Calculus
posted by John on .
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_

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(d²w²)
Let the strength of the resulting rectangular beam be
S(w)=k*w*h²
=k*w*(sqrt(d²w²)²
=k*w*(d²w²)
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(wd²w³))/dw
=k(d²3w²)
Equating dS/dw=0 and solving for w:
w=sqrt(d²/3)
and therefore
h=sqrt(d²w²)
=sqrt(2d²/3)