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Exercise 304
A rectangular box $P$ has distinct edge lengths $a,b,c$. The sum of all $12$ edges is $44$, the sum of the areas of all $6$ faces is $72$, and the volume is $36$. What is the length of the longest interior diagonal of $P$?
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Solution
The given sums imply $a+b+c=44/4=11$, $ab+bc+ca=72/2=36$, and $abc=36$.
Therefore $a,b,c$ are the roots of $x^3-11x^2+36x-36=(x-2)(x-3)(x-6)$, i.e. $2,3,6$.
The longest diagonal has length\[\sqrt{a^2+b^2+c^2}=\sqrt{(a+b+c)^2-2(ab+bc+ca)}=\sqrt{121-72}=7.\]
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