Question

The dimensions of a closed rectangular box are measured as 60 centimeters, 60 centimeters, and 90 centimeters, respectively, with the error in each measurement at most .2 centimeters. Use differentials to estimate the maximum error in calculating the surface area of the box.

Answer 1

The maximum error in calculating the surface area of the box is
`168\:cm^2`

Explanation:

The differential df of a function
`f=f(x,y,z)` is related to the differentials dx, dy, and dz by

`\begin{equation*}df = f_x(x_0,y_0,z_0) dx + f_y(x_0,y_0,z_0)dy+f_z(x_0,y_0,z_0)dz\end{equation*}`

We can use this relationship to approximate small changes in f that result from small changes in x, y and z.

Let the dimensions of the box be
`l`,
`w`, and
`h` for length, width, and height, respectively.

The surface area of a box is the total area of each side and is given by

`S=2(lw+wh+lh)`

The change in area can be written as:

`\Delta S\approx dS = (\partial S)/(\partial l) dl+(\partial S)/(\partial w) dw+(\partial S)/(\partial h) dh`

From the information given the partial derivatives are evaluated at
`l =60`,
`w=60`, and
`h=90`, and
`dl=dw=dh=0.2`.

The partial derivatives are

`(dS)/(dl)=2(w+h)=2(60+90)=300\\\\(dS)/(dw)=2(l+h)=2(60+90)=300\\\\(dS)/(dh)=2(l+w)=2(60+60)=240`

Substituting these in for
`dS`,

`dS = 300\cdot 0.2+300\cdot 0.2+240\cdot 0.2\\\\dS=2\cdot \:300\cdot \:0.2+240\cdot \:0.2\\\\dS=120+48=168`

Thus, the maximum error in calculating the surface area of the box is
`168\:cm^2`