Section 14.4: Tangent Planes

SOLUTION
If the dimensions of the box are \(x\), \(y\), and \(z\), its volume is \(V = xyz\) and so

\[ dV = \frac{\partial V}{\partial x} dx + \frac{\partial V}{\partial y} dy + \frac{\partial V}{\partial z} dz = yz\,dx + xz\,dy + xy\,dz \]

We are given that \(|\Delta x| \leq 0.2\), \(|\Delta y| \leq 0.2\), and \(|\Delta z| \leq 0.2\).
To estimate the largest error in the volume, we therefore use \(dx = 0.2\), \(dy = 0.2\), and \(dz = 0.2\) together with
\(x = 75\), \(y = 60\), and \(z = 40\):

\[ \Delta V \approx dV = (60)(40)(0.2) + (75)(40)(0.2) + (75)(60)(0.2) = 1980 \]

Thus an error of only 0.2 cm in measuring each dimension could lead to an error of approximately 1980 cm³ in the calculated volume! This may seem like a large error, but it’s only about 1% of the volume of the box.