Constructing a Cubical Box to Within 3 cm3

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SUMMARY

The problem involves constructing a cubical box with a volume of 125 cm³, requiring precision in the edge length to maintain the volume within a tolerance of 3 cm³. The volume formula V = a³ leads to the differential equation dV = 3a²da, where a represents the edge length. By substituting a = 5 cm and dV = 3 cm³, the solution for da reveals the acceptable variation in edge length to ensure the volume remains between 122 cm³ and 128 cm³. This approach provides a clear method for determining the necessary precision in the box's dimensions.

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Homework Statement


a) A cubical box is to be built so that it holds 125 cm3. How precisely should the edge be made so
that the volume will be corrected to within 3 cm3?

The Attempt at a Solution



I am not sure even what this is asking exactly could someone decode if they have seen a similar problem before?
 
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Yeah, it's just a differential problem, I think. V=a^3, so dV=3*a^2*da. If a=5cm then V=125cm^3. You are given dV=3cm^3. Solve for da. It's an approximate solution for how large the variation in a can be and still give you a V between 128cm^3 and 122cm^3. Approximate, mind you.
 
I thought that the question might be asking something like that but it seemed kind of strange to me. thanks
 

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