Answer:
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
<em>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)</em>
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.
![C'(x)=\dfrac{8x^3-32}{x^2}\\$Setting C'(x) to zero\\8x^3-32=0\\8x^3=32\\x^3=4\\x=\sqrt[3]{4}\\ x=1.59$ m](https://tex.z-dn.net/?f=C%27%28x%29%3D%5Cdfrac%7B8x%5E3-32%7D%7Bx%5E2%7D%5C%5C%24Setting%20C%27%28x%29%20to%20zero%5C%5C8x%5E3-32%3D0%5C%5C8x%5E3%3D32%5C%5Cx%5E3%3D4%5C%5Cx%3D%5Csqrt%5B3%5D%7B4%7D%5C%5C%20x%3D1.59%24%20m)
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