Problem 3

A box whose shape is a rectangular parallelepiped can be completely filled with cubes of side

1

. If one places inside it the maximum possible number of cubes, each of volume

2

, with their sides parallel to those of the box, then exactly

40\%

of the volume of the box is occupied. Determine the possible dimensions of all such boxes.

Step 5 of 6: Determine the second edge

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In plain words

The same shrinking-bound idea from Step 3 is reapplied one level down, now to a product of only two ratios equal to $5/2$ instead of three equal to $5$ — the argument recurses cleanly onto a smaller sub-problem.

\frac by\cdot\frac cz=\frac52,\qquad b=5,\ \frac by=\frac53

Detailed analysis

From Step 4, $(b/y)(c/z)=5/2$ , and since $b/y\ge c/z$ this gives $b/y\ge\sqrt{5/2}$ . If $b=2$ then $c/z=5/4<\sqrt[3]2$ , contradicting $c/z>\sqrt[3]2$ ; if $b\ge6$ then $y\ge5$ and $b/y<\sqrt[3]2(1+1/5)<\sqrt{5/2}$ , also a contradiction (since $2^2\cdot5^6<(5/2)^3\cdot4^6$ , i.e. $2\cdot(5/4)^6<5^3/2^3$ as $5^3<2^7$ ). Checking $b=3,4$ gives ratios $3/2,4/3$ , both below $\sqrt{5/2}$ ; only $b=5$ works, with $y=3$ and $b/y=5/3\ge\sqrt{5/2}$ .