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

if the person is shown to have +90n, then ans is -10n.

if the person is shown to have -90n, then ans is +10n.