How To Calculate Maximum Dimensions Given A Set Girth

so the maximum is when 2(l) = 2(w) = h

if h must be less than w or l, then the max is
when l=w=h=165/5

ChargingPaladin:
Thank you so much for a simple, easy to understand explanation and solution to my problem.  You truly deserve the points for this question.

Ozo:
I do sincerely appreciate your attempt to assist me with this problem, however you should know that I could not find much basis for your arguments, especially given the very succinct discussion given by ChargingPaladin.  I was able to verify that from other sources that a cube would indeed yield the maximum volume for any box and therefore contradicts the formulas you provided me.  I have thus chosen to accept his arguments over yours.  I hope that you do not find my decision offensive, as that is the furthest from my intentions, my fullest respect to you as one of EE's leading experts.

Volume and girth are not linear.  A box with a given girth will have maximum volume as a cube.  The volume will decrease as you change the variables of L W H.

Provided measurements are constrained to whole inch units:
A box that is 33x33x33 has a girth of 165 and volume of 35,937
A box that is 1x1x81 has a girth of 165 and volume of 81

You would need to constrain the box sizes on two dimensions and limit the last to get your 165 (used to be 88?)

A cube has the maximum volume under the constraint
l + w + h = 165
then (165/3)^3= 166375
But if the constraint is
2(l) + 2(w) + h = 165
then it's better to let h be larger, since the l and w use up more of the 165
a cube under those constraints would have volume
(165/5)^3 = 35937
whereas
(165/6) * (165/6) * (165/3) = 41593.75
Given the additional constraint that h is the smallest dimension
then the maximum is when h as close to (165/3) as possible.  i.e.
when l=w=h=165/5
as I said in http:#a25137848
and affirmed by ChargingPaladin
If you find no basis for this, then you should reject ChargingPaladin's answer as well, until you can verify that it indeed has very solid basis.
There is no contradiction to formulas I provided

Although I answered the question as asked as a strictly Math & Science zone question,
it looks like what UPS describes here:
http://www.ups.com/content/us/en/resources/prepare/weight_size.html#How+To+Measure+Your+Package+Size
is in contradiction to what is described in the question.
It looks like it should be the largest dimension that is multiplied by one, and the two smaller dimensions that are multiplied by two, in which case it would be possible for the largest dimension to reach its optimal value value of 165/3 and exceed the volume of a 165/5 cube.

Thank you so much ozo, for your feedback.  It appears to me that there has been a breakdown in communication between a number of parties on this issue.  The UPS webpage you directed me to is indeed at variance with what I was told by one of its technicians.  This is something I will need to clear up with them later on.  Regardless of that mixup, as far as this specific question is concerned, it appears that you took a strict mathematical approach to the problem and defined girth as being l+w+h=girth.  However, the definition that I had received was that of two times the longest sides plus the length of the third dimension.

The contradictions regarding this problem occurred, I believe, because of the difference in assumptions.  Whereas you stated in http:#a25137753 that maximum volume would be achieved at l=w=165/6, h=165/3, according to the definition I provided which was relayed to me by one of UPS's technicians, the maximum volume would be achieved at l=w=h=165/5 (in agreement with ChargingPaladin's response), and which also contradicts your formula in http:#a25137805.

It also appears that ChargingPaladin's formula will also work given the formula from the UPS webpage, considering the volume of the box is still at its maximum when all sides are equal.

Once again, I do thank you for your kind assistance with this issue.

I did not define l+w+h=girth
that would have given a max at l=w=h=girth/3
You specified  2(l) + 2(w) + h = 165
which has a maximum at l=w=165/6, h = 165/3
If in addition to that, h (the dimension without the factor of 2) is specified to be smaller than the other dimensions, rather than larger than the other dimensions as seems to be the UPS practice,
then that maximum is unattainable, and the closest you can get is  l=w=h=165/5
as I have said three times now.
However, if you use the convention from the UPS webpage, then the maximum volume is attained when the larger dimension is twice length of the smaller dimensions
(2x Width =2x Height = Length, if we rename the dimensions to follow the webpage)
which I have now also stated three times, and which contradicts ChargingPaladin's answer, which would be incorrect in that case, although it was the correct maximum for the question as it was stated,  as was my answer of l=w=h=165/5

Thank you once again for your comments, ozo.  I do stand corrected, you did not define l+w+h=girth, however in response to my question in your very first post you stated that maximum volume would be attained where l=w=165/6 and h=165/3.  This would give a box that is twice as high as it is long or wide.  As I also stated previously, sources other than ChargingPaladin have confirmed that a cube, where all sides are equal (see also mikelfritz's belated comment above), would yield the maximum volume.

The reason I did not give you credit for the correct answer of l=w=h=165/5 is because in http:#a25137848, you stated, "if h must be less than w or l, then the max is
when l=w=h=165/5", and h, in accordance with all our previous discussions, must not be less than w or l, but equal.

ozo, forgive me, as I am not of myself a mathematician.  Please explain how you can say that in absence of the condition "h must be less than w or l", the maximum volume is attained when h is greater than w and l?  In either case, if h is less than or greater than l and w, you do not have a cube.  Do you not agree that a cube has the greatest volume given a set girth?

> Do you not agree that a cube has the greatest volume given a set girth?
It depends on your definition of girth
if you want to be pedantic the UPS website defines girth as (2x Width + 2x Height) where Width and Height exclude the longest side, and the constraint is length + girth = 165, in which case, I do not agrree,
A cube would not have the largest volume, as you can verify that 27.5 * 27.5 * 55 > 33 * 33 * 33
If you define it as 2(l) + 2(w) + h, where h is the shortest side, then yes, a cube would be the greatest volume, but only because you are restricting h to to not get any larger to increase the volume.

? Wow, ozo, I can certainly see where you got your Genius ranking.  I am going to have to get with the moderators to see if there could be some reassigning of points.

Thank you for your time and patience to clear up this matter.

I think there is some further misinterpretation going on here. To clear things up a bit, Ozo, according to the website you linked earlier the girth is in fact a completely different measurement than defined in the original question. This however was merely an error in the name given to what OmniUnlimited was looking for. According to the UPS website, we're actually looking for "Package Size," (PS). The real reason for the discrepancy here is, as you said, which side of the inequality the lone measurement falls on.

Using a mix of the UPS site's terminology and what their representative told OmniUnlimited, the PS, not the girth, must equal 165.

With the largest measurement being that which is not multiplied by anything (L according to the UPS site) then yes (27.5 x 27.5 x 55) is the maximum volume. However, it appears that the UPS representative misled OmniUnlimited by saying the opposite, that the smallest measurement is the lonely one, which gave rise to the answer for the question that was asked, (33 x 33 x 33).

The moral of the story is that while both answers are correct in different contexts, I would check with your UPS representative again, OmniUnlimited, as they are the ones that will be charging you!

:) Thank you ChargingPaladin!  I will be sure to check with UPS to clear up the descrepancy.  However, depending on what the moderators tell me, I think it only fair to divide points among the both of you.  I have asked for the ability to open up this question once more and increase points due to its unforeseen complexity so that the loss to you will be minimum.

Thanks once again.  You have all that it takes to be one of EE's top leaders.

This comment is to test and see if the point value of this question will increase with its submission.

Outstanding job to the both of you.  Thank you very much, especially to you ozo, for all the time and effort spent in assisting my needs.  Best of luck to you both!