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

Explanation:

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

as technology advances, new experiments often expose problems in accepted theories. c

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-lavira: )

energy can be converted from one form to another, but it cannot be created or destroyed in ordinary chemical or physical means. do not understand what you are asking but hope this