Arectangular storage container with open top is to have a volume of 50 ft3 . the length of the base of the container is to be twice its width. material for the sides costs $5 per square foot, and materials for the base cost $8 per square foot. find the dimensions that minimize the cost of the material for the box.
Length = 2.862 ft
Width = 1.431 ft
Height = 3.05 ft
Let L be lenght, W be width, and H be height of the storage container. Since length is to be twice the width, it means L = 2W. The volumn of the container would also be:
SInce the volumn constraint is 50, that means:
The cost for the base would be its area times unit price
Likewise, the cost for the sides would be
We can substitute H from the equation above:
Therefore the total material cost
To find the minimum of this function, we can take first derivative then set to 0:
So W = L/2 = 1.43ft and
If we take the 2nd derivative and substitute L = 2.862 we would have
Hence L = 2.862 would yield the minimum material cost
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