrt{3}}$
    Is this problem even correct? First of all I would think that depth is the dimension that goes inside the log; then if depth^2 times width = strength then since depth isn't bounded by anything then it follows that I should just take the biggest possible width.

    And if the depth is this vertical side of the beam as shown in the picture, then why does the answer they give is in terms of w and h? What is h (height I assume?) then? Is it the dimension that "goes inside" the log? But then it isn't bounded by anything, so I can make it arbitrarily big and it wouldn't matter.

    Can someone clear this up?
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  2. #2
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    Re: Word problem?

    You are to maximise $f(d,w)=wd^2$ subject to the constraint $(d/2)^2+(w/2)^2=r^2$ .. which is the requirement that the beam fit in the log.
    Last edited by zzephod; August 28th 2014 at 12:20 AM.
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  3. #3
    Member maxpancho's Avatar
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    Re: Word problem?

    Why do we divide by two?
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    Re: Word problem?

    Quote Originally Posted by maxpancho View Post
    Why do we divide by two?
    Draw a line segment from the centre to the top right corner. Label the centre O and the top right corner A. Draw a normal from the centre to the right hand side of the beam and label the point where it meets the right hand edge B. Now OBA is a right triangle with right angle at B, and sides r, the radius, and w/2 and d/2.

    Alternatively draw in the diagonal of the beam. This passes through the centre of the circle, and so is a diameter of length 2r. So now we have a right triangle of sides 2r, w, d.

    Now apply Pythagoras' theorem to these sides.

    .
    Last edited by zzephod; August 28th 2014 at 06:20 AM.
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