Volume of a box from its face areas

Text

A rectangular box has total surface area $108$ cm$^2$. Of the three faces meeting at one vertex, one is $\dfrac23$ of another and half of the third. Find the volume.

Solution

Let $F_1,F_2,F_3$ be the three faces at a vertex, with $F_1=\dfrac23F_2$ and $F_1=\dfrac12F_3$, i.e. $F_2=\dfrac32F_1$ and $F_3=2F_1$. The total surface is $2(F_1+F_2+F_3)=108\Rightarrow F_1+F_2+F_3=54$: $F_1\left(1+\tfrac32+2\right)=\tfrac92F_1=54\ \Rightarrow\ F_1=12,\ F_2=18,\ F_3=24.$ The volume of a rectangular box is the square root of the product of its three faces: $V=\sqrt{F_1F_2F_3}=\sqrt{12\cdot18\cdot24}=\sqrt{5184}=72\ \text{cm}^3.$ $\boxed{V=72\ \text{cm}^3}$