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Valentin Fadeev

Calculus of Warehouses. Part 2

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Edited by Valentin Fadeev, Sunday, 14 Mar 2010, 14:43

Another surprising appearance of a fine method in a clumsy problem is offered when calculating optimal parameters for a pile of unitary cargo (bales, cases, crates etc).

Say we have bales of cotton stowed in several layers. According to safety regulations the pile must have a pyramid-like structure with "stairs" after certain amount of layers. Suppose we have 1 layer in the upper tier x bales long and y bales wide:d7609c6675f972b63794bef3b2665403.jpg

The next tier has 3 layers and is 2 bales longer (x+2) and 1 bale wider (y+1). What should be the values of x and y so that area occupied by the pile is minimized? Assume we have a fixed and sufficiently large total amount of bales, say 60.

If we behave negligently enough to give x and y continuous values (which is not so bad when total cargo count is high), the problem yields a nice solution using.. Lagrange multipliers:

x times y plus three times open x plus two close times open y plus one close equals 60

g of x comma y equals sum with, 3 , summands four times x times y plus three times x plus six times y

multirelation f of x comma y equals open x plus two close times open y plus one close right arrow m times i times n

g of x comma y equals zero

normal cap phi of x comma y equals f of x comma y plus lamda times g of x comma y

equation sequence normal partial differential times normal cap phi divided by normal partial differential times x equals sum with, 3 , summands y times open four times lamda plus one close plus three times lamda plus one equals zero

equation sequence normal partial differential times normal cap phi divided by normal partial differential times y equals x times open four times lamda plus one close plus two times open three times lamda plus one close equals zero

equation sequence normal partial differential times normal cap phi divided by normal partial differential times lamda equals x times open four times y plus three close plus six times open y minus nine close equals zero

x almost equals four reverse solidus full stop y almost equals two

Really, if one can use a scalpel as a screwdriver, that's what I am doing, but like in the previous case, it just looks nice

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