Homework Help: Calculus

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

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 - MathMate, Friday, February 25, 2011 at 10:45pm
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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