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For a fixed $n$, what is the shape with the smallest volume, such that by rotation and translation, it can cover any $n$-box with dimension $b_1\times \ldots \times b_n$, where $b_1+\ldots+b_n=1$.

I had this problem when I learned some Chinese airlines have restriction on the sum of the dimensions for carry-on baggage, but no restriction on any individual dimension. It is natural to ask how small they can make the compartment, so the baggage still fits.

Let $B(b_1,\ldots,b_n) = \{(x_1,\ldots,x_n)| 0\leq x_i\leq b_i \}$, then clearly the following shape can cover the $n$-boxes.

$$ \bigcup_{b_1\leq \ldots \leq b_n, b_1+\ldots +b_n = 1}B(b_1,\ldots,b_n)$$

The hard part is to prove it has smallest volume. Are there any shape can do this using up even smaller volume? What if I only require the shape to be convex?

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    In two dimensions, your shape looks like [this](http://i.stack.imgur.com/f05ff.png) (red is $b_2 \in \big[\!\frac12,\frac34\!\big]$, blue is $b_2 \in \big[\!\frac34,1\big]$), while [this arrangement](http://i.stack.imgur.com/NPPBW.png) has smaller area, so the former is not optimal.2012-07-22
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    Vaguely related: http://mathworld.wolfram.com/KakeyaNeedleProblem.html, http://en.wikipedia.org/wiki/Kakeya_set2012-07-22

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