3615.51 – Parallelepiped

A rectangular parallelepiped has edges with integral lengths x, y, and z. The sum of the lengths of all twelve edges is 72 inches. The sum of the areas of all six faces is 212 square inches. The volume of the solid is 144 cubic inches. Find the length in inches of a diagonal of this solid. And think of a reason why you might like to know this length.

Solution

We want to find the quantity L=x2+y2+z2L = \sqrt{{x^2}+{y^2}+{z^2}}.

We know that 4(x+y+z)=72⇝x+y+z=184(x + y + z) = 72 \leadsto x + y + z = 18. And that

2(xy+yz+xz)=2122(xy + yz + xz) = 212. And that xyz=144xyz = 144.

We can get all of these numbers together by considering--aha!--(x+y+z)2{(x + y + z)^2} (which equals 182=32418^2 = 324). Here we go!

(x+y+z)(x+y+z)=x2+xy+xz+yx+y2+yz+xz+zy+z2⇝x2+y2+z2+2(xy+yz+xz)=324⇝x2+y2+z2+212=324⇝x2+y2+z2=112⇝L=x2+y2+z2=112=16⋅7=47\begin{aligned} (x + y + z)(x + y + z) &=& {x^2} + xy + xz + yx + {y^2} + yz + xz + zy + {z^2} \\ &\leadsto& {x^2}+{y^2}+{z^2} + 2(xy + yz + xz) = 324 \\ &\leadsto& {x^2}+{y^2}+{z^2} + 212 = 324 \\ &\leadsto& {x^2}+{y^2}+{z^2} = 112 \\ &\leadsto& L = \sqrt{{x^2}+{y^2}+{z^2}} = \sqrt{112} = \sqrt{16\cdot7} = 4\sqrt{7} \end{aligned}

Done.

And if you think of our parallelepiped as a box, you have just found the length of the longest cane, or sword, or stick, that will fit into it (disregarding its thickness).

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