Question

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?

Answer 1

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.