Answer
a. $1\%$
b. $1.5\%$
Work Step by Step
Given $dx=0.005x$ for a cube edge, we have
a. surface area: $A=6x^2, dA=12xdx=0.06x^2$, thus $\frac{dA}{A}=\frac{0.06x^2}{6x^2}=0.01$ or $1\%$
b. volume: $V=x^3, dV=3x^2dx=0.015x^3$, thus $\frac{dV}{V}=\frac{0.015x^3}{x^3}=0.015$ or $1.5\%$
