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August 28, 2014

August 28, 2014

Posted by **John** on Friday, February 25, 2011 at 10:22pm.

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 -
**MathMate**, Friday, February 25, 2011 at 10:45pmAssume 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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