A company plan to manufacture closed rectangular boxes that have a volume of 8m3 . If the

material for the top and bottom costs twice as much as the material for the sides. Find the

dimensions that will minimize the cost.

Base Side Length =1.59m

Height = 3.16 m

Step-by-step explanation:

Volume of the box =

Let the base dimensions = x and y

Let the height of the box =h

However, for any optimal configuration, Width = Length as varying the length and width to be other than equal reduces the volume for the same total(w+l)

Volume, V=xyh=8

Since x=y

Surface Area of the box

The material for the top and bottom costs twice as much as the material for the sides.

Let the cost of the sides =$1 per square meter

Cost of the material for the sides = 4xh

Cost of the material for the top and bottom =

Therefore:

Total Cost,

Substitution of    into C

To minimize C(x), we find its derivative and solve for the critical points.

To verify if it is a minimum, we use the second derivative test

Since C''(x) is greater than zero, it is a minimum point.

Recall:

Therefore, the dimensions that minimizes the cost are:

Base Side Lengths of 1.59m; and

Height of 3.16 m

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step-by-step explanation:

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