MHB Maximum volume using AM GM inequality

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The discussion revolves around finding the maximum volume of a carry-on bag under the airline's restriction that the sum of its dimensions must not exceed 90 cm. The optimal dimensions for maximum volume are determined to be 30 cm for length, width, and height, resulting in a maximum volume of 27,000 cm³. The use of the AM-GM inequality confirms that a cubic shape provides the largest volume for given constraints. Participants emphasize the importance of specifying units when presenting answers, as this can affect perceptions of size. The conclusion highlights that understanding the relationship between dimensions and volume is crucial in such optimization problems.
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Hi everyone,

I'm a bit confused with this question.

An airline demands that all carry-on bags must have length + width + height at most 90cm. What is the maximum volume of a carry-on bag? How do you know this is the maximum?

[Note: You can assume that the airline technically mean "all carry on bags must fit inside some rectangular prism with length + width + height at most 90cm". Remember that the volume of a rectangular prism is given by length x width x height.]

My attempt at the question:

View attachment 2497I thought my answer was to big for a volume. Any help would be greatly appreciated!
 

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I have moved this thread since this is a better fit.

Your answer looks correct to me (in $\text{cm}^3$), as I find the same value using cyclic symmetry, which implies the maximum will occur for:

$$\ell=w=h=30\text{ cm}$$
 
Thanks!
 
One way of looking at this is that a cubic centimetre is a very small volume. If you had given the result in cubic metres then it would have been $0.027\,\text{m}^3$, and you might have thought that the answer was too small.

In problems that use physical units, you should always specify the units when giving the answer.
 
That is true, I probably would have thought it was too small if the units was in m^3. Thanks!
 
I have been insisting to my statistics students that for probabilities, the rule is the number of significant figures is the number of digits past the leading zeros or leading nines. For example to give 4 significant figures for a probability: 0.000001234 and 0.99999991234 are the correct number of decimal places. That way the complementary probability can also be given to the same significant figures ( 0.999998766 and 0.00000008766 respectively). More generally if you have a value that...

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