Box Constant

Geometry Level 3

k = F ⋅ S V k = \dfrac{F\cdot S}{V}

Let F F be the Harmonic Mean of the width, length, & height of a box; S S be the total surface area of the box; and V V be the volume of the box.

What is the value of the constant k k satisfying the equation above?


The answer is 6.

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Worranat Pakornrat by Worranat Pakornrat , 36, Thailand

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1 solution

The Harmonic Mean of width w w , length l l , and height h h = 3 1 w + 1 l + 1 h \dfrac{3}{\dfrac{1}{w} + \dfrac{1}{l} +\dfrac{1}{h}} .

The total surface area S S = 2 ( w l + l h + h w ) = 2 w l h ( 1 w + 1 l + 1 h ) 2(wl + lh + hw) = 2wlh(\dfrac{1}{w} + \dfrac{1}{l} + \dfrac{1}{h}) , and the volume V V = w l h wlh .

Thus, S V = 2 ( 1 w + 1 l + 1 h ) \dfrac{S}{V} = 2(\dfrac{1}{w} + \dfrac{1}{l} + \dfrac{1}{h}) .

Then, F ⋅ S V = 3 ⋅ 2 = 6 \dfrac{F\cdot S}{V} = 3\cdot 2 = \boxed{6} .

Note: Alternatively, if we let A A be the arithmetic mean of the 6 6 faces' areas, then A = S 6 A = \dfrac{S}{6} .

Therefore, V = F ⋅ A V = F\cdot A .

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