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?

1 Answer
Mar 9, 2015

The answer is 184184 cm^2cm2, but here's how:

The surface area of a closed rectangular box is A = 2LW+2WH+2LHA=2LW+2WH+2LH,
where LL=length, WW=width, HH=height.

The differential of area is used as the approximate maximum error.
A = 2[LW+WH+LH]A=2[LW+WH+LH].

dA=2*[((dL)W+L(dW))+((dW)H+W(dH))+((dL)H+L(dH))]dA=2[((dL)W+L(dW))+((dW)H+W(dH))+((dL)H+L(dH))].

Each measurement has same maximum error, all of the differentials are the same. Namely, 0.20.2 cmcm, so
dL=dW=dH=0.2dL=dW=dH=0.2 cmcm.
L=60L=60 cmcm , W=80W=80 cmcm and H=90H=90 cmcm.

Substituting into:
dA=2*[((dL)W+L(dW))+((dW)H+W(dH))+((dL)H+L(dH))]dA=2[((dL)W+L(dW))+((dW)H+W(dH))+((dL)H+L(dH))].
gives:
dA=2*[((0.2)80+60(0.2))+((0.2)90+80(0.2))+((0.2)90+60(0.2))]dA=2[((0.2)80+60(0.2))+((0.2)90+80(0.2))+((0.2)90+60(0.2))]
=0.4[80+60+90+80+90+60]=0.4[80+60+90+80+90+60]
=0.4(460)=184=0.4(460)=184 cm^2cm2.