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Earlier last week I realized I needed to ship a large volume of things domestically. Of course, I decided that I wanted to do so as cheaply as possible.

I first looked at USPS standard post rates. I noted that if the "combined length and girth" of a box exceeds 108", then the package becomes significantly more expensive, as it uses the "oversized" price bracket, so I resolved not to exceed this size limit. The length of a box is defined as its longest dimension, and its girth is defined as the perimeter around the rectangle to which the length is orthogonal – in other words, the girth is $2w + 2h$.

However, because I have much to ship, I also would like to maximize the volume of such a box while not exceeding this limit. This turns into a maximization problem with a constraint:

Maximize $V(x) = lwh$, while satisfying $108 >= l + 2w + 2h$.

This seems pretty simple, but I'm not sure how to handle the three nominally independent variables $l$, $w$, and $h$. Because I did not want to wait or rely on my faulty math skills, I ultimately asked Wolfram|Alpha to solve this for me, yielding a box of size 36"x18"x18". This makes intuitive sense, to a degree, but I would like a hint as to how I could have proceeded to solve this on my own.

ravron
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2 Answers2

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You asked for a hint. Here are two.

(1) You certainly want $l+2w+2h=108$.

(2) If you temporarily fix $l$, you have a certain amount to make a rectangle with $w$ and $h$. The maximum area would be to make a square out of $w$ and $h$ (i.e., $w=h$).

paw88789
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  • Thanks! Using this same scheme, I can get this to a one-variable maximization problem, and then it's trivial. I knew it couldn't be that hard! – ravron Apr 06 '15 at 16:44
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The AM/GM inequality shows immediately that the maximum volume occurs uniquely when $l=2w=2h$.

Bill Kleinhans
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