Answer
$22.5 \text{ tons}$
Work Step by Step
The variation model described by the problem is $
W=\dfrac{kwh^2}{l}
,$ where $W$ is the maximum weight, $w$ is the width, $h$ is the height, and $l$ is the length.
Substituting the known values of the variables results to
\begin{array}{l}\require{cancel}
W=\dfrac{kwh^2}{l}
\\\\
12=\dfrac{k(1/2)(1/3)^2}{10}
\\\\
12(10)=k(1/2)(1/3)^2
\\\\
120=k\left(\dfrac{1}{2}\right)\left( \dfrac{1}{9} \right)
\\\\
120=k\left(\dfrac{1}{18}\right)
\\\\
120=\dfrac{k}{18}
\\\\
120(18)=k
\\\\
k=2160
.\end{array}
Hence, the equation of variation is
\begin{array}{l}\require{cancel}
W=\dfrac{2160wh^2}{l}
.\end{array}
Using the variation equation above, then
\begin{array}{l}\require{cancel}
W=\dfrac{2160wh^2}{l}
\\\\
W=\dfrac{2160(2/3)(1/2)^2}{16}
\\\\
W=\dfrac{2160\left(\dfrac{2}{3}\right)\left( \dfrac{1}{2}\right)^2}{16}
\\\\
W=\dfrac{2160\left(\dfrac{2}{3}\right)\left( \dfrac{1}{4}\right)}{16}
\\\\
W=\dfrac{\left(\dfrac{4320}{12}\right)}{16}
\\\\
W=\dfrac{360}{16}
\\\\
W=22.5
.\end{array}
Hence, the maximum weight, $W,$ is $
22.5 \text{ tons}
.$
