Hi, recently i was told to help set the minimum and maximun points for the number of pallets in the warehouse?

Question

Hi, recently i was told to help set the minimum and maximun points for the number of pallets in the warehouse and i found out that by using the formula taught in additional maths-derivatives where, y = ax2 + bx + c. however i do not know how to get the value for "a", "b", "c", "x" and "y"...can anyone help

Answer 1

Guessing the situation here...

Assume the warehouse has dimensions W and L. I don't see how the quadratic relationship you cite could result if the warehouse had an irregular perimeter. Then the area of the warehouse is WL. Similarly, if the pallets are rectangular, with dimensions, w and l, then the area of a single pallet would be wl. If pallets cannot be stacked, then the maximum number of pallets would be WL/wl rounded down to the nearest even number.

Unfortunately, there is no assurance that the pallets will fit evenly into the dimensions of the warehouse, so lets set some boundaries. The number of pallets that will fit lengthwise into the warehouse is L/l rounded down [L div l, if you prefer] and the number that will fit in the other direction would be W/w, rounded down. In that case, the
number would be (W/w rounded down)(L/l rounded down), so the number is in the range:

[(W/w) - 1] [(L/l) -1] < N ≤ (W/w)(L/l)

A larger lower limit is possible by rotating all the pallets, when a similar argument results in

[(W/l) - 1] [(L/w) - 1] < N ≤ (W/l)(L/w).

Still larger lower limits are possible by rotating some pallets, but not others.

None of these approximations derive from the quadratic formula. Hope this is of some help.

Answer 2

Hi
you should do measurements fill them into your model and then calculate one by one each parameter.

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