Pkg Solutions

MOST ECONOMIC SIZE RATIOS FOR VARIOUS DESIGNS

The formulas below were calculated using partial differentiation. Naturally, it is impossible to include bend allowances or manufacturing joints in such formulas, as they bear no direct relationship to board area in terms of length, width and depth.

L = length of pack.
W = width of pack.
D = depth of pack.
V = volume of pack.
C = the proportion of the width which is to cover by the outer flaps, expressed as a fraction of the width.

STYLE	MOST ECONOMIC PROPORTIONS	FORMULA FOR CALCULATING DEPTH
RSC / FEFCO 0201 FEFCO 0226	L = D = 2W	D = ³ $\sqrt{2 * V}$
AFM / FEFCO 0204	2L = D = 2W	D = ³ $\sqrt{4 * V}$
RSC Inner flaps meet, outer flaps overlap / FEFCO 0206	2L = D = 4W	D = ³ $\sqrt{8 * V}$
RSC Gapping Flaps / FEFCO 0209	CL = D = 2CW or L = D/C = 2W	D = ³ $\sqrt{2C2V}$ or W = ³ $\sqrt{V / (4 * C)}$
0209-01/0201	L/2(C+1) = D = W(C+1)	D = ³ $\sqrt{(V * (C+1)2) / 2}$
Slotted Tray / FEFCO 0300 0301 0302 0303 0304 0307 0308 0309	L = 4D = W	D = ³ $\sqrt{16 * V}$
0411 0416	L = 4D = 2W	D = ³ $\sqrt{V / 8}$

Naturally, these formula will be of most help when deciding on the arrangement of retail packs in an outer case. However, they should be kept in mind at all times.

A size calculator is available in Boxcomp Help

MATHEMATICAL PROOF OF AN 0201's MOST ECONOMIC SIZE RATIOS BY PARTIAL DIFFERENTIATION

     _______________________________
W/2 |       |       |       |       | Board area = A
    |_______|_______|_______|_______|
    |       |       |       |       | A = 2DL + 2DW + 2WL + 2W2
 D  |       |       |       |       |
    |   L   |   W   |   L   |   W   | Volume = V = LWD
    |_______|_______|_______|_______|
    |       |       |       |       | Therefore - D = V
W/2 |_______|_______|_______|_______|                LW

By substitution:- A = 2V + 2V + 2WL +2W2

W L

By Partial differentiation:-

 dA    -2V 
 --  =  -- + 2W
 dL     L2

dA    -2V
--  =  -- + 2L + 4W 
dW     W2

For max or min value of A:- dA = 0 and dA = 0
                            --         --
                            dL         dW

Therefore:-

-2V + 2W = 0 -2V + 2L + 4W = 0

L2 W2

2W = 2V 2L + 4W = 2V

L2 W2

2WL2 = 2V 2LW2 + 4W3 = 2V

L2 = V 2LW2 + 4W3 = 2WL2

LW + 2W2 = L2

Since V = LWD Therefore:- L2 - LW - 2W2 = 0

L2 = LWD (L - 2W)(L + W) = 0

Therefore:- L = 2W or L = -W

L2 = LD

Therefore:- L = D

Therefore, for maximum board economy L = D = 2W