Since the load L varies directly with the product of width and square of the height h, and inveresly as the length l, so
[tex]\begin{gathered} L=k(\frac{wh^2}{l}) \\ OR \\ \frac{L_1}{L_2}=\frac{w_1}{w_2}\times\frac{h^2_1}{h^2_2}\times\frac{l_2}{l_1} \end{gathered}[/tex]We will use the second rule
Since L is 8740 pounds when w is 5 in., h is 7 in. and l is 144 in.
[tex]\begin{gathered} L_1=8740 \\ w_1=5 \\ h_1=7 \\ l_1=144 \end{gathered}[/tex]We need to find L when w is 6 in., h is 9 in. and l is 216 in.
[tex]\begin{gathered} L_2=? \\ w_2=6 \\ h_2=9 \\ l_2=216 \end{gathered}[/tex]Let us substitute them in the second rule
[tex]\begin{gathered} \frac{8740}{L_2}=\frac{5}{6}\times\frac{7^2}{9^2}\times\frac{216}{144} \\ \frac{8740}{L_2}=\frac{5}{6}\times\frac{49}{81}\times\frac{216}{144} \\ \frac{8740}{L_2}=\frac{245}{324} \end{gathered}[/tex]By using cross multiplication
[tex]\begin{gathered} 245\times L_2=8740\times324 \\ 245L_2=2831760 \end{gathered}[/tex]Divide both sides by 245
[tex]\begin{gathered} \frac{245L_2}{245}=\frac{2831760}{245} \\ L_2=11558.20408 \end{gathered}[/tex]Round it to the nearest integer
[tex]L_2=11558\text{ pounds}[/tex]The load is 11558 pounds
