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Section 4.12 : Differentials

6. The sides of a cube are found to be 6 feet in length with a possible error of no more than 1.5 inches. What is the maximum possible error in the volume of the cube if we use this value of the length of the side to compute the volume?

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Start Solution

Let’s get everything set up first.

If we let the side of the cube be denoted by \(x\) the volume is then,

\[V\left( x \right) = {x^3}\]

We are told that \(x = 6\) and we can assume that \(dx \approx \Delta x = \frac{{1.5}}{{12}} = 0.125\) (don’t forget to convert the inches to feet!).

Show Step 2

We want to estimate the maximum error in the volume and so we can again assume that \(\Delta V \approx dV\).

The differential is then,

\[dV = 3{x^2}dx\]

The maximum error in the volume is then,

\[\require{bbox} \bbox[2pt,border:1px solid black]{{\Delta V \approx dV = 3\left( {{6^2}} \right)\left( {0.125} \right) = 13.5\, \mbox{ft}^{3} }}\]