# How do you use differentials to estimate the maximum error in calculating the surface area of the box if the dimensions of a closed rectangular box are measured as 60 centimeters, 80 centimeters, and 90 centimeters, respectively, with the error in each measurement at most .2 centimeters?

Mar 9, 2015

The answer is $184$ $c {m}^{2}$, but here's how:

The surface area of a closed rectangular box is $A = 2 L W + 2 W H + 2 L H$,
where $L$=length, $W$=width, $H$=height.

The differential of area is used as the approximate maximum error.
$A = 2 \left[L W + W H + L H\right]$.

$\mathrm{dA} = 2 \cdot \left[\left(\left(\mathrm{dL}\right) W + L \left(\mathrm{dW}\right)\right) + \left(\left(\mathrm{dW}\right) H + W \left(\mathrm{dH}\right)\right) + \left(\left(\mathrm{dL}\right) H + L \left(\mathrm{dH}\right)\right)\right]$.

Each measurement has same maximum error, all of the differentials are the same. Namely, $0.2$ $c m$, so
$\mathrm{dL} = \mathrm{dW} = \mathrm{dH} = 0.2$ $c m$.
$L = 60$ $c m$ , $W = 80$ $c m$ and $H = 90$ $c m$.

Substituting into:
$\mathrm{dA} = 2 \cdot \left[\left(\left(\mathrm{dL}\right) W + L \left(\mathrm{dW}\right)\right) + \left(\left(\mathrm{dW}\right) H + W \left(\mathrm{dH}\right)\right) + \left(\left(\mathrm{dL}\right) H + L \left(\mathrm{dH}\right)\right)\right]$.
gives:
$\mathrm{dA} = 2 \cdot \left[\left(\left(0.2\right) 80 + 60 \left(0.2\right)\right) + \left(\left(0.2\right) 90 + 80 \left(0.2\right)\right) + \left(\left(0.2\right) 90 + 60 \left(0.2\right)\right)\right]$
$= 0.4 \left[80 + 60 + 90 + 80 + 90 + 60\right]$
$= 0.4 \left(460\right) = 184$ $c {m}^{2}$.