Answer
(a) The maximum possible error is $270~cm^3$
The relative error is $0.01$
The percentage error is $1\%$
(b) The maximum possible error is $36~cm^2$
The relative error is $0.0067$
The percentage error is $0.67\%$
Work Step by Step
(a) Let $x$ be the length of the edge.
$x = 30~cm$
$dx = 0.1~cm$
We can find the maximum possible error:
$V = x^3$
$dV = 3x^2~dx$
$dV = 3(30~cm)^2~(0.1~cm)$
$dV = 270~cm^3$
We can find the relative error:
$\frac{dV}{V} = \frac{3x^2~dx}{x^3} = 3~(\frac{dx}{x})$
$\frac{dV}{V} = 3~(\frac{0.1~cm}{30~cm})$
$\frac{dV}{V} = 0.01$
The percentage error is $1\%$
(b) Let $x$ be the length of the edge.
$x = 30~cm$
$dx = 0.1~cm$
We can find the maximum possible error:
$A = 6x^2$
$dA = 12x~dx$
$dA = 12(30~cm)~(0.1~cm)$
$dA = 36~cm^2$
We can find the relative error:
$\frac{dA}{A} = \frac{12x~dx}{6x^2} = 2~(\frac{dx}{x})$
$\frac{dA}{A} = 2~(\frac{0.1~cm}{30~cm})$
$\frac{dA}{A} = 0.0067$
The percentage error is $0.67\%$