# A shape that covers all box with certain side lengths

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 (red is $b_2 \in \big[\!\frac12,\frac34\!\big]$, blue is $b_2 \in \big[\!\frac34,1\big]$), while this arrangement has smaller area, so the former is not optimal. –  Rahul Jul 22 '12 at 10:59
–  joriki Jul 22 '12 at 11:30