If the width of a rectangular bean is increased by 11%, the depth decreased by 8%, and the length increased by 6%, by what % will the safe load change?

Solution:

We assume the width is W, the length is L, the depth is D

Safe load = volume = WLD

Here

W'= 1.11W

L = 1.06L

D' = 0.92D

So

V' = 1.11W*1.06L*0.92D = 1.0825 (WDL)

So the safe load is increased by 8.25%