Calculus

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I need help with this question:
The strength of a beam with a rectangular cross section varies directly as x and as the square of y. What are the dimensions of the strongest beam that can be sawed out of a round log with diameter d?

What am I supposed to do?

Thx in advance_

  • Calculus -

    Assume the log to have a perfect circular section, of diameter d.
    The radius is therefore r=d/2.

    We have a choice of cutting a beam out of the log of width w, and height h, as long as sqrt(w²+h²)≤d.
    We can eliminate "h" at the source using equality and the above Pythagoras relation, i.e.
    h=sqrt(d²-w²)

    Let the strength of the resulting rectangular beam be
    S(w)=k*w*h²
    =k*w*(sqrt(d²-w²)²
    =k*w*(d²-w²)

    where k is a constant of proportionality.

    We look for the maximum value of S(w) by varying w, so we set dS/dw=0:
    dS/dw=d(k(wd²-w³))/dw
    =k(d²-3w²)
    Equating dS/dw=0 and solving for w:
    w=sqrt(d²/3)
    and therefore
    h=sqrt(d²-w²)
    =sqrt(2d²/3)

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