packed hexagons gave the minimal extent of boundary in a plane, so the actual solid figure, as determined by Maraldi, might be that which, for a given solid content, gives the minimum of surface: or which, in other words, would hold the most honey for the least wax. He set this problem before Koenig, and the geometer confirmed his conjecture, the result of his calculations agreeing within two minutes (109° 26′ and 70° 34′) with Maraldi’s determination. But again, Maclaurin[370] and Lhuilier[371], by different methods, obtained a result identical with Maraldi’s; and were able to shew that the discrepancy of 2′ was due to an error in Koenig’s calculation (of tan θ = √2),—that is to say to the imperfection of his logarithmic tables,—not (as the books say[372]) “to a mistake on the part of the Bee.” “Not to a mistake on the part of Maraldi” is, of course, all that we are entitled to say. [Illustration: Fig. 132.] The theorem may be proved as follows: _ABCDEF_, _abcdef_, is a right prism upon a regular hexagonal base. The corners _BDF_ are cut off by planes through the lines _AC_, _CE_, _EA_, meeting in a point _V_ on the axis _VN_ of the prism, and intersecting _Bb_, _Dd_, _Ff_, at _X_, _Y_, _Z_. It is evident that the volume of the figure thus formed is the same as that of the original prism with hexagonal ends. For, if the axis cut the hexagon _ABCDEF_ in _N_, the volumes _ACVN_, _ACBX_ are equal. {331} It is required to find the inclination of the faces forming the trihedral angle at _V_ to the axis, such that the surface of the figure may be a minimum. Let the angle _NVX_, which is half the solid angle of the prism, = θ; the side of the hexagon, as _AB_, = _a_; and the height, as _Aa_, = _h_. Then, _AC_ = 2_a_ cos 30° = _a_√3. And _VX_ = _a_/sin θ (from inspection of the triangle _LXB_) Therefore the area of the rhombus _VAXC_ = (_a_^2 √3)/(2 sin θ). And the area of _AabX_ = (_a_/2)(2_h_ − ½_VX_ cos θ) = (_a_/2)(2_h_ − _a_/2 ⋅ cot θ). Therefore the total area of the figure = hexagon _abcdef_ + 3_a_(2_h_ − (_a_/2) cot θ) + 3((_a_^2 √3)/(2 sin θ)). Therefore _d_(Area)/_d_θ = (3_a_^2/2)((1/sin^2 θ) − (√3 cos θ)/(sin^2 θ)). But this expression vanishes, that is to say, _d_(Area)/_d_θ = 0, when cos θ = 1/√3, that is when θ = 54° 44′ 8″ = ½(109° 28′ 16″). This then is the condition under which the total area of the figure has its minimal value. ――――――――――