Calculus
posted by Andrew .
If the tolerance on the volume of a cube is a 2% error then what should the tolerance be fore each side?
So what i did was
V=x^3
dV = 3x^2 dx
dV/V = 2%
What should I do next? the answer is 2/3% but I don't know how it works.
Thanks.

Let the three cube dimensions be w, l and h
V = w l h
dV/V = (lh dw + wh dl + wl dh)/(wlh)
= dw/w + dl/l + dh/h = 2%
If you apply the same relative tolerance to width, length and height, each should be 1/3 of 2%
There is no reason that different tolerances could not be applied to different dimensions, however.