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, 100 centimeters, and 90 centimeters, respectively, with the error in each measurement at most .2 centimeters?

Let's call the measurements #x,y,z# for now.

Then the total surface area will be
#A_0=2*xy+2*xz+2*yz=2(xy+xz+yz)#

If we include the differences and set these to #d#
#A_1=2((x+d)(y+d)+(x+d)(z+d)+(y+d)(z+d))#

#=2((xy+xd+yd+d^2)+(xz+xd+zd+d^2)+(yz+yd+zd+d^2))#

Subtract the original surface area #A_0#

#DeltaA=2(xd+yd+d^2+xd+zd+d^2+yd+zd+d^2)#
#=2(2xd+2yd+2zd+3d^2)#

Since #d^2# is very small compared to the rest, we can ignore it.
#DeltaA=4d(x+y+z)# now fill in the numbers:
#DeltaA=4*0.2(60+100+90)=0.8*250=200cm^2#
For maximum error #-0.2# the answer would be #-200cm^2#

Answer : the maximum error in surface area is #200cm^2#
(On a calulated area of #40800cm^2#, less than #0.5%#)

Remark : if we had taken the #d^2# into account the error would be #0.24cm^2# greater (for positive error) or #0.24cm^2# smaller (for negative error).
This is way below the significance range.