**8/15/2014
**The
maximum safe load for a horizontal beam varies jointly as the width
of the beam and the square of the thickness of the beam, and
inversely with its length. How will the thickness of the beam have
changed if the maximum safe load were halved while the width of the
beam was doubled, and its length increased by a factor of 3?

**Scroll
down for the solution.**

**8/15/2014
**Solution.**
**Let
S denote the maximum safe load for the beam, *w*,
the width of the beam, T, its thickness and *l*,
its length.

Then,
S = *kw*T^{2}/*l*
where *k*
is the constant of proportionality (*)

Initially,
let *w* = S = *l*
= 1unit.

Then,
T = 1/√*k*
[Substituting *w* = S
= *l* = 1 in (*) and
solving]

Next,
as indicated, let S = ½, *w*
= 2 and *l* = 3.

Then, T = √(3/4k) [Substituting these values in (*) and solving]

= ½ √(3/k)

T = ≈0.85/√k [√3 ≈ 1.7]

Therefore,
**the
thickness would have dropped by 15%**.